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\title{\huge\JSLight B\'{e}zout’s Theorem and Algebraic Geometry}
\author{Noah Scott Goldman}
\date{2019 February 25}
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\maketitle
\begin{center}
\textsf{Senior Integrative Exercise}
\end{center}
\begin{abstract}
\noindent Anyone familiar with systems of polynomial equations (whether they majored in math or just had to solve a train-related word problem once) knows solving them can get complicated. Even when we can find solutions, who knows if there are more? How many ought we to expect? The answer to this question, Bézout’s theorem, lies at the “intersection” of geometry and algebra. To address the \emph{algebraic} problem presented by a polynomial system, Bézout’s theorem solves the equivalent \emph{geometric} problem of finding points of intersection between hypersurfaces, and vice versa – simultaneously generalizing both the fundamental theorem of algebra and the concept of linear independence. To prove it, we will embark on a whirlwind tour through algebraic geometry, exploring projective sets, homogeneous ideals, Hilbert polynomials, and the great Nullstellensatz, guided only by that simplest of links between pictures and equations: the coordinate.
\end{abstract}
\tableofcontents*
\chapter{Introduction}
The problem of finding zeros of polynomials is as old as algebra itself. Unfortunately, when it was discovered that not all zeros could be written in terms of radicals, mathematicians had to settle for counting them. In the single-variable case, both tasks can be accomplished by a single theorem.
\begin{theorem}[Fundamental theorem of algebra]
Let \(f \in \mathbb{C}[X]\) be a polynomial of degree \(d > 0\) with complex coefficients. Then \(f\) is of the form
\begin{equation}
f(X) = c\,(X - z_1)^{m_1} \dotsm (X - z_r)^{m_r}
\end{equation}
where the \(z_k\) are distinct complex numbers and the \(m_k\) are positive integers (the \textbf{multiplicities}) with \(m_1 + \dotsb + m_r = d\).
\end{theorem}
How might we generalize this result to multiple dimensions? In the 2D case, the algebraic problem of counting the solutions of a system of two polynomial equations in two variables is equivalent to counting the intersections of two curves. It seems simple, but in fact there are quite a few subtleties involved. One issue is defining factorization or decomposition in multiple dimensions; another is defining multiplicities. This is why, although B\'{e}zout's theorem~-- a generalization of the fundamental theorem of algebra, and the focus of this paper~-- bears the name of Etienne B\'{e}zout, he was unable to prove the theorem satisfactorily.
Nevertheless, B\'{e}zout's insight~-- that the product of the degrees has something to do with the number of shared zeros~-- is the starting point of our excursion into algebraic geometry. We will aim for a comprehensive overview of one approach to proving B\'{e}zout's theorem. Because I found proofs of B\'{e}zout's theorem without any abstract algebraic notions confusing (a problem that hampered B\'{e}zout himself), I have elected to follow the modern development using ideals, projective sets, and Hilbert polynomials. Not only do I find this more pleasing mathematically; I believe it more fruitful to explore a wide variety (pun intended) of definitions, theorems, and concepts which may be useful elsewhere than to hyper-analyze one long proof. Furthermore, this has allowed me to prove the theorem at a greater level of generality than I otherwise would be able to.
Finally, I could not resist inserting some lattice theory. Actually, the lattices I invoke are more than that; they are \emph{topologies}, and this appears to be the gateway to more advanced algebraic geometry. While the Grothendieck theory remains (mostly) opaque to me at this time, I find that the lattice approach (which are a lot like simpler versions of finitely complete categories) illustrates the modern ideas I \emph{can} grasp without straying too far into the ones I can't.
\chapter{Affine \& projective space}
For the rest of the paper, we fix a field \(K\) of characteristic not equal to 2. The example to keep in mind will usually be either \(\mathbb{R}\), the real numbers, or \(\mathbb{C}\), the complex numbers. \parencite{Shahriari2017}
The classical form of B\'{e}zout's theorem has two settings. The first, affine space, is essentially the same space studied in linear algebra.
\begin{definition}[Affine space]
Following \textcite{Holme2012}, we define the \(n\)\textbf{-dimensional affine space} over \(K\) to be the vector space \(\Af^n(K) = K^n\) consisting of all vectors \([x_1 \com \dotsc \com x_n]^{\top}\) such that each \(x_j \in K\).
\end{definition}
The second is projective space. For any vector \(\mathbf{v} \in K^{n}\), denote by \(\langle \mathbf{v} \rangle\) the subspace of \(K^{n}\) generated by \(\mathbf{v}\). When \(\mathbf{v} \neq \textbf{0}\) and \(K = \mathbb{R}\), geometrically \(\langle \mathbf{v} \rangle\) is a line through the origin.
\begin{definition}[Projective space]
Again following \textcite{Holme2012}, we define the \(n\)\textbf{-dimensional projective space} over \(K\) to be the collection of lines
\begin{equation}
\Pro^n (K) = \left\{ \langle \mathbf{v} \rangle : \mathbf{v} \in K^{n+1} \text{ and } \mathbf{v} \neq \textbf{0}\right\}\text{.}
\end{equation}
Elements of \(\Pro^n(K)\) are called \textbf{projective points} (or just \textbf{points} if no confusion will arise). We can represent projective points by \textbf{homogeneous coordinates}~-- that is, if \(\mathbf{v} = [v_0 \com \dotsc \com v_n]^{\top} \in K^{n+1}\) is nonzero, we write \(( v_0 : v_1 : \dotsb : v_n )\) to mean the projective point \(\langle \mathbf{v} \rangle \in \Pro^n(K)\). Note that for any nonzero \(\lambda \in K\), we have \(\langle \mathbf{v} \rangle = \langle \mathbf{v} \lambda \rangle\), so the homogeneous coordinates of any projective point are only unique up to scaling by a unit.
\end{definition}
A differential-geometric sidenote: the space \(\Pro^n(\mathbb{C})\) admits the structure of an \(n\)-dimensional complex manifold. \parencite{Wells2008} Let
\begin{gather*}
U_k = \{(z_0 : \dotsb : z_n) : z_k \neq 0\} \text{ and}\\
\phi_k (z_0 : \dotsb : z_n) = \left[\frac{z_{k+1}}{z_k} \com \frac{z_{k+2}}{z_k} \com \dotsc \com \frac{z_{k+n+1}}{z_k}\right]^{\top}\text{,}
\end{gather*}
where \(z_m = z_{m-(n+1)}\) if \(m > n\). (Intuitively, \(\phi_k\) cycles through all the homogeneous coordinates except \(z_k\), starting with \(z_{k+1}\) and wrapping around to \(z_0\) if necessary.) The family \([(U_0 \com \phi_0) \com \dotsc \com (U_n \com \phi_n)]\) forms a holomorphic atlas for \(\Pro^n(\mathbb{C})\); \textcite{Wells2008} covers the technical details. By a similar construction, \(\Pro^n(\mathbb{R})\) is an \(n\)-dimensional real smooth manifold.\footnote{The same is true of the corresponding affine spaces, simply because manifolds are characterized by their similarity to affine spaces.}
\chapter{Polynomials and algebraic sets}
In algebraic geometry, we are not interested in affine and projective spaces \emph{per s\'{e}}, but rather in particular subsets of these spaces~-- namely, the sets of zeros of polynomials. We will focus on two types of these \textbf{algebraic sets}: affine and projective.
\begin{definition}[Affine subset]
Let \(F \subseteq K\,[\mathbf{X}]\) be a set of polynomials, where \(\mathbf{X} = [X_1 \com \dotsc \com X_n]\). We say that \(\mathbf{v} \in \Af^n(K)\) is an \textbf{affine zero} of \(F\) iff every \(f \in F\) satisfies \(f(\mathbf{v}) = 0\). The set of affine zeros of \(F\) is called its \textbf{(affine) vanishing set} and written as \(\V(F)\) or \(\V_{\Af}(F)\). If \(V = \V_{\Af}(F)\) for some \(F\), we call \(V\) an \textbf{affine subset} of \(\Af^n (K)\).
