## Some Notions of Dimension

It is time to pull together some ideas we’ve built up, and show that they actually correlate how we want them to.

Recall that we have a notion of dimension for a ring called the Krull dimension. Review this if necessary, but essentially you just take the sup of the heights of all prime ideals in your ring.

For a topological space, we first define “irreducible.” Irreducible simply means that you can’t express the space as the union of two nonempty closed sets. To familiarize yourself with this definition some more you can try to prove that it is equivalent to the definition that any two non-empty open sets intersect non-trivially, or any non-empty open set is dense. So note right away that irreducible is a pretty rough condition. Almost none of the spaces I usually talk about are irreducible, since there are tons of non-dense open sets. And by the other criterion, any Hausdorff space is reducible.

Moving on, we now can define the dimension of a topological space to be $sup\{n : Z_0\supsetneq Z_1\supsetneq \cdots \supsetneq Z_n\}$ where the $Z_i$‘s are irreducible subspaces.

Naturally, we are now interested to see if the topological notion of dimension on the topological space $Spec(R)$ is the same as the Krull dimension of the ring R.

First, we’ll need a quick lemma:

A subspace $E\subset Spec(R)$ is irreducible if and only if $I(E)$ is a prime ideal. Note that this is really exactly what we wanted to happen, since prime ideals are points in Spec(R). When we say something is “irreducible,” what we are talking about are the smallest things that cannot be broken apart, i.e. points. I confess I am far oversimplifying this idea of “points” as we will see before the week is over if all goes as planned. (At this point you might want to review spec).

Proof of Lemma: I promise to fill this in later in the week. I just realized that it is an assignment to be turned in on Friday, and not all the readers of this blog that are in my commutative algebra class have this done yet.

Now let’s actually check dimensions. Theorem: $Krulldim(R)=dim(Spec(R))$. Just write it out now:

$dim(Spec(R))=sup\{n: Z_0\supsetneq Z_1\supsetneq \cdots \supsetneq Z_n , \ Z_i \ irreducible\}$
$=sup\{n: p_0\subsetneq p_1\subsetneq \cdots \subsetneq p_n , \ p_i=I(Z_i)\}$
$= Krulldim(R)$ where that switching comes from the Lemma. The correspondence is 1-1.

So for some future posts, I want to clarify some more on dimension and what “points” really are. I also want to do the Nullstellensatz and talk about why it is so important, but I may not. I’ll do some hunting to see what other math bloggers have done on it. I’m pretty sure it has been posted on extensively already, in which I’ll just point people in those directions and add some things that I personally find interesting.

## Krull Dimension

I didn’t actually want to take that long of a break before this post, but I had to do a final exam and give/grade a final, so that ate up lots of time. The next natural thing to move on to is something called Krull dimension. This is sort of annoying to define, but highly useful. I’ve also decided I’m going to stop “fraking” my $\frak{p}$‘s, since it is annoying to type and just use capital P’s for prime ideals.

First we need to define something I’ll call “height.” A prime chain is a strictly decreasing chain of prime ideals: $P_0\supsetneq P_1 \supsetneq \cdots \supsetneq P_n$. Now we define the height of a prime ideal P, ht(P), to be the length of the longest prime chain with $P=P_0$.

Some quick examples: It is easy to check that ht(P)=0 if and only if P is minimal, and hence in an integral domain ht(P)=0 if and only if $P=\{0\}$. Let $R=k[x_1, x_2, \ldots]$ where k is a field. Then let $P_i=(x_i, x_{i+1}, \ldots)$ be the prime ideal generated by those indeterminants (check that it is prime easily by noting $R/P_i\cong k[x_1, \ldots , x_{i-1}]$ which is clearly an integral domain). Then for any n, we can make a chain $P_1\supsetneq P_2\supsetneq \cdots \supsetneq P_{n+1}$. Thus $ht(P_1)=\infty$.

Now for the actual definition we want to work with. I’ll denote the Krull dimension simply by “dim” rather than “Krulldim”. Then we define: $dim(R)=\sup\{ht(P) : P\in Spec(R)\}$. So our quick example here is that for integral domains, $dim(R)=0$ if and only if $R$ is a field.

My goal for the day is to characterize all Noetherian rings of dim 0. The claim is that dim(R)=0 if and only if every finitely generated R-module M has a composition series. Since R is Noetherian, there are only finitely many minimal prime ideals. Since dim(R)=0, every prime ideal is minimal and hence there are only finitely many. Let’s call them $P_1, \ldots, P_n$.

