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Commutative Algebra and Algebraic Geometry

Throughout this section, rings $R$ are commutative with unity.

Noetherian Rings

Definition: A ring is Noetherian if it satisfies either of the following equivalent conditions:

  1. Every ideal is finitely generated.
  2. Every ascending chain $I_0 \subset I_1 \subset \cdots$ of ideals eventually stabilizes.

Theorem: (Hilbert’s Basis Theorem) If $R$ is Noetherian, then so is $R[x]$.

Proof: The proof of Hilbert’s theorem is given in detail in DF, Theorem 21 of Section 9.6. The idea is something like this. Suppose $I$ is an ideal of $R[x]$. Then each element $f$ of $I$ is a polynomial of degree $n$: $f=a_nx^n+\ldots$ where $a_n\in R$ is not zero. Here $a_n$ is called the leading term of $f$ The ideal $\ell(I)$ of leading terms of $I$ is the ideal of $R$ generated by the leading terms of all elements of $I$. Since $R$ is Noetherian, $\ell(I)$ is finitely generated. If $b_1,\ldots, b_N$ generate $\ell(I)$, we can find elements $f_1,\ldots, f_N$ so that $f_i=b_i x^k_{i}+\ldots$ in $I$.

Now let $g$ be any element of $I$ of degree $m$. Suppose $m$ is larger then all of the $k_{i}$. Since $\ell(g)$ is in $\ell(I)$, we can find $r_i\in R$ so that

\[\ell(g) =\sum r_{i} b_{i}\]

and therefore

\[g-\sum r_{i}f_{i}x^{m-k_{i}}+\ldots\]

This $g$ has degree smaller than $m$ (since we killed off its leading term). It follows that the ideal generated by the $f_i$ contains all elements of $I$ of degree at least $N=\max{k_{i}}$.

To finish the proof, we consider all the leading terms of all elements of $I$ of degree less than $N$ and consider the ideal they generate. This again is finitely generated, by, say, $c_1,\ldots, c_{M}$ and there are corresponding elements $h_{j}$ of $R[x]$ where $h_{j}=c_{j}x^{r}+\ldots$. An argument similar to what we used above says that we can use the $h_{j}$ to systematically eliminate the leading terms of $g$ until eventually we show that $g$ is in the ideal generated by the $h_{j}$ and the $f_{i}$. In other words, our original ideal $I$ is finitely generated, and thus $R[x]$ is Noetherian.

Definition: If $K$ is a field, a $K$-algebra $S$ is finitely generated if there is a surjection $K[x_1,\ldots, x_n]\to S$ for some $n$.

Warning: Note the difference between being finitely generated as a $K$-module and as a $K$-algebra.

Affine space, points, and ideals

Affine $n$-space over a field $k$, written $\mathbb{A}k^{n}$ is the set of points $(x_1,\ldots, x_n)$ with coordinates in $k$. The ring $k[x_1,\ldots, x_n]$ is the ring of polynomial functions on $\mathbb{A}{k}^{n}$.

Sets and ideals:

  • If $X\subset\mathbb{A}^{n}{k}$ is a subset, then $I(X)={f: f(x)=0 \forall x\in \mathbb{A}^{n}{k}|$.

  • If $f_1,\ldots, f_k$ are a collection of elements in $k[x_1,\ldots, x_n]$ then $Z(f_1,\ldots, f_k)={x\in \mathbb{A}^{n}{k}: f_1(x)=\cdots=f_k(x)=0}$. Notice that $Z(f_1,\ldots, f_k)=Z(I)$ where $I$ is the ideal of functions generated by the $f{i}$.

Both operations are inclusion reversing: $X\subset Y$ implies $I(Y)\subset I(X)$ and $J\subset L$ implies $Z(L)\subset Z(J)$.

The sets $Z(I)$ satisfy the axioms for the closed sets of a topology, called the zariski topology. A set $X$ of the form $Z(I)$ is called an (affine) algebraic set.

Lemma: If $X\subset\mathbb{A}^{n}_k$, then $X\subset Z(I(X))$. If $J\subset k[x_1,\ldots, x_n]$ is an ideal, then $J\subset I(Z(J))$.


  • \(Z(I(Z(J)))=Z(J)\).