\end{definition}
\begin{definition}[Projective subset] \label{def-projective-subset}
Let \(F \subseteq K[\mathbf{X}]\) be a set of polynomials, where \(\mathbf{X} = [X_0 \com X_1 \com \dotsc \com X_n]\). We say that \(v \in \Pro^n(K)\) is a \textbf{projective zero} of \(F\) iff every \(f \in F\) is identically zero when restricted to the elements of \(v\). The set of projective zeros of \(F\) is called its \textbf{(projective) vanishing set} and written as \(\V(F)\) or \(\V_{\Pro}(F)\). If \(V = \V_{\Pro}(F)\) for some \(F\), we call \(V\) a \textbf{projective subset} of \(\Pro^n(K)\).
\end{definition}
A polynomial \(h\) is \textbf{homogeneous of degree \(k\)} if \& only if, for every scalar \(\lambda \in K\), we have \(h(\mathbf{X} \lambda) = h(\mathbf{X})\lambda^k\). It turns out that, in the study of projective zeros, these polynomials are the only ones that really matter. Every polynomial that is zero on a set \(S \subseteq \Pro^n(K)\) can be expressed as a sum of homogeneous polynomials, each of which is also zero on that set \(S\).
From here on out, we'll concentrate mostly on the projective case.
\chapter{Ideals and algebraic sets}
While the statement that algebraic geometry links algebra and geometry might not seem very enlightening (since it's in the name), the link itself is tremendously so. That link, it turns out, is a relationship between \emph{lattices}.
\begin{definition}[Lattice]
A \textbf{lattice} is a partially-ordered set \((L \com \preceq)\) in which every finite subset has a \textbf{meet} (\(\land\)), or greatest lower bound, and a \textbf{join} (\(\lor\)), or least upper bound. If there are elements\footnote{The convention is to pronounce \(\bot\) as ``bottom'' and \(\top\) as ``top'', respectively.} \(\bot \com \top \in L\) such that \(\bot \preceq x \preceq \top\) for all \(x \in L\), we say that the lattice is \textbf{bounded}.
\end{definition}
It turns out that the \emph{ideals} of any ring form a lattice.
\begin{definition}[Ideal]
Let \(R\) be a ring. A \textbf{subring} of \(R\) is a set \(S \subseteq R\) that is closed under the operations of \(R\). An \textbf{ideal} \(I\) is a subring with the additional property that \(Ir \subseteq I\) and \(rI \subseteq I\) for all \(r \in R\).
\end{definition}
If \(G \subseteq R\), we write \(\langle G \rangle\) for the intersection of all ideals of \(R\) containing \(G\). More concretely, \(\langle G \rangle\) consists of all the elements of \(G\), along with the combinations thereof under sums and ring-element-products. We can define operations on ideals as follows:
\begin{align*}
I_1 + I_2 &= \{r_1 + r_2 : r_k \in I_k\}\\
I_1 \cap I_2 &= \{r \in R : r \in I_k\}\\
I_1 \, I_2 &= \langle r_1 r_2 : r_k \in I_k\rangle\\
I : B &= \{r \in R : r B \subseteq I\}
\end{align*}
\begin{proposition}[Lattice of ideals]
The ideals of a ring \(R\) form a bounded lattice under inclusion, with meet and join given by
\begin{equation*}
I_1 \meet I_2 = I_1 \cap I_2 \quad \text{and} \quad I_1 \join I_2 = I_1 + I_2
\end{equation*}
respectively, bounded below by \(\bot = \{0\}\) and above by \(\top = R\).
\end{proposition}
\begin{definition}[Lattice homomorphism]
Let \(L \com M\) be lattices and \(\phi: L \leftarrow M\) a function between them. We say that \(\phi\) is an \textbf{order-reversing lattice homomorphism} iff
\begin{align*}
\phi(m_1 \meet m_2) &= \phi(m_1) \join \phi(m_2)\\
\phi(m_1 \join m_2) &= \phi(m_1) \meet \phi(m_2)
\end{align*}
and \textbf{order-preserving} iff it obeys the same identities with join and meet interchanged on one side. A \textbf{lattice isomorphism} must satisfy all of the above \emph{and} have an inverse, which must also be a homomorphism.\footnote{It turns out that the last condition is not necessary; the inverse of a lattice homomorphism is always another lattice homomorphism.}
\end{definition}
Consider a set \(S \subseteq \Pro^n(K)\) of (projective) points. Following \textcite{Fulton2008}, we define the \textbf{(projective) ideal of} \(S\) to be the ideal in \(K[\mathbf{X}]\) of polynomials that are identically zero on the elements of \(S\), and write it as \(\I_{\Pro}(S)\) or just \(\I(S)\). The \textbf{affine ideal} is defined similarly. In set-builder notation:
\begin{align}
\I_{\Af}(S) = \{p(&X_1 \com \dotsc \com X_n) : S \subseteq \V_{\Af}\{p\}\}\\
\I_{\Pro}(S) = \{h(X_0 \com &X_1 \com \dotsc \com X_n) : S \subseteq \V_{\Pro}\{h\}\}
\end{align}
We need one more definition to prove the main result of this section.
\begin{definition}[Prime ideal]
Let \(P\) be an ideal of a ring \(R\). We say that \(P\) is \textbf{prime} iff whenever \(H_1\) and \(H_2\) are ideals such that \(H_1 H_2 \subseteq P\), we have \(H_1 \subseteq P\) or \(H_2 \subseteq P\).
\end{definition}
\begin{example}
Suppose \(v \in \Pro^n(K)\) is a single point. Then \(\I\{v\}\) is a prime ideal, since if \(f_1(v) f_2(v) = 0\) then \(f_1(v) = 0\) or \(f_2(v) = 0\).
\end{example}
\noindent This gives us all we need to prove the main result of this section:
\begin{theorem}[V is a lattice homomorphism]
The vanishing set operator \(\V\), when restricted to the ideals of \(K\,[\mathbf{X}]\):
\begin{enumerate}
\item Is surjective onto the projective (or affine) sets: every projective (or affine) set can be expressed as \(\V(H)\) for some ideal \(H \subseteq K\,[\mathbf{X}]\).
\item Is an order-reversing lattice homomorphism: for any ideals \(I_1\) and \(I_2\),
\begin{align}
\V(I_1 \cap I_2) &= \V(I_1) \cup \V(I_2) \label{eq-V-intersection} \\
\V(I_1 + I_2) &= \V(I_1) \cap \V(I_2) \label{eq-V-sum}
\end{align}
\end{enumerate}
\begin{proof}
First, we'll prove surjectivity. Let \(\V(G)\) be a projective set for some set \(G \subseteq K\,[\mathbf{X}]\). Consider the ideal \(H = \langle G \rangle\) generated by \(G\). Wherever \(H\) vanishes, so do its subsets, including \(G\). Thus \(\V(H) \subseteq \V(G)\).
Next, let \(v \in \V(G)\). Then every polynomial in \(G\) vanishes at \(v\), so \(G \subseteq \I\{v\}\). However, because \(H = \langle G \rangle\) is the smallest ideal containing \(G\), we have \(H \subseteq \I\{v\}\) and thus \(v \in \V(H)\). It follows that \(\V(G) = \V(H)\).
Second, we'll prove equation \eqref{eq-V-intersection}, because it shows all the techniques required to prove equation \eqref{eq-V-sum} and then some.
