# Zero divisor graph of a commutative ring

Posted by Jason Polak on 18. January 2018 · Write a comment · Categories: commutative-algebra · Tags: ,

Let $R$ be a commutative ring. The zero divisors of $R$, which we denote $Z(R)$ is the set-theoretic union of prime ideals. This is just because in any commutative ring, the set of subsets of $R$ that can be written as unions of prime ideals is in bijection with the saturated multiplicatively closed sets (the multiplicatively closed sets that contain the divisors of each of their elements).

Istvan Beck in 1986* introduced an undirected graph (in the sense of vertices and edges) associated to the zero divisors in a commutative ring. Recall that an undirected graph is just a set of vertices (points) and edges connecting the point. What is his graph? His idea was to let the vertices correspond to points of $R$, and the edges correspond to the relation than the product of the corresponding elements is zero.

There is a slightly different definition due to Anderson and Livingston, which is the main one used today. Let $Z(R)^*$ denote the nonzero zero divisors. Their graph is $\Gamma(R)$, which is defined as the graph whose vertices are the elements of $Z(R)^*$, and whose edges are defined by connecting two distinct points if and only if their product is zero. Naturally, if $Z(R)^*$ is not empty then the resulting graph $\Gamma(R)$ will have some edges. The actual information contained in $\Gamma(R)$ is pretty much the same as the information contained in Beck’s version and so we’ll just stick with $\Gamma(R)$.

For this idea to be more than just a curiosity, the graph theoretic properties of $\Gamma(R)$ should tell us something about hte ring theoretic properties of $R$. Does it? Anderson and Livingston showed in 1998 that there exists a vertex of $\Gamma(R)$ adjacent to every other vertex if and only if either $R = \Z/2\times A$ where $A$ is an integral domain or $Z(R)$ is an annihilator. They also showed that for $R$ a finite commutative ring, if $\Gamma(R)$ is complete, then $R\cong \Z/2\times \Z/2$ or $R$ is local with characteristic $p$ or $p^2$.
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# Working with group rings in Sage

Posted by Jason Polak on 16. January 2018 · Write a comment · Categories: commutative-algebra, ring-theory · Tags:

Let $\Z[\Z/n]$ denote the integral group ring of the cyclic group $\Z/n$. How would you create $\Z[\Z/n]$ in Sage so that you could easily multiply elements?

First, if you’ve already assigned a group to the variable ‘A’, then

will give you the corresponding group ring and store it in the variable ‘R’. The first argument of ‘GroupAlgebra(-,-)’ is the group and the second is the coefficient ring. Sage uses ‘ZZ’ to denote the integers, ‘QQ’ to denote the rationals, etc.

So how do you specify the cyclic group $A$? The first posibility is to use the construction:

where you’d replace ‘n’ by the actual number that you want there. This is useful if you want to work with other permutation groups, because the elements of ‘A’ are stored as permutations:

The output to this snippet is:

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# What’s in all powers of a principal prime?

Posted by Jason Polak on 05. January 2018 · Write a comment · Categories: commutative-algebra · Tags:

Let $R$ be a commutative ring and $(p)$ be a principal prime ideal. What can be said about the intersection $\cap_{k=1}^\infty (p)^k$? Let’s abbreviate this $\cap (p)^k$ (I like to use the convention that when limits are not specified, then the operation like intersection is taken over all possible indices).

Let’s try an example. For the integers, every principal prime is of the form $(p)$ where $p$ is a prime number or zero. And $(p)^k = (p^k)$ so $\cap (p)^k = (0)$. In fact if $R$ is any Noetherian integral domain then $\cap (p)^k = 0$.

If $R$ is not an integral domain then $\cap (p)^k$ is not necessarily zero. For example, let $S$ be an integral domain and let $R = S\times S$. In $S\times S$, the prime ideal generated by the single element $p = (1,0)$ is its own $k$-th power for all $k$. So $\cap (p)^k = p$.

Of course, it is impossible that in an integral domain to have $(p) = (p)^2$ for some principal prime $p$ unless $p = 0$. Of course, it is possible in an integral domain to have $P = P^2$ for a nonzero prime ideal $P$ that is necessarily not principal. Just take a “polynomial” ring over a field where the powers are allowed to be all nonnegative rationals; that is, a ring of the form $k[\Q^+]$ where $\Q^+$ is the monoid of all nonnegative rational numbers under multiplication. In the case of $k[\Q^+]$, a prime such that $P^2 = P$ would be the prime $P$ generated by all elements of the form $x^q$ where $q \gt 0$ is a rational number.

I will leave the reader with the following question:

Does there exist an integral domain, necessarily non-Noetherian, that contains a principal prime $(p)$ with $\cap (p)^k\not= 0$?

# Poset of prime ideals

Posted by Jason Polak on 04. January 2018 · Write a comment · Categories: commutative-algebra

For a commutative ring, what does the partially ordered set (=poset) of primes look like? I already talked a little about totally ordered sets of primes, but what about in general?

