Some Classic Affine Algebraic Geometry

This will serve as background for a post I wanna write on Dedekind domains. The idea here is to blaze through 1 some geometry so I can use words like “Artinian ring” and “Krull dimension” when discussing Dedekind domains. The plan is to talk about integeral extensions of rings (+ other stuff), spectra of rings, dimensions of rings, and then Artinian rings. In a sense, this might be more commutative algebra than (classic) algebraic geometry, but what’s the difference? Note that all rings in this post are commutative with unity (and all ring maps preserve the unit).

A Mess of Words I Don’t Think I’ve Defined on this Blog Before

For completeness, before getting into integral extensions I want to really quickly fill a couple gaps in this blog by saying what a module and an algebra are.

Let $R$ be a ring. A (left) $R$-module $M$ is an abelian group with an $R$-action $R\by M\to M$ satisfying the following
  1. $r(m_1+m_2)=rm_1+rm_2$ always.
  2. $(r_2r_1)m=r_2(r_1m)$ always.
  3. $(r_1+r_2)m=r_1m+r_2m$ always.
  4. $1m=m$ where $1\in R$ is the multiplicative unit
An $R$-module structure of a group $M$ is the same thing as a ring homomorphism $R\to\End(M)$ into $M$'s endomorphism group.
If $R$ is a field, then an $R$-module is an $R$-vector space. If $R=\Z$, then an $R$-module is an abelian group.
Let $R$ be a ring. An $R$-algebra $A$ is a ring with an $R$-module structure such that $$r(a_1a_2)=(ra_1)a_2=a_1(ra_2).$$
An $R$-algebra is the same thing as a ring $A$ with a ring map $R\to Z(A)$ into the center of $A$.

Modules, and in particular algebras, will feature heavily in this post. Modules in general can be rather poorly behaved, but we can maintain our sanity if we restrict ourselves to modules over a commutative ring that satisfies a nice finiteness property.

An $R$-module $M$ is called Noetherian if it satisfies the ascending chain condition on submodules; that is, any chain $$M_1\subseteq M_2\subseteq M_3\subseteq\cdots\subseteq M$$ on submodules eventually stabilizes (i.e. $M_n=M_{n+1}=\cdots$ for some $n\ge1$). If $R$ is Noetherian as an $R$-module, then we call it a Noetherian ring.
I never defined what a submodule is, but I think it's not that hard to figure out. However, while I'm on the subject of submodules, two things: (1) quotient modules always exist (i.e. given any submodule $N\subseteq M$, the natural quotient $M/N$ can be given an $R$-module structure) and (2) an $R$-submodule of $R$ is the same thing as an ideal.

The Noetherian property is a kind of natural generalization being a PID. This is maybe not immediately obvious from the given definition, but this next theorem will help.

Let $M$ be an $R$-module. Then $M$ is Noetherian iff every submodule of $M$ is finitely generated.
$(\to)$ Assume $M$ is Noetherian, and let $N\subset M$ be a submodule. Fix some $x_1\in N$, and let $N_1=Rx_1$ be the module it generates. Inductively choose $x_n\in M$ ($n>1$) such that $x_n\in N\sm N_{n-1}$ if $N_{n-1}\neq N$, and $x_n=x_{n-1}$ otherwise; set $N_n=\sum_{i=1}^nRx_i$. This gives an ascending chain $$N_1\subseteq N_2\subseteq N_3\subseteq\cdots\subseteq M$$ of submodules of $M$ (submodules of $N$ even), and so must stabalize at the $m$th step for some $m$. Then, $N_m=N$ (otherwise, $N_{m+1}$ would be bigger by construction), so $N$ is generated by $m<\infty$ elements.
$(\from)$ Exercise.
A ring is Noetherian iff every ideal is finitely generated.
This proposition makes it clear that subrings/submodules of a Neotherian ring/module are also noetherian (e.g. since submodules of $N\subset M$ are still submodules of $M$ and submodules of $M/N$ are submodules of $M$ containing $N$).

Another nice thing to know about the Noetherian property is that almost every ring/module you will ever care about is Noetherian.