Let \(v \in \V(I_1)\). Then every polynomial in \(I_1\) is zero at \(v\), including those in \(I_1 \cap I_2\). So \(\V(I_1) \subseteq \V(I_1 \cap I_2)\). The same argument shows \(\V(I_2) \subseteq \V(I_1 \cap I_2)\) as well. Therefore, \(\V(I_1) \cup \V(I_2) \subseteq \V(I_1 \cap I_2)\).
Now let \(v \in \V(I_1 \cap I_2)\). Then \(I_1 \cap I_2\) vanishes at \(v\), so \(I_1 \cap I_2 \subseteq \I\{v\}\). In particular, because \(I_1 I_2 \subseteq I_1 \cap I_2\), we have \(I_1 I_2 \subseteq \I\{v\}\). But \(\I\{v\}\) is a prime ideal, so \(I_1 \subseteq \I\{v\}\) or \(I_2 \subseteq \I\{v\}\). Either way, \(v \in \V(I_1)\cup\V(I_2)\). Therefore, \(\V(I_1 \cap I_2) = \V(I_1)\cup\V(I_2)\), and we're done.
\end{proof}
\end{theorem}
We conclude this section with a few properties of the ideal operator I.
\begin{proposition}[Properties of projective ideals] \label{prop-proj-ideals}
The following hold.
\begin{enumerate}
\item For every \(S \subseteq \Pro^n(K)\), the ideal \(\I(S)\) is \textbf{homogeneous}. In other words, the polynomials in \(\I(S)\) are precisely those expressible as sums of homogeneous polynomials in \(\I(S)\). \label{proj-ideals-homo}
\item If \(R \subseteq S\), then \(\I(R) \supseteq \I(S)\).
\item The ideal of the empty set is \(K[\mathbf{X}]\);\footnote{Note that \(\V\langle \mathbf{X} \rangle\) is also the empty set, where \(\langle \mathbf{X} \rangle\) is the irrelevant ideal. See definition \ref{def-irrelevant}.} the ideal of \(\Pro^n(K)\) is \(\{0\}\). \label{proj-ideals-empty-set}
\item The ideal of a single point \((v_0 : \dotso : v_n)\), where \(v_k \neq 0\) for some fixed \(k\), is generated by the polynomials of the form \(v_j X_k - X_j v_k\) for \(j = 0 \com \dotsc \com n\). \label{proj-ideals-single-pt}
\item For a set \(P\) of polynomials, we have \(P \subseteq (\I \circ \V)(P)\). For a set \(S\) of points, we have \(S \subseteq (\V \circ \I)(S)\).
\item The operators \(\I\) and \(\V\) are weak functional inverses of each other: \(\V \circ \I \circ \V = \V\) and \(\I \circ \V \circ \I = \I\). In particular, if \(W\) is an algebraic set, \((\V \circ \I)(W) = W\); if \(J\) is the ideal of an algebraic set, \((\I \circ \V)(J) = J\).
\end{enumerate}
\end{proposition}
Essentially the same properties hold in the affine case (with the proper adjustments), except for property \ref{proj-ideals-homo}. Also, the ideal of a point \(\mathbf{v} = [v_1 \com \dotsc \com v_n]^{\top}\) in affine space is not just prime, but maximal, and takes the form
\[\I\{\mathbf{v}\} = \langle X_1 - v_1 \com \dotsc \com X_n - v_n\rangle\text{.}\]
\chapter{The Nullstellensatz and the lattice isomorphism}
The German word \emph{Nullstellensatz} means, literally, ``zero-locus theorem''. Discovered by David Hilbert, it fine-tunes the lattice homomorphism we derived in the last section. It shows, in other words, that V is an \emph{iso}morphism under the proper conditions.
What are these proper conditions? They begin with some generalizations of prime ideals.
\begin{definition}[Semiprime ideal, radical]
The \textbf{radical} of an ideal \(I\) is the intersection \(\sqrt{I}\) of all prime ideals containing \(I\). An ideal \(I\) is \textbf{semiprime} iff \(I\) is an intersection of prime ideals~-- or, equivalently, \(\sqrt{I} = I\).
\end{definition}
\begin{proposition}
Every ideal of the form \(\I(E)\), where \(E \subseteq \Pro^n(K)\) is any set, is semiprime.
\begin{proof}
Observe that
\[\I(E) = \bigcap_{e \in E} \I\{e\}\text{.}\]
Since each \(\I\{e\}\) is prime, \(\I(E)\) is semiprime.
\end{proof}
\end{proposition}
\begin{proposition}[Lattice of semiprime ideals]
The semiprime ideals of a ring \(R\) form a bounded lattice under inclusion, with meet and join given by
\begin{equation}
I_1 \meet I_2 = I_1 \cap I_2 \quad \text{ and } \quad I_1 \join I_2 = \sqrt{I_1 + I_2}
\end{equation}
respectively, bounded below by \(\bot = \{0\}\) and above by \(\top = R\).
\end{proposition}
We would like I to be \(\V^{-1}\), but unfortunately that isn't true in general; multiple ideals can give the same projective set. For example, suppose \(\V(H) = \emptyset\) for some ideal \(H\). What could \(H\) be? Perhaps the most readily apparent possibility is that \(H = K\,[\mathbf{X}]\), but there are alternatives~-- for example, that \(H\) equals the \emph{irrelevant ideal:}
\begin{definition}[Irrelevant ideal] \label{def-irrelevant}
The ideal \(\langle \mathbf{X} \rangle = \langle X_0 \com \dotsc \com X_n \rangle\) is called the \textbf{irrelevant ideal}.
\end{definition}
\begin{definition}[Primary ideals]
An ideal \(I\) is \textbf{primary} if \& only if, whenever \(H_1\) and \(H_2\) are ideals such that \(H_1 H_2 \subseteq I\), either \(H_1 \subseteq \sqrt{I}\) or \(H_2 \subseteq \sqrt{I}\) for some integer \(m>0\). When \(I\) is primary and \(P = \sqrt{I}\), we say that \(I\) is \(P\)\textbf{-primary}.
\end{definition}
Note that being primary and semiprime is equivalent to being prime.
\begin{definition}[Relevant]
An ideal \(I\) is \textbf{relevant} iff \(I\) is \emph{not} \(\langle \mathbf{X} \rangle\)-primary.
\end{definition}
Now that we have the necessary concepts, it is time to state and prove Hilbert's Nullstellensatz. It turns out that affine space is better suited to proving the Nullstellensatz, while projective space is the natural setting for B\'{e}zout's theorem. So we will return to affine space for now.
Much like the laws of large numbers, there many theorems called ``Nullstellensatz'', the top two are separated into ``weak'' and ``strong'', and the weak helps to prove the strong. This aids our intuition somewhat by splitting the result into manageable chunks, but even those are best broken into subchunks~-- that is, lemmas.
\begin{lemma}[Zariski lemma, one variable] \label{lem-Zariski}
Let \(K\,[\epsilon]\) be an integral domain generated by \(\epsilon\) over the field \(K\). If \(K\,[\epsilon]\) is a field, then \(\epsilon\) is algebraic over \(K\); that is, \(p(\epsilon) = 0\) for some nonzero polynomial \(p\) over \(K\). In particular, if \(K\) is algebraically closed, then \(K = K\,[\epsilon]\).
\begin{proof}
Inspired by \textcite{Azarang2015}, we will prove the contrapositive: if \(\epsilon\) is not algebraic over \(K\), then \(K\,[\epsilon]\) is not a field. Assuming \(\epsilon\) is not algebraic, the infinite list \((1 \com \epsilon \com \epsilon^2 \com \epsilon^3 \com \dotsc)\) forms a basis of \(K\,[\epsilon]\) as a vector space over \(K\).
