Commutative Algebra and Geometry

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:

Every ideal is finitely generated.
Every ascending chain $I_{0} \subset I_{1} \subset \dots$ 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 $n$ : $f = a_{n} x^{n} + \dots$ where $a_{n} \in R$ is not zero. Here $a_{n}$ is called the leading term of $f$ The ideal $ℓ (I)$ of leading terms of $I$ is the ideal of $R$ generated by the leading terms of all elements of $I$ . Since $R$ is Noetherian, $ℓ (I)$ is finitely generated. If $b_{1}, \dots, b_{N}$ generate $ℓ (I)$ , we can find elements $f_{1}, \dots, f_{N}$ so that $f_{i} = b_{i} x_{i}^{k} + \dots$ in $I$ .

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

ℓ (g) = \sum r_{i} b_{i}

and therefore

g - \sum r_{i} f_{i} x^{m - k_{i}} + \dots

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, $c_{1}, \dots, c_{M}$ and there are corresponding elements $h_{j}$ of $R [x]$ where $h_{j} = c_{j} x^{r} + \dots$ . 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 $K [x_{1}, \dots, x_{n}] \to S$ 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} $i s t h e s e t o f p o i n t s$ (x_1,\ldots, x_n) $w i t h c o o r d i n a t e s i n$ k $. T h e r i n g$ k[x_1,\ldots, x_n] $i s t h e r i n g o f p o l y n o m i a l f u n c t i o n s o n$ \mathbb{A}{k}^{n}$.

Sets and ideals:

If $X\subset\mathbb{A}^{n}{k} $i s a s u b s e t, t h e n$ I(X)={f: f(x)=0 \forall x\in \mathbb{A}^{n}{k}|$.
If $f_{1}, \dots, f_{k}$ are a collection of elements in $k [x_{1}, \dots, x_{n}]$ then $Z(f_1,\ldots, f_k)={x\in \mathbb{A}^{n}{k}: f_1(x)=\cdots=f_k(x)=0} $. N o t i c e t h a t$ Z(f_1,\ldots, f_k)=Z(I) $w h e r e$ I $i s t h e i d e a l o f f u n c t i o n s g e n e r a t e d b y t h e$ 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 $X \subset A_{k}^{n}$ , then $X \subset Z (I (X))$ . If $J \subset k [x_{1}, \dots, x_{n}]$ is an ideal, then $J \subset I (Z (J))$ .

Also:

$Z (I (Z (J))) = Z (J)$ .

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 $k [X] = k [x_{1}, \dots, x_{n}] / I (X)$ 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}, \dots, x_{n}) = (ϕ_{1} (x_{1}, \dots, x_{n}), \dots, ϕ_{m} (x_{1}, \dots, 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 $ψ : K [Y] \to K [X]$ yields a morphism $X \to Y$ . Choose generators $x_{1}, \dots, x_{m}$ for $K [Y]$ and let $F_{i} = ψ (x_{i})$ . 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 $H (F_{1}, \dots, F_{m})$ 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 $ϕ : X \to Y$ is an arbitrary map of algebraic sets. Then $ϕ$ is a morphism if and only if, for every $f \in k [Y]$ , the composition $f \circ ϕ$ is in $k [X]$ . If $ϕ$ is a morphism, then $ϕ (v) = w$ if and only if ${\tilde{ϕ}}^{- 1} (I (v)) = I (w)$ where $\tilde{ϕ}$ is the pullback of $ϕ$ .

Proof: The second part first. If $ϕ (v) = w$ , and $f \in I (w)$ , then $f (ϕ (v)) = 0$ so $\tilde{ϕ} (I (w)) \subset I (v)$ and $(\tilde{ϕ})^{- 1} I (v) \supset I (w)$ . Since $I (w)$ is maximal, we have $(\tilde{ϕ})^{- 1} (I (v)) = I (w)$ . Conversely, if $(\tilde{ϕ})^{- 1} (I (v)) = I (w)$ , then $f (ϕ (v)) = 0$ if and only if $f (w) = 0$ so $I (ϕ (v)) = I (w)$ and therefore $ϕ (v) = w$ . 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 $\tilde{ϕ} : k [Y] \to k [X]$ . Consequently there is a morphism from $X$ to $Y$ coming from $\tilde{ϕ}$ ; call it $\tilde{Φ}$ where $Φ$ 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 $Φ$ and $ϕ$ agree with one another.