Let $R$ be a noetherian ring. Then every finitely generated (f.g.) $R$-module is noetherian.
The main part of the proof is to show that noetherianess is preserved under extensions (i.e. if $N\subset M$ and $M/N$ are noetherian, then so is $M$), so we only prove this (to finish, show f.g. free modules $R^{\oplus m}$ are noetherian, and every f.g. module is a quotient of a f.g. free module). Consider a short exact sequence of $R$-modules $$0\too N\too M\too M/N\too 0$$ where $N$ and $M/N$ are Noetherian. Let $M'\subset M$ be any $R$-submodule, so we get another short exact sequence $$0\too N\cap M'\xtoo fM'\xtoo gM'/(N\cap M')\too0$$ where $N\cap M'\subseteq N$ and $M'/(N\cap M')\simeq (N+M')/N\subseteq M/N$. Hence, these modules are finitely generated, say by $\{e_1,\dots,e_n\}\subseteq N\cap M'$ and $\{\bar f_1,\dots,\bar f_m\}\subseteq M/(N\cap M')$, respectively. Now, pick any $m\in M'$, and write (non-uniquely) $g(m)=r_1\bar f_1+\dots+r_m\bar f_m$ with $r_i\in R$. Because $g$ is surjective, we can pick (non-unique) lifts $f_1,\dots,f_m\in M'$ of $\bar f_1,\dots,\bar f_m$, so $m-(r_1f_1+\dots+r_mf_m)\in\ker g=\im f$. Hence, $$m=(r_1f_1+\dots+r_mf_m)+(s_1e_1+\dots+s_ne_n)$$ for some $s_j\in R$ (where we've identified $e_i\in N\cap M'$ with $f(e_i)\in M'$ since $f$ is injective). Thus, $M'$ is generated by $\{e_1,\dots,e_n\}\cup\{f_1,\dots,f_m\}$, so every submodule of $M$ is finitely generated.
Let $R$ be a Noetherian ring. Then, $R[x]$ is Noetherian as well.
Exercise (hint: given an ideal $I\subseteq R[x]$, first consider the ideal of leading coefficients of elements of $I$)
Let $R$ be a Noetherian ring. Then every finitely generated $R$-algebra is Noetherian.

I think that’s all the highlights about Noetherianess. Let’s get into integral extensions.

Let $\phi:R\to S$ be a ring map (e.g. an injection of rings). Then, $s\in S$ is integral over $R$ if it is the root of a monic polynomial in $R[x]$. If all $s\in S$ are integral over $R$, then $S$ is integral over $R$ or an integral extension of $\phi(R)$.
Let $\phi:R\to S$ be a ring map. The integral closure of $R$ in $S$ is the set $$\bar R=\{s\in S:s\text{ is integral over }R\}.$$ We say that $R$ is integrally closed in $S$ if $\bar R=R$, and we say that a domain $R$ is integrally closed if it is integrally closed in $\Frac(R)$.
Integral closures are rings.
Let $\phi:R\to S$ be a ring map, and let $a,b\in S$ be integral over $R$. We need to show that $ab$ and $a+b$ are integral over $R$ as well (I guess we also need that $-a$ is integral over $R$). This is a consequence of the following lemma (consider the ring $R[a,b]$).
Let $\phi:R\to S$ be a ring map. Then, $s\in S$ is integral over $R$ iff there exists a a subring $S'\subseteq S$ containing $s$ which is finitely generated as an $R$-module.
$(\to)$ Assume that $s\in S$ is integral over $R$, and let $d$ be the degree of a monic polynomial in $R[x]$ vanishing at $s$. Then, $R[s]\subseteq S$ is a ring generated as an $R$-module by $\{s^k\}_{k=0}^{d-1}$.
$(\from)$ Let $S'\subseteq S$ be a subring containing $s$ which is finitely generated as an $R$-module. Write $S'=\sum_{i=1}^nRe_i$ with $e_i\in S'$. Note that the multiplication by $s$ map $m_s:S'\to S', x\mapsto sx$ is $R$-linear, and so can be represented (non-uniquely) by some matrix. That is, $m_s(e_j)=\sum_ia_{ij}e_i$ for some $a_{ij}\in R$, so $$\Mat s\dots0\vdots\ddots\vdots0\dots s\vVec{e_1}\vdots{e_n}=\Mat{a_{11}}\dots{a_{1n}}\vdots\ddots\vdots{a_{n1}}\dots{a_{nn}}\vVec{e_1}\vdots{e_n}.$$ Subtracting now gives $$\Mat {s-a_{11}}\dots{-a_{1n}}\vdots\ddots\vdots{-a_{n1}}\dots{s-a_{nn}}\vVec{e_1}\vdots{e_n}=\vVec0\vdots0.$$ Call the matrix on the left $M$. By Cramer's rule, we can form its adjugate $M^{\mrm{adj}}$ so that $M^{\mrm{adj}}M=\det M\cdot I$ where $I$ is the identity matrix. Multiplying the above equation by the adjugate gives $(\det M)e_i=0$ for all $i$. Since the $e_i$'s generate $S'$, we conclude $(\det M)x=0$ for all $x\in S'$. Taking $x=1$, gives $\det M=0$. Finally, $\det M$ literally is a monic polynomial in with $R$-coefficients vanishing at $s$, so we win.
A finitely generated $R$-algebra $S$ is integral over $R$ iff $S$ is finitely generated as an $R$-module.