Now, define the ring homomorphism \(\epsilon^{*} : K\,[\epsilon] \leftarrow K\,[X]\), where \(K\,[X]\) is the polynomial ring over \(K\), by \(\epsilon^{*}p(X) = p(\epsilon)\). Notice that, since every element of \(K\,[\epsilon]\) takes the form \(p(\epsilon)\), \(\epsilon^{*}\) is surjective. We will show \(\epsilon^{*}\) is injective as well. Suppose that \(\epsilon^{*}p(X) = 0\) for some \(p(X)\). Then
\begin{align*}
\epsilon^{*}(p_0 + p_1 X + \dotsb + p_m X^m) &= 0\text{;}\\
p_0 + p_1 \epsilon + \dotsb + p_m \epsilon^m &= 0\text{.}
\end{align*}
But since the list \((1 \com \epsilon \com \dotsc \com \epsilon^m)\) is linearly independent, we have \(p_k = 0\) for all \(k\). Thus \(p(X) = 0\), and the kernel of \(\epsilon^{*}\) is trivial. This shows \(\epsilon^{*}: K\,[\epsilon] \leftarrow K\,[X]\) is an isomorphism, and \(K\,[\epsilon]\) is thus not a field.
\end{proof}
\end{lemma}
Combining lemma \ref{lem-Zariski} with induction yields the following generalization.
\begin{corollary}[Zariski lemma, multivariable] \label{cor-Zariski}
Let the integral domain \(R = K\,[\epsilon_1 \com \dotsc \com \epsilon_r]\) be finitely generated over the field \(K\). If \(R\) is a field, then each \(\epsilon_k\) is algebraic over \(K\). In particular, if \(K\) is algebraically closed, then \(K = R\).
\end{corollary}
Although we will not repeat the argument, \textcite{Azarang2015} provides a proof of corollary \ref{cor-Zariski}.
\begin{theorem}[Weak Nullstellensatz] \label{thm-weak-nullst}
Suppose \(K\) is algebraically closed. Let \(I\) be a proper ideal of \(K\,[\mathbf{X}]\), where \(\mathbf{X} = [X_1 \com \dotsc \com X_n]\). Then the (affine) zero set \(\V(I)\) is nonempty.
\begin{proof}
\parencite{Fulton2008} Let \(M\) be a maximal ideal containing \(I\). It suffices to show \(\V(M)\) is nonempty, since \(\V(M) \subseteq \V(I)\). Because \(M\) is maximal, the quotient ring \(Q = K\,[\mathbf{X}]/M\) is a field. Under the quotient map \(K[\mathbf{X}] \rightarrow Q\), the constant polynomials modulo \(M\) form a subfield of \(Q\) which may be regarded as \(K\).
Since \(K\) is algebraically closed, and \(Q\) is finitely generated over \(K\) by the elements \(X_k + M\), corollary \ref{cor-Zariski} applies. Thus \(K = Q\), so for every polynomial \(p(\mathbf{X}) \in K[\mathbf{X}]\) there is a representative \(a \in K\) such that \(a + M = p(\mathbf{X}) + M\), i.e. \(p(\mathbf{X}) - a \in M\).
For each \(X_k\), let \(v_k \in K\) be such that \(X_k - v_k \in M\). Then \(M\) contains the ideal \(\langle X_1 - v_1 \com \dotsc \com X_n - v_n \rangle\). But the latter is also maximal, implying it equals \(M\). Since \([v_1 \com \dotsc \com v_n]^{\top}\) lies in \(\V(M)\), it follows that \(\V(M)\) is nonempty.
\end{proof}
\end{theorem}
\begin{theorem}[Strong Nullstellensatz]
Suppose \(K\) is algebraically closed. Let \(H\) be an ideal of \(K\,[\mathbf{X}]\). Then \((\I_{\Af} \circ \V_{\Af})(H) = \sqrt{H}\).
\begin{proof}
\parencite{Fulton2008} First we must show \(\sqrt{H} \subseteq (\I \circ \V)(H)\). Let \(p \in \sqrt{H}\). Then there exists an integer \(m > 0\) such that \(p^m \in H\). By the definition of affine zero set, \(p(\mathbf{v})^m = 0\) for all \(\mathbf{v} \in \V(H)\). Since \(K\) has no zero divisors, this implies \(p(\mathbf{v}) = 0\). Hence \(p\) qualifies for membership in the ideal \((\I \circ \V)(H)\).
Most of the work goes into showing the reverse inclusion \((\I \circ \V)(H) \subseteq \sqrt{H}\). Let \(p \in (\I \circ \V)(H)\), and define an ideal \(J\) of the \((n+1)\)-variable polynomial ring \(K[\mathbf{X} \com T]\) by
\[J = \langle H \rangle + \langle p(\mathbf{X})T - 1\rangle\text{.}\]
Consider the zero set \(\V(J) \subseteq \Af^{n+1}(K)\). Any point \((\mathbf{v} \com t) \in \V(J)\) must satisfy \emph{at least} the following two equations, which are inconsistent:
\begin{equation*}
p(\mathbf{v}) = 0 \text{ and } p(\mathbf{v})t - 1 = 0\text{.}
\end{equation*}
Thus \(\V(J)\) is empty, so by the weak Nullstellensatz (theorem \ref{thm-weak-nullst}), the ideal \(J\) is in fact the whole ring \(K\,[\mathbf{X} \com T]\). In particular, \(1 \in J\), so there exist polynomials \(h \in H\) and \(a_1 \com b_1 \in K\,[\mathbf{X} \com T]\) such that
\begin{equation*}
a_1(\mathbf{X} \com T) h(\mathbf{X}) + b_1(\mathbf{X} \com T)(p(\mathbf{X})T - 1) = 1\text{.}
\end{equation*}
Let \(Y = T^{\;-1}\), and let \(m\) be greater than the degrees of \(a_1\) and \(b_1\) as polynomials in \(T\). Multiplying both sides by \(Y^{\,m}\) yields an equation of the form
\begin{equation*}
a_2(\mathbf{X} \com Y) h(\mathbf{X}) + b_2(\mathbf{X} \com Y)(p(\mathbf{X}) - Y) = Y^{\,m}\text{.}
\end{equation*}
If we substitute \(p(\mathbf{X})\) for \(Y\), we end up with \(p(\mathbf{X})^m = a_2(\mathbf{X} \com p(\mathbf{X})) h(\mathbf{X})\), which entails that \(p^m \in H\) and \(p \in \sqrt{H}\).
\end{proof}
\end{theorem}
As a corollary, we have the projective Nullstellensatz:
\begin{corollary}[Projective Nullstellensatz]
Suppose \(K\) is algebraically closed. Let \(H\) be a homogeneous ideal of \(K\,[\mathbf{X}]\) which is not \(\langle \mathbf{X} \rangle\)-primary. Then \((\I_{\Pro} \circ \V_{\Pro})(H) = \sqrt{H}\).
\end{corollary}
The utility of the Nullstellensatz, for us, is that it implies V, when properly restricted, is a lattice isomorphism. In the affine case, the correspondence is particularly nice.
\begin{theorem}[\(\V_{\Af}\) is a lattice isomorphism]
The affine vanishing set operator \(\V = \V_{\Af}\) is a lattice isomorphism when restricted to the semiprime ideals of \(K\,[\mathbf{X}]\), where \(\mathbf{X} = [X_1 \com \dotsc \com X_n]\). That is:
\begin{enumerate}
\item If \(H\) is a semiprime ideal and \(U\) an affine set, then \(U = \V(H)\) if \& only if \(H = \I(U)\).