For a general partially ordered set $S$ there are two immediate questions that come to mind:

1. Does there exist a commutative ring whose poset of primes is $S$?
2. Does there exist a commutative ring whose poset of primes contains an embedded copy of $S$?

For example, consider this partially ordered set:

I draw the partially ordered sets so that “higher” is larger. This partially ordered set can be embedded into the poset of prime ideals of the integers

What about the totally ordered set $\Z$ itself? It cannot exist in any poset of primes, because it has no minimal or maximal element, whereas the both the intersection and union of a chain of primes are also primes.

Can the closed interval $[0,1]$ be embedded in a poset of primes? Alas, no. Even though $[0,1]$ now has a lower and upper bound, it is a dense ordered set, and a poset of primes cannot contain a “dense part”. More precisely, suppose that $P\subset Q$ are two distinct prime ideals and let $\{P_i\}$ be a maximal chain of prime ideals between $P$ and $Q$. Let $x\in Q – P$ and let
$$P’ = \cup \{ P_i : x\not\in P_i\}\\ Q’ = \cap \{ P_i : x\in P_i\}$$
Then $P’$ and $Q’$ are two distinct prime ideals such that $P’\subset Q’$ and such that there is no prime between $P’$ and $Q’$. So, $[0,1]$ indeed cannot appear in any poset of prime ideals of a commutative ring.

# Non-Noetherian domain but finitely generated ideals principal

Posted by Jason Polak on 02. January 2018 · Write a comment · Categories: commutative-algebra · Tags: ,

A finitely-generated module over a principal ideal domain is always isomorphic to $R^n\oplus R/a_1\oplus\cdots\oplus R/a_n$ where $n$ is a nonnegative integer and $a_i\in R$ for $i=1,\dots,n$. This is called the structure theorem for modules over a principal ideal domain. Examples of principal ideal domains include fields, $\Z$, $\Z[\sqrt{2}]$, and the polynomial ring $k[x]$ when $k$ is a field.

If $a\in R$ is not a unit, then $R/a$ is not projective, since $a$ annihilates any element of $R/a$ and therefore $R/a$ cannot be the direct summand of any free module. Therefore, we can conclude from the structure theorem that any finitely-generated projective module over a principal ideal domain is a free module. Don’t get your hopes up though: there are many examples of non-free projective modules.

But let’s stick with principal ideal domains. It is actually true that every projective module over a principal ideal domain is free. Kaplansky in [1] proved the following even stronger theorem:

Theorem. If $R$ is an integral domain in which every finitely generated ideal is principal, then every projective $R$-module is free.

# When the set of prime ideals is linearly ordered

Posted by Jason Polak on 30. December 2017 · Write a comment · Categories: commutative-algebra · Tags: ,

Imposing structure on the poset of prime ideals of a ring $R$ is one way to gain a hold onto its structure. The poset of prime ideals of $R$ is simply a fancy term for the set of prime ideals of $R$, partially ordered by inclusion. Usually this set is not totally ordered: in the ring of integers $\Z$ for instance, the prime ideals $(2)$ and $(3)$ cannot be compared by inclusion. It seems to me that requiring the poset of primes to be totally ordered is a strict condition indeed.

Here is one type of domain in which the prime ideals are totally ordered: the valuation domain.
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# Replacing two idempotents with one

Posted by Jason Polak on 27. December 2017 · 2 comments · Categories: commutative-algebra · Tags:

Let $R$ be a commutative ring. Two idempotents $e$ and $f$ are called orthogonal if $ef = 0$. The archetypal example is $(0,1)$ and $(1,0)$ in a product ring $R\times S$.

Let $e$ and $f$ be orthogonal idempotents. Then the ideal $(e,f)$ is equal to the ideal $(e + f)$. To see, this first note that $(e + f)\subseteq (e,f)$. On the other hand:
$$(1-e)(e + f) = e + f – e – ef = f$$
Therefore $f \in (e + f)$. Switching $e$ and $f$ in this calculation shows that $e\in (e + f)$. Using the fact that $e + f$ is also an idempotent, we see that by induction, if $e_1,\dots,e_n$ are pairwise orthogonal idempotents, then the ideal $(e_1,\dots,e_n)$ is generated by the single element $e_1 + \dots e_n$.

Now suppose $e$ and $f$ are idempotents that are not necessarily orthogonal. Then $(e,f)$ is still a principal ideal. To see this, consider the element $e – ef$. The calculation
$$(e – ef)^2 = e – 2ef + ef = e – ef$$
shows that $e – ef$ is an idempotent. Furthermore, $(e,f) = (e – ef,f)$ and $e-ef$ and $f$ are orthogonal idempotents. By what we discussed in the previous paragraph, $(e,f) = (e-ef,f)$ is generated by $e – ef + f$.

Everything we did assumed $R$ was commutative. But what if we foray into the land of noncommutative rings? Is it still true that a left-ideal generated by finitely many idempotents is also generated by a single idempotent? Any ideas?