Integral extensions are the ring-theoretic analogue of algebraic extensions from field theory. Indeed, if $R$ is a field, then $x$ is integral over $R$ iff it is algebraic over $R$. Integral extensions have many nice properties, some of which are collected below.

Integral closures are integrally closed.
Exercise.
UFDs are integrally closed.
Exercise (Hint: this is basically the rational root theorem).
Let $\phi:A\to B$ be a ring map, and let $S\subset A$ be a multiplicative set. If $B$ is integral over $A$, then $\phi(S)^{-1}B$ is integral over $\sinv A$.
Exercise.
Let $S\subset R$ be a multiplicative set, and suppose that $R$ is integrally closed. Then, $\sinv R$ is integrally closed as well.
Let $A\subset B$ be an integral extension of rings. Then, $A$ is a field iff $B$ is a field.
$(\to)$ Suppose that $A$ is a field, and fix any $b\in B$. Then, $b$ is the root of some $$f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in A[x].$$ This means that $$-a_0=b(b^{n-1}+a_{n-1}b^{n-2}+\cdots+a_1),$$ and hence $\inv b=\frac{-1}{a_0}(b^{n-1}+a_{n-1}b^{n-2}+\dots+a_1)\in B$.
($\from$) This direction is proved similarly using that $\inv a\in B$ for any nonzero $a\in A$.
Let $A\subset B$ be an integral extension of rings. Let $\mfa\subsetneq B$ be an ideal, and let $\mfa'=\mfa\cap A$. Then, $B/\mfa$ is integral over $A/\mfa'$.
Just take a monic polynomial and reduce it mod $\mfa'$.
Let $A\subset B$ be an integral ring extension, and let $\mfp\subset B$ be a prime ideal. If $\mfp\cap A$ is maximal, then so is $\mfp$.
$B/\mfp$ is integral over the field $A/(\mfp\cap A)$.
Let $A\subset B$ be an integral ring extension, and let $\mfp\subset A$ be prime. Then, there's some prime ideal $\mfp'\subset B$ with $\mfp'\cap A=\mfp$.
Let $S=A\sm\mfp$, so $\sinv B$ is integral over the local ring $\sinv A=A_{\mfp}$ with unique maximal ideal $\sinv\mfp$. Let $\mfm\subset\sinv B$ be any maximal ideal. Then, $\mfm'=\mfm\cap\sinv A$ is prime, so $\sinv B/\mfm$ is a field integral over $\sinv A/\mfm'$; hence, $\mfm'$ is maximal and so must be $\sinv\mfp$. Thus, $\mfp'=\mfm\cap B$ is a prime of $B$ laying over $\mfp$ since $$(\mfm\cap B)\cap A=\mfm\cap A=(\mfm\cap\sinv A)\cap A=\sinv\mfp\cap A=\mfp.$$

Alright. I think that’s more than enough about integral extensions for now. Bottom line: they’re preserved by reasonal operations and they play nicely with primes.

Finally, Some Geometry

Let’s actually justify the use of the word geometry in the title of this post by talking of $\spec$. Hilbert’s Nullstellensatz (which we’ll prove later) shows that, for an algebrically closed field $k$, points in $k^n$ are in bijection with maximal ideals of $k[x_1,\dots,x_n]$. Since we’d like to do geometry in purely algebraic settings (i.e. over an arbitrary ring $R$), we may think that a good replacement for $k^n$ is the set of maximal ideals of $R$. However, there’s a better choice; the set of prime ideals of $R$.