\item For ideals \(I_1\), \(I_2\) and affine sets \(V_1\), \(V_2\):
\begin{align*}
\begin{array}{cc}
\V\left(\sqrt{I_1 + I_2}\right) = \V(I_1) \cap \V(I_2) & \I(V_1 \cup V_2) = \I(V_1) \cap \I(V_2)\\
\V(I_1 \cap I_2) = \V(I_1) \cup \V(I_2) & \I(V_1 \cap V_2) = \sqrt{\I(V_1) + \I(V_2)}
\end{array}
\end{align*}
\end{enumerate}
\end{theorem}
It's not as nice in the projective case, but still pretty good.
\begin{theorem}[\(\V_{\Pro}\) is a lattice isomorphism] \label{thm-lattice-iso}
The projective vanishing set operator \(\V = \V_{\Pro}\) is a lattice isomorphism when restricted to the relevant, homogeneous, semiprime ideals of \(K\,[\mathbf{X}]\), where \(\mathbf{X} = [X_0 \com X_1 \com \dotsc \com X_n]\). That is:
\begin{enumerate}
\item If \(H\) is a relevant, homogeneous, semiprime ideal and \(U\) a projective set, then \(U = \V(H)\) if \& only if \(H = \I(U)\).
\item For ideals \(I_1\), \(I_2\) and projective sets \(V_1\), \(V_2\):
\begin{align*}
\begin{array}{cc}
\V\left(\sqrt{I_1 + I_2}\right) = \V(I_1) \cap \V(I_2) & \I(V_1 \cup V_2) = \I(V_1) \cap \I(V_2)\\
\V(I_1 \cap I_2) = \V(I_1) \cup \V(I_2) & \I(V_1 \cap V_2) = \sqrt{\I(V_1) + \I(V_2)}
\end{array}
\end{align*}
\end{enumerate}
\end{theorem}
\chapter{The primary decomposition}
Many properties of algebraic subsets can be formulated in terms of ideals and vice versa. Here is a definition that, at first glance, has nothing to do with ideals.
\begin{definition}[Irreducible set]
We say that an algebraic subset \(V\) is \textbf{irreducible} or a \textbf{variety} iff, whenever \(V_1\) and \(V_2\) are algebraic subsets with \(V_1 \cup V_2 = V\), either \(V_1\) or \(V_2\) equals \(V\). (Compare the notion of connectedness in topology.)
\end{definition}
\noindent Yet, it turns out, irreducibility is expressible in terms of ideals:
\begin{theorem}
An algebraic set \(W\) is irreducible if \& only if \(\I(W)\) is a prime~ideal.
\begin{proof}
\parencite{Fulton2008} First assume \(W = \V(P)\) is irreducible. Let \(h_1\) and \(h_2\) be polynomials such that \(h_1 h_2 \in \I(W)\); we will show \(h_1\) or \(h_2\) is in \(\I(W)\). Let \(V_1 = \V\{h_1\} \cap W\) and \(V_2 = \V\{h_2\} \cap W\); these are algebraic sets with \(W = V_1 \cup V_2\). Since \(W\) is irreducible, \(W = V_k\) for some \(k\). By definition of \(V_k\), this means \(W \subseteq \V\{h_k\}\), so \(h_k\) is identically zero on \(W\); thus \(h_k \in \I(W)\).
Now suppose \(P = \I(W)\) is a prime ideal. Write \(W\) as a union of algebraic sets \(W_1 \cup W_2\). By theorem \ref{thm-lattice-iso}, \(P = \I(W_1) \cap \I(W_2) \supseteq \I(W_1)\I(W_2)\). Because \(P\) is a prime ideal, we know \(P \supseteq \I(W_k)\) for some \(k\). The operator \(\V\) reverses this inclusion to \(W \subseteq W_k\), which essentially means \(W = W_k\).
\end{proof}
\end{theorem}
There is a dual concept of irreducibility for ideals. A homogeneous ideal \(I \subseteq K\,[\mathbf{X}]\) is \textbf{irreducible} if \& only if, whenever \(I_1\) and \(I_2\) are homogeneous ideals with \(I_1 \cap I_2 = I\), either \(I_1\) or \(I_2\) equals \(I\). Every prime ideal is irreducible: if a nontrivial ideal \(I\) is reducible as \(I = I_1 \cap I_2\) where \(I_1 \com I_2 \supset I\), then there exist \(p_1 \in I_1 \setminus I\) and \(p_2 \in I_2 \setminus I\). Though neither \(p_1\) nor \(p_2\) is in \(I\), the product \(p_1 p_2\) is by definition; thus \(I\) is not prime.
The following lemma generalizes the ``factor tree'' algorithm used in pen-and-paper integer factorization (as its proof will show).
\begin{lemma} \label{lem-ideal-irred-decomp}
Every homogeneous ideal of \(K\,[\mathbf{X}]\) can be written as a finite intersection of (one or more) irreducible homogeneous ideals.
\begin{proof}
\parencite{Arrondo2017} Let \(H\) be a homogeneous ideal of \(K\,[\mathbf{X}]\). Apply the following procedure: If \(H\) is irreducible, leave it alone. Otherwise, \(H\) is reducible, so choose homogeneous ideals \(H_0\) and \(H_1\) properly containing \(H\) such that \(H = H_0 \cap H_1\), then apply this procedure to \(H_0\) and \(H_1\), in turn.
This process yields an outward binary tree rooted at \(H\), where any bifurcation \(A_0 \leftarrow A \rightarrow A_1\) signifies that \(A = A_0 \cap A_1\) and \(A \neq A_0 \com A_1\). Every directed path in this tree is an ascending chain of ideals, which must terminate because \(K\,[\mathbf{X}]\) is a Noetherian ring; thus the tree is finite. The nodes with no children are homogeneous, irreducible ideals.
Now, in the expression for \(H\), replace every node with the intersection of its (immediate) children, or itself if it has none; for example,
\begin{equation*}
H \quad = \quad H_0 \cap H_1 \quad = \quad (H_{00}\cap H_{01}) \cap (H_{10} \cap H_{11}) \quad = \quad \dots
\end{equation*}
Continue this process until the expression doesn't change anymore. At that point, what remains is
\[H = I_1 \cap \dotsb \cap I_{m} \text{,}\]
where each \(I_k\) has no children and is thus a homogeneous, irreducible ideal.
\end{proof}
\end{lemma}
Among ideals, being primary and semiprime is equivalent to being prime. Also, the radical of a primary ideal is always prime.
\begin{lemma} \label{lem-irred-implies-primary}
Every homogeneous irreducible ideal of \(K\,[\mathbf{X}]\) is primary.
\begin{proof}
\parencite{Arrondo2017} Let \(I\) be homogeneous and irreducible, and fix \(p_1 p_2 \in I\). Define a sequence \((I_k)_k\) of ideals by
\begin{equation}
I_k = \left\{h \in K\,[\mathbf{X}] : h \, p_1^k \in I \right\}\text{.}
\end{equation}
This sequence forms a chain, since \(I_k \subseteq I_{k+1}\) for each \(k \geq 0\), which eventually stabilizes because \(K\,[\mathbf{X}]\) is Noetherian. Let \(m\) be the least integer such that \(I_m = I_{m+1}\).
We will show that
\[I = \langle I \com p_1^m \rangle \cap J\com\]
where \(J = \left\{h \in K\,[\mathbf{X}] : p_1 h \in I \right\} = I_1\).
By definition, \(I\) is contained in \(\langle I \com p_1^m \rangle\), while \(I\) is contained in \(J\) because \(I\) is an ideal. Consequently, \(I\) must be contained in the intersection \(\langle I \com p_1^m \rangle \cap J\).
Now suppose \(a \in \langle I \com p_1^m \rangle \cap J\). Membership in \(J\) implies \(p_1 \, a \in I\), while membership in \(\langle I \com p_1^m \rangle\) implies \(a = p_1^m q + r\) for some \(q \in K\,[\mathbf{X}]\), \(r \in I\). Multiplying these yields \(p_1^{m+1} q + p_1 r = p_1 \, a\), which can be rearranged as \(p_1^{m+1} q = p_1 a - p_1 r \in I\). Thus \(q \in I_{m+1} = I_m\), so \(p_1^m q \in I\). Together with \(a = p_1^m q + r\), this implies that \(a \in I\).