Let’s look at the nilradical: $\sqrt{R}=\cap P_i$. Since the radical is nilpotent, there is some m such that $(\sqrt{R})^m=\{0\}$. So we define $N=P_1\cdots P_n\subset P_1\cap\cdots\cap P_n=\sqrt{R}$, so $N^m=\{0\}$.

Let M be a finitely generated R-module. Then we have the chain $M\supset P_1 M\supset P_1P_2M\supset\cdots\supset NM$. Now note that as a module $\frac{P_1\cdots P_{i-1}M}{P_1\cdots P_i M}$ is an $R/P_i$-module. But $P_i$ is maximal and so $R/P_i$ a field, so it is a vector space. But M is finitely generated, so finite-dimensional, thus we can refine the chain so that all factors are simple.

Now we do this same trick on each of the chains $j=1, \ldots, m$: $N^jM\supset P_1N^jM\supset\cdots \supset N^{j+1}M$. Since at the m stage we get $N^m=\{0\}$, we have a composition series for M.

For the converse suppose every finitely generated R-module has a composition series. Dimension zero is equivalent to showing that R has no prime ideals P, and Q such that $P\supsetneq Q$. Suppose such exist. Let’s pass to the quotient, $R/Q$, and reinterpret our hypothesis. Then R is an integral domain that has a nonzero prime ideal and a composition series $R\supset I_1\supset \cdots \supset I_d\neq \{0\}$. So $I_d$ is minimal. Let $x\in I_d$ be any nonzero element. Then since $x I_d\subset I_d$ and $xI_d\neq \{0\}$ (we’re in a domain), then by minimality we have $xI_d=I_d$. So there is a $y\in I_d$ such that $xy=x$, i.e. $y=1\in I_d$. And hence $I_d=R$. Thus R is a field which contradicts our having a nonzero prime ideal.

Well, I think that is enough fun for one day. I may post again tomorrow, since my final is Wed.

## Lying Over and Going Up Part II

I realized there was one more result I probably should have included last time. Oh well. Here goes:

Let $R^*/R$ be integral, $\frak{p}$ a prime ideal in R and $\frak{p}^*, \frak{q}^*$ prime ideals in $R^*$ lying over $\frak{p}$. If $\frak{p}^*\subset \frak{q}^*$, then $\frak{p}^*=\frak{q}^*$.

Proof: Recall that $S^{-1}R^*$ is integral over $S^{-1}R$ by last time for any multiplicative set, and also that prime ideals are preserved in rings of fractions. Thus the hypotheses still hold if we localize at $\frak{p}$. Thus $R_\frak{p}^*$ is integral over $R_\frak{p}$, and $\frak{p}^*R_\frak{p}^*\subset \frak{q}^*R_\frak{p}^*$ are prime ideals. Thus we can WLOG replace $R^*$ and $R$ by their localizations and hence assume they are local. So now $\frak{p}$ is a maximal ideal in $R$. Thus by last time $\frak{p}^*$ is maximal. Since $\frak{p}^*\subset \frak{q}^*$, we have $\frak{p}^*=\frak{q}^*$.

Now we are ready for the two big theorems. Here is the “Lying Over” Theorem. Let $R^*/R$ be an integral extension. If $\frak{p}$ is a prime ideal in R, then there is a prime ideal $\frak{p}^*$ in $R^*$ lying over $\frak{p}$, i.e. $\frak{p}^*\cap R=\frak{p}$.

Proof: First note that $R \stackrel{i}{\longrightarrow} R^* \stackrel{h^*}{\longrightarrow} S^{-1}R^*$ and $R \stackrel{h}{\longrightarrow} R_\frak{p} \stackrel{j}{\longrightarrow} S^{-1}R^*$ form the two sides of a commutative diagram. By last time $S^{-1}R^*$ is integral over $R_\frak{p}$. Choose a maximal ideal $\frak{m}^*$ in $S^{-1}R^*$. Thus $\frak{m}^*\cap R_\frak{p}$ is maximal in $R_\frak{p}$. But $R_{\frak{p}}$ is local with unique max ideal $\frak{p}R_\frak{p}$, so $\frak{m}^*\cap R_\frak{p}=\frak{p} R_\frak{p}$. But the preimage of a prime ideal is prime, so $\frak{p}^8=(h^*)^{-1}(\frak{m}^*)$ is a prime ideal in $R^*$.

Now we just diagram chase: $(h^*i)^{-1}(\frak{m}^*)=i^{-1}(h^*)^{-1}(\frak{m}^*)=i^{-1}(\frak{p}^*)=\frak{p}^*\cap R$. And also: $(jh)^{-1}(\frak{m}^*)=h^{-1}j^{-1}(\frak{m}^*)=h^{-1}(\frak{m}^*\cap R_\frak{p})=h^{-1}(\frak{p}R_\frak{p})=\frak{p}$.