Let $R$ be a ring. Its spectrum is $$\spec R=\bracks{\mfp\subset R:\mfp\text{ is prime}}.$$

Now, these spectres or whatever are supposed to be our replacements for things like $\C^n$, so they better be geometric in some sense. At the very least, they better have a topology.

Given an ideal $I\subseteq R$, its "vanishing set" is $V(I)=\bracks{\mfp\in\spec R:\mfp\supseteq I}$. The Zariski topology on $\spec R$ is given by having closed sets be $V(I)$ for $I$ an ideal.
Prove that this is an actual topology. Also prove that $V(I)$ is homeomorphic to $\spec R/I$. If you haven't had enough of this exercise, show that the open set $\spec R\sm V(r)$ (where $r\in R$) is homeomorphic to $\spec R_r=\spec R[1/r]$.
Given a ring map $f:R\to S$, we get an induced map on spectra $\spec S\to\spec R$ given by sending $\mfp\subset S$ to $\inv f(\mfp)\subset R$. This map is easily seen to be continuous, and so $\spec$ is a contravariant functor from Rng to Top.

The idea behind the Zariski topology is that “zero sets of polynomials” should be closed 2, so how do we get from that to this? Well, think of $\spec R$ as some variety (e.g. if $R=\C[x,y]/(y^2-x^3-x)$, then $\spec R$ is the curve given by $y^2=x^3-x$) and we want elements of $R$ to be functions on $\spec R$ (i.e. intuitively, elements of $R$ should give well-defined polynomials on $\spec R$ (see previous parenthetical)). The natural way of realizing this is to say that $r(\mfp)= r\pmod\mfp$ where $r\in R$ and $\mfp\in\spec R$ 3, so $r(\mfp)=0$ precisely when $r\in\mfp$. With this in mind, the points of $\spec R$ vanishing on each function in an ideal $I$ 4 are exactly those that contain $I$.

We now have a topological space associated to an arbitrary ring $R$. How should we think about it geometrically? Well, motivated by Hilbert’s Nullstellensatz, we should think of the maximal ideals of $R$ as the (closed) points of $\spec R$ (point here used geometrically as a 0-dimensional thing. Any element of $\spec R$ could be reasonably called a point since it’s a space). Motivated by the exercise, the non-maximal prime ideals should correspond to higher-dimensional subvarieties: curves and hyperplanes and whatnot. This does beg the question though: what do I mean by dimension here? To answer that, I first need to define irreducible sets. 5

A subvariety $V(I)\subset\spec R$ (note: $I$ arbitrary) is called irreducible if writing $V(I)=V(J_1)\cup V(J_2)$ requires that $I=J_1$ or $I=J_2$.
Show that $V(I)$ is irreducible iff $V(I)=V(\mfp)$ with $\mfp\subset R$ prime.

Irreducible varieties are the ones that we care about; these are things like 1 point, 1 curve, 1 plane, etc. Intuitively, an irreducible variety cannot contain two things of the same dimension (otherwise you could write it as the union of those two things), so if $V(I)\subset V(J)$ with both irreducible, you would expect that $\dim V(I)<\dim V(J)$.

Let $X=\spec R$. Then, $\dim X$ is defined to be the length of the longest chain $X_1\subsetneq X_2\subsetneq\dots\subsetneq X_n=X$ of irreducible subvarieties of $X$. Similarly, $\dim R$ is defined to be $\dim(\spec R)$.
By the previous exercise, the dimension of a ring is the length of its longest chain of prime ideals.
It is fairly easy to see that $\dim A\ge\dim A_\mfp+\dim A/\mfp$ for all $\mfp\in\spec A$. It turns out that this is always an equality. I won't prove this in this post, but after finishing this section, you should be able to do this by inducting on $\dim A$ and making use of Noether normalization.
Let $\A_k^n=k[x_1,\dots,x_n]$ where $k$ is a field. Then, $\dim\A_k^n=n$. In general, $\dim R[x]=1+\dim R$.
$\dim\Z=1$. In general, $\dim R\le1$ if $R$ is a PID.
$\dim k=0$ if $k$ is a field (this is not an iff!).
$\dim R=0$ iff all prime ideals are maximal. For $R$ a domain, $\dim R=1$ iff all nonzero prime ideals are maximal.