# Number of irreducible polynomials over a finite field

Posted by Jason Polak on 20. December 2017 · Write a comment · Categories: commutative-algebra · Tags: ,

Over a finite field, there are of course only finitely many irreducible monic polynomials. But how do you count them? Let $q = p^n$ be a power of a prime and let $N_q(d)$ denote the number of monic irreducible polynomials of degree $d$ over $\F_q$. The key to finding $N_q(d)$ is the following fact: the product of all the monic, irreducible polynomials of degree $d$ with $d \mid n$ in the finite field $\mathbb{F}_q$ is the polynomial
$$x^{q^n} – x.$$
So let’s say $f_1,f_2,\dots, f_k$ are all the irreducible monic polynomials of degree $d$ with $d\mid n$. By taking degrees on both sides of the equation $x^{q^n} -x = f_1f_2\cdots f_k$, we get the formula
$$q^n = \sum_{d\mid n} dN_q(d).$$
Hey this is pretty good! For example, if $q = 2$ and $n = 3$ then the formula reads
$$8 = N_2(1) + 3N_2(3)$$
Now, $N_q(1)$ is always easy to figure out. All monic linear polynomials are irreducible so $N_q(1) = q$. Therefore, $N_2(3) = 2$. In fact, these two polynomials are: $x^3 + x + 1$ and $x^3 + x^2 + 1$. Okay, what about if $q = 3$ and $n = 6$? Then our formula tells us that
$$3^6 = N_3(1) + 2N_3(2) + 3N_3(3) + 6N_3(6).$$
So we now have to recursively compute: if we did that we would get that $N_3(1) = 1$, and this gives $2N_3(2) = 6$. Finally, $3N_3(3) = 24$. Therefore, we would calculate that $N_3(6) = 116$. It would not be too hard to write such a recursive algorithm and I encourage the reader to try it.
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# First-order characterisations of free and flat…projective?

Posted by Jason Polak on 28. September 2017 · 3 comments · Categories: commutative-algebra, homological-algebra, modules

Here is an interesting question involving free, projective, and flat modules that I will leave to the readers of this blog for now.

First, consider free modules. If $R$ is a ring, then every $R$-module is free if and only if $R$ is a division ring. The property of $R$ being a division ring can be expressed in terms of first-order logic in the language of rings: $\forall x[x\not=0 \rightarrow \exists y(xy = 1)]$.

The meat of this first-order statement is the equation $xy = 1$. Now, multiply by $x$ on the right to get the equation $xyx = x$. Now we can put this in a first-order sentence: $\forall x\exists y[xyx = x]$. Notice how we removed the condition $x\not=0$ from this one. That’s because $x=0$ satisfies $xyx = x$ for any $y$ in all rings. Rings that model $\forall x\exists y[xyx = x]$ are called von Neumann regular. More importantly, these are exactly the rings for which every $R$-module is flat.

By weakening the statement that $R$ is a division ring, we got a statement equivalent to the statement that every $R$-module is flat. One might wonder: where did the projective modules go? Is there a first-order sentence (or set of sentences perhaps) in the language of rings whose models are exactly those rings $R$ for which every $R$-module is projective? Diagrammatically:

Can we replace the question mark with a first-order sentence, or a set of them?

My initial thoughts are no because of ultraproducts, but I have not yet come up with a rigorous argument.

# Extensions of Finite Rings are Integral

Posted by Jason Polak on 19. July 2017 · Write a comment · Categories: commutative-algebra · Tags: ,

Here’s a classic definition: let $R\subseteq S$ be commutative rings. An element $s\in S$ is called integral over $R$ if $f(s)=0$ for some monic polynomial $f\in R[x]$. It’s classic because appending the solutions of polynomials to base rings goes way back to the ancient pasttime of finding solutions to polynomial equations.

For example, consider $\Z\subseteq \Z[\sqrt{2}]$. Every element of $\Z[\sqrt{2}]$ is integral over $\Z$, which essentially comes down to the fact that $\sqrt{2}$ satisfies $x^2 – 2$. On the other hand, the only elements of $\Q$ integral over $\Z$ are the integers themselves.

The situation is much different for finite commutative rings. If $R\subseteq S$ are finite rings, then every element of $S$ is integral over $R$. Proof: suppose $s\in S$ and set $T = \{ f(s): f\in R[x]\}$. For each $t\in T$ fix a polynomial $f$ such that $f(s) = t$. The set of all such polynomials is finite so we can define $m$ as the maximum degree of all these polynomials. Then $s^{m+1}\in T$ and so there is an $f$ of degree at most $m$ such that $s^{m+1} – f(s) = 0$. Thus $s$ satisfies the monic polynomial $x^{m+1} – f(x)$. QED.

Cool right? However, this is just a more general case of the following theorem: let $R\subseteq S$ be commutative rings. Then $S$ is finitely generated as an $R$-module if and only if $S$ is finitely generated as an $R$-algebra and every element of $S$ is integral over $R$.