We have shown \(I = \langle I \com p_1^m \rangle \cap J\). Since \(I\) is irreducible, either \(I = \langle I \com p_1^m \rangle\) (which contains \(p_1^m\)) or \(I = J\) (which contains \(p_2\)). We derived this from assuming \(p_1 p_2 \in I\), so \(I\) is primary.
\end{proof}
\end{lemma}
\begin{definition}[Primary decomposition]
A \textbf{primary decomposition} of an ideal \(I\) is a finite collection of primary ideals \(C = \{I_1 \com \dotsc \com I_m\}\) such that \(I = I_1 \cap \dotsb \cap I_m\). The \textbf{associated primes} of \(C\) are the (prime) ideals in the set \(\sqrt{C} = \{\sqrt{I_1} \com \dotsc \com \sqrt{I_m}\}\). We say that \(C\) is \textbf{irredundant} if \& only if the radicals \(\sqrt{I_k}\) are pairwise distinct and no \(I_k\) contains the intersection of the others.
\end{definition}
\begin{theorem}[Primary decomposition theorem]
Every homogeneous ideal \(I\) of \(K\,[\mathbf{X}]\) admits a irredundant primary decomposition, and any two such decompositions share the same set of associated primes.
\begin{proof}
\emph{Existence.} By lemma \ref{lem-ideal-irred-decomp}, there exists a finite collection \(C_0\) of irreducible ideals whose intersection is \(I\). Irreducible ideals are primary by lemma \ref{lem-irred-implies-primary}, so \(C_0\) is a primary decomposition. After removing any ideals from \(C_0\) that contain the intersection of the others, we are left with an \emph{irredundant} primary decomposition \(C = \{I_1 \com \dotsc \com I_m\}\).
\emph{Uniqueness.} \parencite{Cox2007} Let \(C = \{I_1 \com \dotsc \com I_m\}\) be an irredundant primary decomposition of \(I\) with associated primes \(P_k = \sqrt{I_k}\). We will show that the \(P_k\) are precisely the \emph{prime} ideals of the form \(\sqrt{I}:f\),\footnote{Really, this is an abuse of notation. If we wanted to be more precise we'd write \(\sqrt{I} : \langle f \rangle\).} where \(f \in K\,[\mathbf{X}]\). This will suffice, as the collection of such ideals does not depend on any particular primary decomposition.
Before we begin element-chasing, observe that for any \(f \in K\,[\mathbf{X}]\),
\begin{align}
\begin{split} \label{eq-ass-primes}
\sqrt{I}:f &= (P_1 \cap \dotsb \cap P_m):f\\
& = (P_1 : f) \cap \dotsb \cap (P_m : f)\text{.}
\end{split}
\end{align}
With this underfoot, suppose \(P = \sqrt{I}:f\) is a prime ideal. Prime ideals are irreducible, so by equation \eqref{eq-ass-primes}, there is some \(k\) such that \(P = P_k : f\). Either \(f\) is in \(P_k\) or it isn't. If \(f\) is in \(P_k\), then \(P_k:f = K\,[\mathbf{X}]\), which is impossible because prime ideals are proper. Thus \(f\) lies outside \(P_k\). But \(P_k\) is prime, meaning that \(P_k : f = P_k\); hence \(P = P_k\).
Now fix an arbitrary \(k\), and consider \(P_k\). Because \(C\) is an irredundant decomposition, the set \(\bigcap_{j \neq k} (P_j \setminus P_k)\) contains at least one element \(f\). By the prime-ideal comment from before, \(P_j : f = K\,[\mathbf{X}]\) (for \(j \neq k\)) and \(P_k : f = P_k\). Substituting these values into equation \eqref{eq-ass-primes} yields that \(\sqrt{I}:f = P_k\).
\end{proof}
\end{theorem}
\chapter{The Hilbert polynomial}
To define the intersection multiplicity, we first need to take a closer look at homogeneous ideals. In what follows, we will consider a fixed homogeneous ideal \(I \subseteq K\,[\mathbf{X}]\) and the associated projective subset \(V = \V_{\Pro}(I) \subseteq \Pro^n(K)\).
\begin{definition}[Coordinate ring]
The \textbf{(homogeneous) coordinate ring} of \(I\), denoted by \(\Gamma_{\Hmg}(I)\) or \(\Gamma(I)\), is the quotient ring \(K\,[\mathbf{X}]/I\).
\end{definition}
Now, \(K\,[\mathbf{X}]\), as a vector space over \(K\), decomposes to the direct sum
\begin{equation}
K\,[\mathbf{X}] = \bigoplus_{d=0}^{\infty} K[\mathbf{X}]_{d} \text{,}
\end{equation}
where \(K[\mathbf{X}]_d\) is the subspace of all homogeneous polynomials of degree \(d\). The elements \(X_0 \com \dotsc \com X_n\) generate the \(K[\mathbf{X}]_d\), each of which has dimension \(\frac{(n+1)_d^{+}}{d!}\), giving \(K\,[\mathbf{X}]\) the structure of a graded ring. Consequently, the quotient map \(K\,[\mathbf{X}] \rightarrow \Gamma(I)\) naturally induces a grading on the coordinate ring:
\begin{equation}
\Gamma(I) = \bigoplus_{d=0}^{\infty} \Gamma(I)_d\text{,}
\end{equation}
where we define \(\Gamma(I)_d\) as the image of \(K[\mathbf{X}]_d\) under the quotient map \(p \mapsto p+I\). That map is linear, so the \(\Gamma(I)_d\) are vector spaces, but in general their dimension need not equal \(\frac{(n+1)_d^{+}}{d!}\). In fact, the dimensions of these subspaces contain valuable geometric information about \(I\) and \(V\). All that data may be encapsulated by a single function.
\begin{definition}[Hilbert function]
Let the commutative graded ring \(A = \bigoplus_{d=0}^{\infty} A_d\) be finitely generated over \(A_0 = K\). The \textbf{Hilbert function} of \(A\) is the map \(h_A : \mathbb{N}_0 \rightarrow \mathbb{N}_0\) defined by
\begin{equation}
h_A(d) = \dim{(A_d)}\text{;}
\end{equation}
that is, \(h_A(d)\) is the dimension of \(A_d\) as a vector space over \(K\). The \textbf{Hilbert function of} \(I\) is that of the algebra \(\Gamma(I)\). \parencite{Arrondo2017}
\end{definition}
\begin{proposition}
Let the commutative graded ring \(A = \bigoplus_{d=0}^{\infty} A_d\) be finitely generated over \(A_0 = K\) by the homogeneous elements \(x_0 \com \dotsc \com x_n\) of degree 1. There exists a polynomial \(P\) with rational coefficients such that, for all sufficiently large \(d\), we have \(h_A(d) = P(d)\).
\begin{proof}
The proof is by induction on the number of generators. If there are no generators, then \(A = K\) and \(h_A(d) = 0\) for all \(d > 0\), so \(P\) can be the zero polynomial.