For the rest of this section, I’ll discuss some alternative definitions of dimension when your ring $R$ is niceish. Fix a field $k$. Unless otherwise stated, assume that $R$ is a finitely generated $k$-algebra for the rest of the section. This means that we can write $R=k[x_1,\dots,x_n]/I$.

An affine irreducible variety is an irreducible set $\spec k[x_1,\dots,x_n]/I=V(I)\subset\A_k^n$.
Let $\phi:\spec B\to\spec A$ be a morphism induced by a $k$-algebra homomorphism $f:A\to B$. We say that $\phi$ is finite if $B$ is integral over $f(A)$.
If $\phi:\spec B\to\spec A$ is finite, then
  1. $\phi$ is a closed map
  2. for any $\mfp\in\spec A$, $\inv\phi(\mfp)$ is finite.
  3. $\phi$ is injective iff it was induced by a surjective map of rings
I'll only prove 2. Assume $\phi$ is finite and induced by $f:A\to B$. Fix any $\mfp\in\spec A$, and let $B_\mfp=f(A\sm\mfp)^{-1}B$. Then, $B_\mfp$ is an integral extension of $A_\mfp=(A\sm\mfp)^{-1}A$, and so a finitely generated $A_\mfp$-module. Note that primes of $B_\mfp$ are in bijection with primes of $B$ away (i.e. disjoint) from $f(A\sm\mfp)$, so in particular, any $\mfq\in\inv\phi(\mfp)$ corresponds to a unique prime of $B_\mfp$. Now, let $\mfm=\mfp A_\mfp$ be the unique maximal ideal of $A_\mfp$, and consider the $A_\mfp/\mfm$-algebra $B_\mfp/f(\mfm)$. Note that $B_\mfp/f(\mfm)$ is a finitely generated $A_\mfm/\mfm$-module (i.e. a finite-dimensional vector space), and so has only finitely many prime ideals (we'll see this when we talk about Artinian rings). However, prime ideals of $B_\mfp/f(\mfm)$ are prime ideals of $B_\mfp$ containing $f(\mfm)$, and so are a superset of $\inv\phi(\mfp)$; thus, we win.
Let $X\subset\A_k^n$ be an irreducible subvariety. Then, there's a map $\pi:\A_k^n\onto\A_k^d$ such that the composition $X\to\A_k^n\onto\A_k^d$ is finite and surjective.
We'll prove this assuming $k$ infinite (the general case is similar but with a non-linear transformation). Note that, by induction, it suffices to prove this in the case that $X=\spec k[x_1,\dots,x_n]/(f)$ is an irreducible hypersurface, so that's what we'll do. For an aribtrary nonzero $c=(c_1,c_2,\dots,c_{n-1})\in k^{n-1}$, consider the projection $\A_k^n\to\A_k^{n-1}$ induced by $y_i=c_ix_n+x_i$ for $i=1,\dots,n-1$. Write $f=f_d+f_{d-1}+\dots+f_0$ where $d=\deg f$ and $f_i$ is homogeneous of degree $i$ for all $i$. Then, $x_n$ satisfies the polynomial $$g(T)=f(y_1-c_1T,y_2-c_2T,\dots,y_{n-1}-c_{n-1}T,T)\in k[y_1,\dots,y_{n-1}][T].$$ Furthermore, the leading term of $g(T)$ is $f_d(-c_1T,-c_2T,\dots,-c_{n-1}T,T)=(-T)^df_d(c_1,c_2,\dots,c_{n-1},1)$. Since $f_d\neq0\in k[x_1,\dots,x_n]$ and $k$ is infinite, we can fix a choice of $r=(r_1,\dots,r_n)\in k^n$ s.t. $f_d(r)\neq0$. Possibly after reordering, we may assume $r_n\neq0$, so $f_d(r/r_n)=f_d(r)/r_n^d\neq0$. Thus, choosing $c=(r_1/r_n,\dots,r_{n-1}/r_n)\in k^{n-1}$ causes the composite $\pi_c:X\to\A_k^n\to\A_k^{n-1}$ induced by the map $$\mapdesc\phi{k[y_1,\dots,y_{n-1}]}{k[x_1,\dots,x_n]/(f)}{y_i}{c_ix_n+x_i+(f)}$$ to be finite; this choice of $c$ makes $x_n$ integral over the image, and the other $x_i$ are also integral over the image since $x_i=\phi(y_i)-c_ix_n$ and being integral is closed under ring operations. Hence, we only need to show that $\pi_c$ is surjective (i.e. that $\phi$ is injective). Fix some $p(y_1,\dots,y_{n-1})\in\ker\phi$ so $$p(c_1x_n+x_1,\dots,c_{n-1}x_n+x_{n-1})=f(x_1,\dots,x_n)q(x_1,\dots,x_n).$$ Now, divide $p(y_1,\dots,y_{n-1})$ by $f(y_1-c_1y_n,\dots,y_{n-1}-c_{n-1}y_n,y_n)$ in $k(y_1,\dots,y_{n-1})[y_n]$ to get $$p(y_1,\dots,y_{n-1})=\st q(y_1,\dots,y_n)f(y_1-c_1y_n,\dots,y_{n-1}-c_{n-1}y_n,y_n)+r(y_1,\dots,y_n).$$ Suppose that $p\neq0$; we can apply $\phi$ to $y_i$ ($i < n$) and send $y_n\mapsto x_n$, resulting in $$p(c_1x_n+x+1,\dots,c_{n-1}x_n+x_{n-1})=r(c_1x_n+x_1,\dots,c_{n-1}x_n+x_{n-1},x_n),$$ so the total degree of $p$ is the total degree of $r$ which is less than the total degree of $f$. This contradicts $p(c_1x_n+x_1,\dots,c_{n-1}x_n+x_{n-1})=f(x_1,\dots,x_n)q(x_1,\dots,x_n)$, so we must have $p=0$. Hence, $\phi$ is injective, and $\pi_c$ is surjective.