Thus $\frak{p}^*$ lies over $\frak{p}$.

Our other big theorem is the one about “Going Up”: If $R^*/R$ is an integral extension and $\frak{p}\subset \frak{q}$ are prime in R, and $\frak{p}^*$ lies over $\frak{p}$, then there is a prime ideal $\frak{q}^*$ lying over $\frak{q}$ with $\frak{p}^*\subset \frak{q}^*$.

Proof: By last time $(R^*/\frak{p}^*)/(R/\frak{p})$ is an integral extension where $R/\frak{p}$ is embedded in $R^*/\frak{p}^*$ as $(R+\frak{p}^*)/\frak{p}^*$. Now we just replace $R^*$ and R by these rings so that both $\frak{p}^*$ and $\frak{p}$ are $\{0\}$. Now we just apply the Lying Over Theorem to get our result.

So as we see here integral extensions behave extremely nicely. These theorems guarantee that se always have prime ideals lying over ones in the lower field. This has some important applications to the Krull dimension that we’ll start looking at next time.

## Lying Over and Going Up

If you haven’t heard the terms in the title of this post, then you are probably bracing yourself for this to be some weird post on innuendos or something. Let’s first do some motivation (something I’m not often good at…remember that Jacobson radical series of posts? What is that even used for? Maybe at a later date we’ll return to such questions). We can do ring extensions just as we do field extensions, but they tend to be messier for obvious reasons. So we want some sort of property that will force an extension to be with respect to prime ideals. Two such properties are “lying over” and “going up.”

Let $R^*/R$ be a ring extension. Then we say it satisfies “lying over” if for every prime ideal $\mathfrak{p}\subset R$ in the base, there is a prime ideal $\mathfrak{p}^*\subset R^*$ in the extension such that $\mathfrak{p}^*\cap R=\mathfrak{p}$. We say that $R^*/R$ satisfies “going up” if in the base ring $\mathfrak{p}\subset\mathfrak{q}$ are prime ideals, and $\mathfrak{p}^*$ lies over $\mathfrak{p}$, then there is a prime ideal $\mathfrak{q}^*\supset \mathfrak{p}^*$ which lies over $\mathfrak{q}$. (Remember that Spec is a contravariant functor).

Note that if we are lucky a whole bunch of posts of mine will finally be tied together and this was completely unplanned (spec, primality, localization, even *gasp* the Jacobson radical). First, let’s lay down a Lemma we will need:

Let $R^*$ be an integral extension of R. Then
i) If $\mathfrak{p}$ a prime ideal of R and $\mathfrak{p}^*$ lies over $\mathfrak{p}$, then $R^*/\frak{p}^*$ is integral over $R/\mathfrak{p}$.
ii) If $S\subset R$, then $S^{-1}R^*$ is integral over $S^{-1}R$.

Proof: By the second iso theorem $R/\frak{p}=R/(\frak{p}^*\cap R)\cong (R+\frak{p}^*)/\frak{p}^*\subset R^*/\frak{p}^*$, so we can consider $R/\frak{p}$ as a subring of $R^*/\frak{p}^*$. Take any element $a+\frak{p}^*\in R^*/\frak{p}^*$. By integrality there is an equation $a^n+r_{n-1}a^{n-1}+\cdots + r_0=0$ with the $r_i\in R$. Now just take everything $\mod \frak{p}^*$ to get that $a+\frak{p}^*$ integral over $R/\frak{p}$. This yields part (i).

For part (ii), let $a^*\in S^{-1}R^*$, then $a^*=a/b$, where $a\in R^*$ and $b\in\overline{S}$. By integrality again we have that $a^n+r_{n-1}a^{n-1}+\cdots + r_0=0$, so we multiply through by $1/b^n$ in the ring of quotients to get $(a/b)^n+(r_{n-1}/b)(a/b)^{n-1}+\cdots +r_0/b^n=0$. Thus $a/b$ is integral over $S^{-1}R$.

I’ll do two quick results from here that will hopefully put us in a place to tackle the two big results of Cohen and Seidenberg next time.

First: If $R^*/R$ is an integral ring extension, then $R^*$ is a field if and only if $R$ is a field. If you want to prove this, there are no new techniques from what was done above, but you won’t explicitly use the above result, so I won’t go through it.