Proof: We know that $J\subset I(Z(J))$ so $Z(I(Z(J)))\subset Z(J)$. On the other hand, if $x\in Z(J)$, then $f(x)=0$ for all $f\in I(Z(J))$. That means that $x$ is a common zero for $I(Z(J))$, so $x$ is in $Z(I(Z(J)))$. Therefore $Z(J)\subset Z(I(Z(J)))$.

  • \(I(Z(I(X)))=I(X)\).

Proof: We know that $X\subset Z(I(X))$ since if $x\in X$, every $f\in I(X)$ satisfies $f(x)=0$, and therefore $x\in Z(I(X))$. So $I(Z(I(X)))\subset I(X)$. On the other hand, if $f\in I(X)$, then $f(x)=0$ for all $x\in Z(I(X))$. This in turn means that $f$ is in $I(Z(I(X)))$ so $I(x)\subset I(Z(I(X)))$.

Coordinate rings and morphisms

If $X$ is an algebraic set, then the ring $k[X]=k[x_1,\ldots, x_n]/I(X)$ is called the coordinate ring of $X$. Notice that elements of this ring give well-defined functions on $X$; they are called “regular functions” on $X$.

Definition: A map $f:X\to Y$ between algebraic sets is called a morphism if there are polynomials so that the map $f$ is given by

\[f(x_1,\ldots, x_n)=(\phi_1(x_1,\ldots, x_n),\cdots, \phi_m(x_1,\ldots, x_n))\]

A morphism whose inverse is also a morphism is called an isomorphism of algebraic sets.

In general, if $f:x\to Y$ is a morphism, and $h\in k[Y]$ is a regular function on $Y$ then the composition $h\circ f$ is a regular function on $X$. This operation is called pullback.

  • The pullback induces a welldefined algebra homomorphism $f^{*}:k[Y]\to k[X]$.

  • Conversely, any algebra homomorphism $\psi:K[Y]\to K[X]$ yields a morphism $X\to Y$. Choose generators $x_1,\ldots, x_m$ for $K[Y]$ and let $F_{i}=\psi(x_i)$. Let $F$ be the corresponding map of algebraic sets. We need to check that the morphism given by $F$ carries $X$ to $Y$. The key is that if $H$ is in $I(Y)$ then $H(F_1,\ldots, F_m)$ is in $I(X)$. So every polynomial in $I(Y)$ vanishes on the image of $F$; and since $Y=Z(I(Y))$ it follows that the image of $F$ is in $Y$.

It follows that there is a bijective correspondence between $k$-algebra homomorphisms $k[X]\to k[Y]$ and regular maps $Y\to X$. This corresondence respects composition of functions and preserves isomorphisms.

Universal property

Theorem: Suppose $\phi:X\to Y$ is an arbitrary map of algebraic sets. Then $\phi$ is a morphism if and only if, for every $f\in k[Y]$, the composition $f\circ\phi$ is in $k[X]$. If $\phi$ is a morphism, then $\phi(v)=w$ if and only if $\tilde{\phi}^{-1}(I(v))=I(w)$ where $\tilde{\phi}$ is the pullback of $\phi$.

Proof: The second part first. If $\phi(v)=w$, and $f\in I(w)$, then $f(\phi(v))=0$ so $\tilde{\phi}(I(w))\subset I(v)$ and $(\tilde{\phi})^{-1}I(v)\supset I(w)$. Since $I(w)$ is maximal, we have $(\tilde{\phi})^{-1}(I(v))=I(w)$. Conversely, if $(\tilde{\phi})^{-1}(I(v))=I(w)$, then $f(\phi(v))=0$ if and only if $f(w)=0$ so $I(\phi(v))=I(w)$ and therefore $\phi(v)=w$. For the first part, pullback automatically respects addition and multiplication, so if it happens to land in $k[X]$ then it will be an algebra homomorphism $\tilde{\phi}:k[Y]\to k[X]$. Consequently there is a morphism from $X$ to $Y$ coming from $\tilde{\phi}$; call it $\tilde{\Phi}$ where $\Phi$ is a morphism from $X\to Y$. However, since the algebra maps are the same, their behavior on the ideals $I(v)$ and $I(w)$ are the same, so $\Phi$ and $\phi$ agree with one another.