This is a nice lemma. For example, we can use it to give another definition of dimension.

Let $R=k[x_1,\dots,x_n]/\mfp$ where $\mfp$ is prime, and let $\spec R\onto\A_k^d$ be the map from Noether normalization. Then, $d=\dim R$.
We have that $R$ is integral over $k[x_1,\dots,x_d]$. Hence, any chain of primes in $R$ restricts to a chain of primes in $k[x_1,\dots,x_d]$, and any chain of primes in $k[x_1,\dots,x_d]$ lifts to a chain of primes in $R$ (Going up), so $\dim R=\dim k[x_1,\dots,x_d]=d$.
Let $R=k[x_1,\dots,x_n]/\mfp$ where $\mfp$ is prime. Then, $\trdeg_k\Frac R=\dim R$.
Noether normalization gives the existence of $d=\dim R$ algebraically independent elements $y_1,\dots,y_d\in R$ such that $R$ is integral of $S=k[y_1,\dots,y_d]$. Let $F=\Frac(R)$ and $K=\Frac(S)$. Since $R$ is integeral over $S$, $F/K$ is algebraic so $\trdeg_kF=\trdeg_kS=d$.

We also (finally) get a nice proof of the Nullstellensatz.

Let $k$ be a field, $A$ a finitely generated $k$-algebra, and $\mfm$ a maximal ideal of $A$. Then, $A/\mfm$ is a finite degree field extension of $k$.
Note that $\dim(A/\mfm)=0$ since it's a field, so Noether normalization gives the existence of a surjective, finite map $\spec(A/\mfm)\to\spec k$. This means that $A/\mfm$ is integral over $k$ and hence an algebraic extension. Since $A$ was finitely generated as a $k$-algebra, we conclude that $A/\mfm$ is a finitely generated algebraic extension of $k$ (i.e. a finite extension of $k$).
Let $k$ be a field. For $S\subset k^n$, let $$I(S)=\bracks{f\in k[x_1,\dots,x_n]:\forall x\in S,f(x)=0}.$$ Conversely, for $J\subset k[x_1,\dots,x_n]$, let $$V(J)=\bracks{x\in k^n:\forall f\in J,f(x)=0}.$$
Let $k$ be an algebraically closed field. Then, maximal ideals of $k[x_1,\dots,x_n]$ are in bijection with points of $k^n$: $(a_1,\dots,a_n)\in k^n\mapsto\mfm_a=(x_1-a_1,\dots,x_n-a_n)$.
Let $\mfm$ be a maximal ideal of $k[x_1,\dots,x_n]$. Then, $k[x_1,\dots,x_n]/\mfm$ is a finite degree extension of the algebrically closed field $k$, and so must itself be $k$. Pick a $k$-isomorphism $\phi:k[x_1,\dots,x_n]/\mfm\to k$, and let $a_i=\phi(x_i)$. Then, $\mfm\supseteq\mfm_a$ where $a=(a_1,\dots,a_n)$, so $\mfm=\mfm_a$ since $\mfm_a$ is maximal and $\mfm$ is proper.