Second: If $R^*/R$ is an integral ring extension, ten if $\frak{p}$ is a prime ideal in R and $\frak{p}^*$ is a prime ideal lying over $\frak{p}$, then $\frak{p}$ is maximal if and only if $\frak{p}^*$ is maximal.

Proof: By part (i) of above, $R^*/\frak{p}^*$ is integral over $R/\frak{p}$ and so as a corollary to “First” we have one is a field if and only if the other is. This is precisely the statement that $\frak{p}$ is maximal iff $\frak{p}^*$ is maximal.

## Spec? You mean like glasses?

So I’ve built up localization starting there, and I’ve built up the theory of prime ideals scattered throughout, but ending here. I also just assume the basics of topology in my posts, so we are in the perfect position to talk about a very fascinating construction and incredibly useful tool that combines all these things.

Warning: I have just started learning about this stuff, so it could be riddled with confusion or error. Luckily, I’m just posting the basics which some readers probably know like the back of their hand and will hopefully point out problems.

Of course what I’m referring to is Spec. As usual let’s assume that R is a commutative ring with 1 (I don’t think we need the 1). Then $Spec(R)=\{P : P \ prime \ ideal \ of \ R\}$. So we have the collection of all (proper) prime ideals of the ring. Other than prime ideals being my favorite type of ideal, this seems to be useless right now.

Let’s put a topology on our set now (the “points” of our space are prime ideals). Let $asubset R$ be any ideal. Define $V(a)=\{\mathfrak{p}\in Spec(R) : a \ subset \ \mathfrak{p}\}$. Then we define the closed sets of the topology to be the family of all such sets, i.e. $\{V(a) : a \ subset \ R \ an \ ideal\}$ are the closed sets. This is known as the Zariski topology.

To check that these really satisfy the right axioms, (I won’t go through it, but) note that $V(0)=Spec(R)$, $V(R)=\emptyset$, $V(\sum a_i)=\cap V(a_i)$ and $V(a \cap b)=V(a)\cup V(b)$ (The last is probably the least trivial, but they all follow in a straightforward from definition way).

Examples:

1. If our ring is a field k, then $Spec(k)=\{*\}$ the spectrum is a point.

2.Another common example would be $Spec(\mathbb{Z})=\{(0), (2), (3), (5), ldots \}$. In other words, the prime ideals can just be identified with the prime number that generates them (and we have (0) as a special circumstance). So open sets are subsets of $\mathbb{Z}$ that are missing finitely many prime numbers. So we see that the Zariski topology is not Hausdorff (and rarely is). It will, however, always be compact.

3. Possibly the most important examples are the ones dealing with polynomial rings. In the nicest case, when k is an algebraically closed field, we have that $Spec(k[x])=\{*\}\cup k$ since the prime ideals are just multiples of linear polynomials, we have the bijection of sending any $c \in k$ to the prime ideal generated by $(x-c)$ (and we still have that pesky “zero” floating around that we’ll talk about later).

Last for today is that Spec is a contravariant functor from rings to topological spaces. We’ve basically done everything we need, since we see how it takes a ring object to a Top object. Also if we have a ring hom $f:R \to S$, then define $Spec(f)=f^* : Spec(S)\to Spec(R)$ in the obvious way, i.e. $\mathfrak{p} \mapsto f^{-1}(\mathfrak{p})$.

I promised some localization and we should be able to get to that next time, but there is just so much going on here that it is nearly impossible to exhaust (well, from my perspective as a newbie to the topic).

## Noetherian Rings

I promised this awhile back. It seems as if the Noetherian condition is really the last major thing I need before being able to move on.

A ring is Noetherian if every ascending chain of ideals stabilizes (or “terminates”). So, this means that given any collection of ideals $\{I_n\}\subset R$ such that $I_1\subset I_2\subset I_3 \subset \cdots$ we have that there exists some $N$ so that $I_n=I_{n+1}=\cdots$ for all $n>N$. This condition seems very strange at first. It is known as the Ascending Chain Condition, or ACC for short, but it turns out that it is equivalent to some other things and makes sure our rings are somewhat well-behaved.

Since for the purpose of this collection of posts we only care about commutative rings, the ACC is equivalent to the condition that every ideal is finitely generated.

Proof) Suppose every ideal is finitely generated. Then let $I_n$ be an ascending chain of ideals. Since $I=\cup I_n$ is an ideal, it is generated by say m elements: $I=$. But each one of these elements come from some specific ideal, so suppose $a_1\in I_{n_1}, \ldots, a_m\in I_{n_m}$. Then just take $N=\max(n_1, \ldots, n_m)$ and we have that the chain stabilizes after that.