Now suppose that the theorem has been proved for \(n\) generators. Let \(A = \bigoplus_{d=0}^{\infty} A_d\) be generated by the \(n+1\) generators \(x_0 \com \dotsc \com x_n\), fix some integer \(d \geq 0\), and consider the linear map \(x_0: A_d \rightarrow A_{d+1}\) defined by \(x_0(a) = x_0 a\). Define \(N_d\) to be the kernel of \(x_0\) in \(A_d\), and \(M_d = A_{d+1} / \text{im}(x_0)\) the cokernel. This gives an exact sequence
\[0 \rightarrow N_d \hookrightarrow A_d \xrightarrow{x_0} A_{d+1} \twoheadrightarrow M_d \rightarrow 0\text{.}\]
Since \(N_d\) and \(M_d\) are both finitely generated \(A\)-modules annihilated by \(x_0\), both are \(K[x_1 \com \dotsc \com x_n]\)-modules and thus vector spaces over \(K\). We have
\begin{equation*}
\dim(N_d) - \dim(A_d) + \dim(A_{d+1}) - \dim(M_d) = 0
\end{equation*}
and hence, by the induction hypothesis, there exist polynomials \(P_M \com P_N\) such that for sufficiently large \(d\),
\begin{equation*}
h_A(d+1) - h_A(d) = P_M(d) - P_N(d)\text{.}
\end{equation*}
We will construct a polynomial \(P\) such that \(P(T+1)-P(T) = P_M(T) - P_N(T)\). Let \(m = \deg(P_M - P_N)\), and using the basis of binomial coefficients in \(T\), write
\[ (P_M - P_N)(T) = \sum_{k=0}^m a_k\frac{(T)_k^{-}}{k!}\text{.}\]
We can choose \(P\) to be
\[P(T) = \sum_{k=0}^{m} a_k\frac{(T+1)_{k+1}^{-}}{(k+1)!}\text{.}\]
Then \(h_A(d) = P(d) + c\) for some constant \(c\).
\end{proof}
\end{proposition}
The polynomial \(P\) is called the \textbf{Hilbert polynomial} of \(A\). By abuse of terminology, we will refer to the Hilbert polynomial of \(\Gamma(I)\) as simply the Hilbert polynomial of \(I\), and write it as \(\HP[I]\) or \(\HP_I\). The Hilbert polynomial of a projective set \(V\) is that of \(\I(V)\); it is written \(\HP_V\).
It turns out that the Hilbert polynomial of an ideal contains geometric information in its coefficients, degree, and values.
\begin{proposition}[Properties of the Hilbert polynomial] \label{prop-Hilbert-polynomial}
Let \(I\), \(I_1\), \(I_2\) be homogeneous ideals and \(V\) a projective set.
\begin{enumerate}
\item The set \(\V(I)\) is empty if \& only if \(\HP_I = 0\).
\item Let \(I = I_0 \cap J\), where \(I_0\) is the \(\langle \mathbf{X} \rangle\)-primary component of \(I\), and \(J\) is the intersection of the other components. Then \(\HP_I = \HP_{J}\).
\item The following are equivalent:
\begin{itemize}
\item \(V\) is a finite set.
\item \(\HP_V\) is constant.
\item \(\HP_V = \# V\), the number of points in \(V\).
\end{itemize}
\item Inclusion-exclusion:
\begin{equation}
\HP[I_1 + I_2] + \HP[I_1 \cap I_2]= \HP[I_1] + \HP[I_2] \label{eq-inc-exc-hp}
\end{equation}
\item If \(f\) is homogeneous of degree \(k\) and not contained in any relevant associated prime of \(I\), then
\begin{equation}
\HP_{\langle I \com f \rangle}(T) = \HP_I(T) - \HP_I(T - k) \label{eq-Hilbert-degree-reduction}
\end{equation}
\end{enumerate}
\end{proposition}
\begin{definition}[Dimension of a projective set]
Let \(W \subseteq \Pro^n(K)\) be a projective set. The \textbf{dimension} of \(W\) is the degree of the Hilbert polynomial of \(W\); that is, \(\dim(W) = \deg(\HP_{W})\).
\end{definition}
\begin{proposition}
Let \(I_1 \com I_2\) be homogeneous ideals of \(K\,[\mathbf{X}]\) such that \(\V(I_1) = \V(I_2)\). Then \(\deg{\HP[I_1]} = \deg{\HP[I_2]}\). In particular, if \(V = \V(I)\), then \(\dim(V) = \deg{\HP_I}\).
\end{proposition}
\chapter{B\'{e}zout’s theorem}
Let \(I_1\) and \(I_2\) be homogeneous ideals of \(K\,[\mathbf{X}]\), and consider the corresponding projective sets \(\V(I_1)\) and \(\V(I_2)\). We want to talk about the intersection \(U = \V(I_1 + I_2)\). If \(I_1\) and \(I_2\) are suitably independent, this intersection consists of finitely many points.
Suppose \(U\) is finite. We can write \(U\) as \(\{u_1\} \cup \dotsb \cup \{u_m\}\). The associated ideal is \(\I(U) = \sqrt{I_1 + I_2} = \I\{u_1\} \cap \dotsb \cap I\{u_m\}\),
and thus \(I_1 + I_2\) has primary decomposition \(\{J_1 \com \dotsc \com J_m\}\)
where for each \(k\), the ideal \(J_k\) is \(\I\{u_k\}\)-primary. Since \(\V(J_k) = \{u_k\}\), the Hilbert polynomial of each \(J_k\) is constant. The next definition captures the value of this constant.
\begin{definition}[Intersection multiplicity]
Let \(u\) be a point in \(U = \V(I)\), and suppose \(\I\{u\}\) is an associated prime of \(I\). The \textbf{intersection multiplicity} of \(I\) at \(u\), denoted \(\IM_u(I)\), is the constant value of the Hilbert polynomial \(\HP[I_u]\), where \(I_u\) is the \(\I\{u\}\)-primary component of \(I\).
For projective sets \(W_1\), \(W_2\), the \textbf{intersection multiplicity} of \(W_1\) and \(W_2\) at \(u\) is assumed to be that of \(\I(W_1) + \I(W_2)\). By abuse of notation, we may write \(\IM_u(W_1 \cap W_2)\) for \(\IM_u(\I(W_1) + \I(W_2))\).
\end{definition}
% Add example of intersection multiplicity
Proving B\'{e}zout's theorem relies on an essential concept: the degree. Consider a projective set \(W\) of dimension \(m\). The Hilbert polynomial of \(W\) is given by
\[\HP_W(T) = aT^{\;m} + r(T)\]
where \(a > 0\) and \(\deg(r) < m\). Let \(A = \V\{\alpha\}\) where \(\alpha \in K\,[\mathbf{X}] \setminus \I(W)\) is a linear functional (making \(A\) a hyperplane). According to equation \eqref{eq-Hilbert-degree-reduction},
\begin{align*}
\HP_{A \cap W}(T) &= \HP_W(T) - \HP_W(T - 1)\\
&= a[T^{\;m} - (T - 1)^{m}] + r(T) - r(T-1)\text{,}
\end{align*}
which has leading term \(amT^{\;m-1}\) by the binomial theorem. We can do this \(m\) times, intersecting \(W\) with independent hyperplanes \(A_1 \com \dotsc \com A_m\), leaving a finite set \(W \cap A_1 \cap \dotsb \cap A_m\) with constant Hilbert polynomial \(am!\).
\begin{definition}[Degree]
Let \(W \subseteq \Pro^n(K)\) be a projective set of dimension \(m\). The \textbf{degree} of \(W\) is defined to be \(am!\), where \(a\) is the leading coefficient of \(\HP_W\).
Similarly, the \textbf{degree} of a homogeneous ideal \(I\) is \(am!\), where \(a\) is the leading coefficient and \(m\) the (polynomial) degree of \(\HP_I\).