Artinian Rings

I think at this point, we’ve had a nice little introduction to algebraic geometry. I mentioned at the beginning that this post was motivated primarily by my desire to give a nice treatment of Dedekind domains. For that, I’ll need to make use of some facts about Artinian rings, so this is what we’ll end on. Artinian rings are dual to Noetherian ones, but as we’ll see, they are much more constrained.

A ring $A$ is called Artinian if it satisfies the descending chain condition on ideals; that is, any chain $$A\supseteq I_0\supseteq I_1\supseteq\dots$$ of ideals stabilizes.
If $k$ is a field, then $k[x]/(x^n)$ is Artinian.

We want to understand the structure of Artininan rings. We first observe that they only have finitely many maximal ideals.

Let $A$ be an Artinian ring. Then, $A$ has finitely many maximal ideals.
Suppose otherwise and let $\mfm_1,\mfm_2,\dots$ be an infinite list of distinct maximal ideals. Then, $$\mfm_1\supseteq\mfm_1\cap\mfm_2\supseteq\mfm_1\cap\mfm_2\cap\mfm_3\supseteq\cdots$$ is a descending chain of ideals, so $\mfm_1\cap\cdots\cap\mfm_n=\mfm_1\cap\cdots\cap\mfm_n\cap\mfm_{n+1}$ for some $n$. Then, Chinese remainder theorem gives $$\prod_{i=1}^n\frac A{\mfm_i}\simeq\prod_{i=1}^{n+1}\frac A{\mfm_i},$$ which is absurd.

This proposition is actually stronger than it may seem at first, since in fact all prime ideals of $A$ are maximal.

Let $A$ be an Artinian ring. Then, $\dim A=0$.
Let $\mfp\subset A$ be prime. Then, $A/\mfp$ is an Artinian domain. Pick any nonzero $x\in A/\mfp$ and consider the chain $(x)\supseteq(x^2)\supseteq(x^3)\supseteq\dots$. This stabilizers, so $x^n=ux^{n+1}$ for some $n$ and some unit $u$. Thus, $$0=x^n(ux-1).$$ Since $A/\mfp$ is a domain and $x\neq0$, we must have $ux=1$, so $x$ is a unit.
$A$ Artinian $\implies\spec A$ finite and discrete.

We’ll next show one of the stranger properties of Artinian rings: they’re isomorphic to the product of their localizations. To do this, we’ll need to make use of Nakayama’s lemma which I’ll state without proof 6. We only need the first version, but the second is also nice.