For the reverse we go by contrapositive. Let $I\subset R$ be some ideal that is not finitely generated. Then we can find $a_1\in I$ such that $\neq I$. We can also find $a_2\in I\setminus $ such that $\neq I$. We can continue this process without termination. If it terminated at some step then, the ideal would be finitely generated. Thus we now just note that we have an ascending chain that doesn’t terminate $\subset \subset \cdots$.

It is easily seen that every PID is Noetherian. Rings tend to stay Noetherian under new constructions. The ring of polynomials (in finitely many indeterminates) and ring of power series where the coefficients come from a Noetherian ring is Noetherian. The former is known as the Hilbert Basis Theorem. Both the quotient $R/I$ and the ring of fractions $S^{-1}R$ are Noetherian if R is Noetherian.

But remember we want to figure out how this works with prime ideals. It turns out that prime isn’t quite what we want to get the best results, but in order to not introduce yet another type of ideal, I’ll leave this out since it won’t appear in anything I do later. So it turns out that if $I\subset R$ an ideal and R Noetherian, every prime ideal $P\supset I$ contains a minimal-over-I prime ideal, say $P_0\supset I$. This is just a standard one step application of Zorn’s Lemma.

So I think I’ve beat primality to death. Next time I’ll do a sort of “history of math” type post on Hilbert’s Zahlbericht to put into the blog carnival. This will give me some time to think of where to go from here. I’m thinking the algebraic number theory side…I just don’t want to have to build Galois theory before I do it.

## More on Primality

I want to wrap up some loose ends on the greatness of prime ideals before moving on in the localization theme. So. Recall that we formed the ring of quotients just like you would form the field of quotients. Only this time your “denominator” can be an arbitrary multiplicative set and this construction only gets us a ring. Moreover, this ring is not necessarily local. If we do the construction on a ring R with and the multiplicative set $R\setminus P$ where P is a prime ideal, then we do get a local ring and we call this the localization.

Definition. Unique Factorization Domain (UFD): An integral domain in which every non-zero non-unit element can be written as a product of primes. (Note that there are equivalent definitions other than this one).

Quick property: Every irreducible element is prime.

Thus, it is instructive to look at some properties of prime ideals. First off, let’s look at the special case of UFD’s. It turns out that if R is a UFD, then for a multiplicative set S, $S^{-1}R$ is also a UFD. This mostly has to do with the fact that $R\hookrightarrow S^{-1}R$ is an embedding and anything in $S^{-1}R$ is associate to something in $R$. This makes a nice little exercise for the reader.

So what’s so special about prime ideals in UFD’s? Well every nonzero prime ideal contains a prime element.

Proof: Suppose $P\neq 0$ and P prime. Then there exists $a\in P$, $a\neq 0$ such that $a=up_1\cdots p_n$ where $u$ a unit and $p_i$ prime. Thus $u\notin P$. But this means that $p_1\cdots p_n\in P$ and since it is prime we have some $p_j\in P$.

Theorem: If R is not a PID, and P an ideal which is maximal with respect to the property of not being principal, then P is prime (and will always exist).

Sketch of existence: Zorn’s Lemma. The proof of this contains lots of nitty gritty element-wise computation and a weird trick, so I don’t see it as beneficial. What is beneficial is that we get this great corollary: A UFD is a PID if and only if every nonzero prime ideal is maximal.

I’ve been kind of stingy on the examples, so I’ll leave you with a pretty common example of a ring of fractions. These are usually called dyadic rational numbers. Take your ring to be $\mathbb{Z}$. Then take your multiplicative set to be $S=\{1, 2, 2^2, 2^3, \ldots\}$. Now $S^{-1}\mathbb{Z}$ are just the rational numbers with denominator a power of 2.

More generally we can form the p-adic integers (although that term is laden with many meanings, so I hesitate to actually use it). Let $R=\times_{i=1}^\infty \mathbb{Z}/p^i$. Where we have the restriction $a\in R$ iff $a=(a_1, a_2, \ldots )$ satisfies $a_i\cong a_{i+1} \mod p^i$. So  elements of the ring are sequences. (Note $\mathbb{Z}$ embeds naturally since $i\mapsto (i, i, i, \ldots)$ satisfies that relation). This is a ring with no zero divisors, so we can take it to be the multiplicative set and we get the field of fractions $\mathbb{Q}_p$. The multiplicative group has a nice breakdown as $\mathbb{Q}_p^{\times}\cong p^{\mathbb{Z}}\times \mathbb{Z}_p^{\times}$.

Next time: Why Noetherian is important. How primality relates to it. And possibly another example.