\end{definition}
The degree of \(V\), as we saw above, may be characterized geometrically: it is the maximum number of intersections between \(W\) and \(\dim(W)\) hyperplanes in general position.
\begin{lemma} \label{lem-bezout-beta}
Let \(H\) be an ideal of degree \(D\) and dimension \(m > 0\), and let \(f\) be a polynomial of degree \(d\) not in any associated prime of \(H\). Then
\begin{equation}
\deg{\langle H \com f\, \rangle} = Dd\text{.}
\end{equation}
\begin{proof}
Express the Hilbert polynomial of \(H\) as \(\HP_H(T) = r(T) + \frac{1}{m!} D T^{\,m}\), where \(\deg(r) < m\). Then
\begin{align*}
\HP_{\langle H \com f\, \rangle}(T) &= \HP_H(T) - \HP_H(T - d)\\
&= r(T) - r(T - d) + \tfrac{1}{m!}D \, \big(T^{\,m} - (T - d)^{\,m}\big)\text{.}
\end{align*}
From this expression, we see \(\HP_{\langle H \com f\, \rangle}\) is a polynomial of degree \(m-1\). Its leading term is the same as that of \(\frac{1}{m!} D\, \big(T^{\,m} - (T-d)^{\,m}\big)\), which (by the binomial theorem) is \(\frac{1}{(m-1)!}Dd\,T^{\;m-1}\). By the definition of degree, \(\deg{\langle H \com f \, \rangle} = Dd\).
\end{proof}
\end{lemma}
At long last, we are ready to prove B\'{e}zout's theorem!
\begin{theorem}[B\'{e}zout]
Suppose \(K\) is algebraically closed. Let \(F\) be a set of \(n\) homogeneous polynomials in \(K\,[\mathbf{X}]\), each of which generates a semiprime ideal. If \(Z = \V(F)\) is finite, then
\begin{equation}
\sum_{z \in Z} \IM_z(H) = \prod_{f \in F} \deg(f)\text{,} \label{eq-bezout}
\end{equation}
where \(H = \langle F \rangle\).
\begin{proof}
Assume \(Z\) is finite. The key is to show that both sides of equation \eqref{eq-bezout} equal \(\deg(H)\). By applying lemma \ref{lem-bezout-beta} once for each \(f \in F\), we get that
\[\deg(H) = \prod_{f \in F} \deg(f)\text{.}\]
Now let \(C\) be an irredundant primary decomposition of \(H\).
\emph{Case 1.} Suppose \(C\) consists only of minimal components. Then \(C = \{I_z : z \in Z\}\) where each \(I_z\) is \(\I\{z\}\)-primary. Hence, we have
\begin{equation*}
H = \bigcap_{z \in Z} I_z \quad \Rightarrow \quad \HP_H = \HP\left[\bigcap_{z \in Z} I_z\right]\text{.}
\end{equation*}
Now, if \(y \neq z\), then \(\V(I_y + I_z) = \{y\}\cap\{z\} = \emptyset \), implying that \(\HP[I_y + I_z] = 0\). So, applying the inclusion-exclusion principle,
\begin{equation*}
\HP_H = \sum_{z \in Z} \HP[I_z] = \sum_{z \in Z} \IM_z(H)\text{.}
\end{equation*}
\emph{Case 2.} Now suppose that \(\{I_z : z \in Z\}\) are the minimal components but \(I_0 \in C\) is not. Then, by definition of minimality, for some \(z \in Z\) the proper inclusion \(\sqrt{I_z} \subset \sqrt{I_0}\) holds. This implies \(\V(I_z) \supset \V(I_0)\), and since \(\V(I_z) = \{z\}\) is a singleton, the set \(\V(I_0)\) is empty. By the Nullstellensatz, \(I_0\) is \(\langle\mathbf{X}\rangle\)-primary, so it is the \emph{only} non-minimal component. Proposition \ref{prop-Hilbert-polynomial} then gives \(\HP_H = \HP\left[\bigcap_{z \in Z} I_z \right]\), from which point the argument is identical to that of case 1.
\end{proof}
\end{theorem}
\printbibliography
\appendix
\chapter{Notation} \label{appendix-notation}
We write \(f: W \leftarrow X\) or \(f: X \rightarrow W\) to mean that \(f\) is a function with codomain \(W\) and domain \(X\). For any subset \(S_W \subseteq W\), we define the \textbf{preimage} of \(S_W\) under \(f\) as
\begin{equation}
f^{\;-\smallsquare}(S_W) = \{x \in X : f(x) \in S_W\}\text{.}
\end{equation}
Similarly, given \(S_X \subseteq X\), we define the \textbf{image} of \(S_X\) under \(f\) as
\begin{equation}
f^{\;\smallsquare}(S_X) = \{w \in W : \exists x \in S_X \sthat w = f(x) \}\text{.}
\end{equation}
For inclusion, we write \(A \subseteq B\) to mean ``\(A\) is a subset of \(B\).'' When \(A\) is a subset of \(B\), but \(A \neq B\), we write \(A \subset B\).
We write \(\mathbb{N}_0 = \mathbb{Z} \cap [0 \com \infty)\) to denote the set of natural numbers starting at 0, and \(\mathbb{N}_1 = \mathbb{Z} \cap [1 \com \infty)\) to denote the set of natural numbers starting at 1. The cardinality of a set \(S\) is written \(\# S\).
The \textbf{rising} and \textbf{falling factorials} are given by
\begin{align}
(z)_k^{+} &= z \, (z + 1) \dotso (z + k - 1) \text{ and}\\
(z)_k^{-} &= z \, (z - 1) \dotso (z - k + 1)\text{, respectively.}
\end{align}
The \textbf{binomial coefficient}, given by
\begin{equation}
\frac{(z)_k^{-}}{k!} = \frac{z \, (z-1) \dotso (z - k + 1)}{k!}\text{,}
\end{equation}
counts the \(k\)-element subsets of a set with cardinality \(z\), when \(z\) is a positive integer. Similarly, the \textbf{multiset coefficient}, given by
\begin{equation}
\frac{(z)_k^{+}}{k!} = \frac{z \, (z+1) \dotso (z + k - 1)}{k!}\text{,}
\end{equation}
counts the ways to buy \(k\) items from a catalog of \(z\) products (buying the same product more than once is allowed).
Let \((A \com \prec)\) be a strict poset. We say that \(m\) is a \textbf{minimal element} of \(A\) if \& only if there does \emph{not} exist \(a \in A\) such that \(a \prec m\).
\chapter{Abstract algebra}
\begin{definition}[Ring]
A \textbf{ring} is an algebraic structure \(\mathbf{R} = [R \com 0 \com - \com + \com \times]\), where 0 is a constant, \(-\) (negative) is unary, and \(+\) (plus) and \(\times\) (times) are binary, satisfying the following conditions.
\begin{itemize}
\item First, \([R \com 0 \com - \com +]\) is an abelian group.
\begin{align*}
(r_1 + r_2) + r_3 &= r_1 + (r_2 + r_3)\\
r_1 + r_2 &= r_2 + r_1\\
0 + r &= r\\
-r + r &= 0
\end{align*}
\item Second, the times operation is associative and distributes over addition.
\begin{align*}
(r_1 \; r_2) \; r_3 &= r_1 \; (r_2 \; r_3)\\
r \, (s_1 + s_2) &= r s_1 + r s_2\\
(r_1 + r_2) \, s &= r_1 s + r_2 s
\end{align*}
\end{itemize}
If there is also an element \(1 \in R\) such that \(1 \, r = r \, 1 = r\), then we call \([R \com 0 \com 1 \com {-} \com {+} \com {\times}]\) a \textbf{ring with identity} or \textbf{unital ring}. A ring that satisfies the law \(r_1 r_2 = r_2 r_1\) is called a \textbf{commutative ring}.
\end{definition}
\begin{definition}[Homogeneous ideal]
A \textbf{homogeneous ideal} \(H\) of the polynomial ring \(K[X_0 \com \dotsc \com X_n]\) is an ideal satisfying the following property:
Let \(h = h_0 + \dotsb + h_d\) be a sum such that each \(h_k\) is homogeneous of degree \(k\). If \(h \in H\), then \(h_k \in H\).
\end{definition}
\end{document}