Let $R$ be a commutative ring, and let $\mf M=\cap\mfm$ where $\mfm$ ranges over maximal ideals of $R$. Let $M$ be a finitely generated $R$-module. Then, $\mf M\cdot M=M\implies M=0$.
Let $R$ be a local ring with maximal ideal $\mfm$, and let $M$ be a finitely generated $R$-module. Them, $M/\mfm M$ is a finite dimensional $R/\mfm$-vector space. A subset $X\subset M$ generates $M$ iff its image $\bar X\subset M/\mfm M$ generates $M/\mfm M$.
Let $A$ be a local Artinian ring. Then, its maximal ideal is nilpotent.
Let $\mfm\subset A$ be maximal. The chain $\mfm\supseteq\mfm^2\supseteq\cdots$ tells us that $\mfm^n=\mfm^{n+1}$ for some $n$. Since $\mfm$ is the only maximal ideal, Nakayama says that $\mfm^n=0$.
Let $A$ be Artinian. Then, $$A\simeq\prod_{\mfm\in\spec A}A_\mfm.$$
Since $\spec A$ is finite, we can pick some $N$ large enough that $(\mfm A_\mfm)^N=\mfm^NA_\mfm=0$ for all $\mfm\in\spec A$. Hence, $(\mfm^NA_\mfm)\cap A=\mfm^N=0$ for all $\mfm\in\spec A$. Fix a prime $\mfm\in\spec A$, and consider the natural map $f:A\to A_\mfm$. It's clear that $\ker f\supseteq\mfm^N$, but we claim that this is an equality. Pick any $a\in\ker f$. Then, $a/1=0/1$ so there's some $s\not\in\mfm$ s.t. $sa=0\in\mfm^N$. As $s\not\in\mfm$, an easy induction argument shows that $a\in\mfm^N$, so $\ker f=\mfm^N$. We also claim that $f$ is surjective. Pick some $s\not\in\mfm$. The image of $s$ in $A/\mfm^N$ is invertible (one stupid way to see this is that $(s)+\mfm^N=A$ by checking locally), so there's some $a\in A$ s.t. $sa-1\in\mfm^N$. Hence, $$sa-1\in\mfm^NA_\mfm\implies sa-1=0\in A_\mfm\implies\frac a1=\frac1s,$$ so $f$ is surjective (for elements with non-unit numerator, use that $f$ is multiplicative). Thus, $A/\mfm^N\simeq A_\mfm$. To finish, let $J=\prod_{\mfm\in\spec A}\mfm$ be the nilradical of $A$. Then, $J^N=0$ since locally, $J^N_\mfm=\mfm^NA_\mfm=0$, and so Chinese remainder theorem gives $$A\simeq A/J^N\simeq\prod_{\mfm\in\spec A}\frac A{\mfm^N}\simeq\prod_{\mfm\in\spec A}A_\mfm$$ as desired.
Let $A$ be Artinian. Then, $A$ is Noetherian.
By the proposition (and the easy fact that finite products of noetherian rings are noetherian), it suffices to prove this when $A$ is local. Let $\mfm\subset A$ be its unique maximal ideal, and write $\mfm^N=0$. Then, $$0=\mfm^N\subseteq\mfm^{N-1}\subseteq\cdots\subseteq\mfm\subseteq A$$ is a finite filtration of $A$ giving rise to the short exact sequences ($0\le n < N$) $$0\too\mfm^{n+1}\too\mfm^n\too\mfm^n/\mfm^{n+1}\too0.$$ Since $0$ is Noetherian and extensions of noetherian rings are noetherian (I don't think I proved this either but it's also easy), it suffices to show that $\mfm^n/\mfm^{n+1}$ is noetherian for all $n$. However, $\mfm^n/\mfm^{n+1}$ is a vector space over $A/\mfm$, and is finite-dimensional because otherwise you could get an infinite descending chain (which you can then pull back to $A$). Since it's a finite dimensional vector space, it's also noetherian (e.g. because it's a free $A/\mfm$-module and so noetherien as an $A/\mfm$-module, but every ideal is an $A/\mfm$-submodule so an ascending chain of ideals is an ascending chain of $A/\mfm$-submodules which then must stabilize). Thus, we win.

In the end, we’ve shown that every Artinian ring is a $0$-dimensional, Noetherian ring. One can in fact show the converese, so a ring is Artinian precisely when it’s 0-dimensional and Noetherian. I’ll leave the reverse direction as an exercise 7.

  1. This means there will be omitted definitions/examples and probably a lot of things left as exercises 

  2. The connection to polynomials comes from the fact that clasically people tend to work with finitely generated $k$-algebras, so your ring $R$ looks like $R=k[x_1,\dots,x_n]/I$. 

  3. Note that this is weird because $r$ doesn’t have a (nice) well-defined domain. The image of every $\mfp\in\spec R$ lies in a different ring. 

  4. Equivalently, on each of the generators of $I$ 

  5. I should mention that I think the term variety is usally reserved for irreducible sets (and general $V(I)$ may instead be called algebraic sets), but oh well. 

  6. Exercise: prove it 

  7. It should be possible to show that the nilradical is nilpotent, and then $A$ is the product of its localizations (each with a nilpotent maximal ideal). From this, it’s not hard to conclude that $A$ is artinian. 

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