# For (most) PIDs: Trace zero matrices are commutators

Posted by Jason Polak on 07. September 2018 · Write a comment · Categories: commutative-algebra · Tags: , ,

Let $R$ be a commutative ring and $M_n(R)$ denote the ring of $n\times n$ matrices with coefficients in $R$. For $X,Y\in M_n(R)$, their commutator $[X,Y]$ is defined by
$$[X,Y] := XY – YX.$$ The trace of any matrix is defined as the sum of its diagonal entries.

If $X$ and $Y$ are any matrices, what is the trace of $[X,Y]$? It's zero! That's because the trace of $XY$ is the same as the trace of $YX$. Therefore:

Any commutator has trace zero.

What about the converse? Is any trace zero matrix also a commutator? In other words, given a trace zero matrix $Z\in M_n(R)$, can we find matrices $X$ and $Y$ such that $Z = [X,Y]$? Albert and Muckenhoupt proved that you can, assuming that $R$ is a field.

What happens if you also want $X$ and $Y$ to have trace zero?

Good question. In general, this is not possible. For example, let's consider the simplest field of all, the field with two elements denoted by $\F_2$. Okay, it's actually debatable whether $\F_2$ really is the simplest field, because so many problems happen in characteristic two. For example, this problem we've been considering: in $\F_2$, if $X$ and $Y$ are $2\times 2$ matrices of trace zero, then $[X,Y]$ will have zero off-diagonal entries. So for example, the matrix
$$\begin{pmatrix}1 & 1\\1 & 1\end{pmatrix}\in M_2(\F_2)$$ cannot be written as a commutator of two matrices with trace zero.

It seems that characteristic two is the only obstruction, in the case of $2\times 2$ matrices. In fact, Alexander Stasinski proved in his paper [1] the following:

Theorem. Let $R$ be a principal ideal domain. If $n\geq 3$ then any matrix in $M_n(R)$ of trace zero can be written as the commutator of two matrices in $M_n(R)$, each having trace zero. The same holds for $n=2$ if two is invertible in $R$.

Notice how the characteristic two problem only happens in the $2\times 2$ case.

[1] Stasinski, A. Isr. J. Math. (2018). https://doi.org/10.1007/s11856-018-1762-5

# Ideal in a union of ideals

Posted by Jason Polak on 24. June 2018 · Write a comment · Categories: commutative-algebra · Tags:

Suppose $I$ is an ideal in a ring $R$ and $J,K$ are ideals such that $I\subseteq J\cup K$. Then either $I\subseteq J$ or $I\subseteq K$. Indeed, suppose that there is some $x\in I$ such that $x\not\in J$. If $y\in I$ is arbitrary and $y\not\in K$ then $x + y$ is in neither $J$ nor $K$. Thus, $y\in K$ and so $I\subseteq K$.

In other words, if there is some element of $I$ that is not in $J$, then $I$ is contained entirely in $K$.

A generalisation for commutative rings is as follows: if $J_1,J_2,\dots,J_n\subseteq R$ are ideals such that at most two of them are not prime ideals, and $I$ is an ideal such that $I\subseteq \cup_i J_i$ then $I\subseteq J_k$ for some $k$. Of course, one does not need the hypothesis that at most two of the $J_1,\dots,J_n$ are not prime if $I$ is principal.

If one drops the hypothesis that at most two of the ideals $J_1,\dots,J_n$ are not prime, then the conclusion no longer holds in general, though.

For example, consider the ring $R = \Z/2[x,y]/(x^2,y^2)$. It is a ring with sixteen elements. In $R$, the ideal $(x,y)$ has eight elements. Furthermore,
$$(x,y)\subseteq (x)\cup (y)\cup (x+y).$$
However, each of the ideals $(x), (y),$ and $(x+y)$—none of which are prime—only has four elements, and so $(x,y)$ is not contained in any of them.

# Degrees of some permutation polynomials

Let $\F_q$ be a finite field. For any function $f:\F_q\to \F_q$, there exists a polynomial $p\in \F_q[x]$ such that $f(a) = p(a)$ for all $a\in \F_q$. In other words, every function from a finite field to itself can be represented by a polynomial.

In particular, every permutation of $\F_q$ can be represented by a polynomial. This isn't true for other finite rings, a fact which is the topic of one of my papers. At any rate, polynomials that represent permutations are called permutation polynomials.

It's not easy to determine which polynomials are permutation polynomials. The exception is for the finite field $\F_2$, and more generally $\Z/2^w$. Then this case Ronald Rivest gave a straightforward criterion for whether a polynomial is a permutation polynomial.

However, there is a classical result that says any permutation may be represented uniquely by a polynomial whose degree is at most $q-2$. Here is an example noted by Charles Wells. Let $a,b\in\F_q$ be two distinct elements. The polynomial
$$f(x) = x + (a-b)(x-a)^{q-1} + (b-a)(x-b)^{q-1}$$
represents the transposition of $a$ and $b$. I suggest that you verify this by direct substitution, using the identity $c^{q-1} =1$ whenever $c\not=0$, which in turn follows from Fermat's little theorem.
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# 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?