Image factorisation in abelian categories

Let $R$ be a ring and $f:B\to C$ be a morphism of $R$-modules. The image of $f$ is of course
$${\rm im}(f) = \{ f(x) : x\in B \}.$$The image of $f$ is a submodule of $C$. It is pretty much self-evident that $f$ factors as
$$B\xrightarrow{e} {\rm im}(f)\xrightarrow{m} C$$where $e$ is a surjective homomorphism and $m$ is an injective homomorphism. In fact, there is nothing special about working in the category of $R$-modules at all. The same thing holds in the category of sets and a proof for the category of sets works perfectly well for the category of $R$-modules. This set-theoretic reasoning is very natural.

However, we can't always work with categories whose objects are sets with additional structure and whose morphisms are set functions that respect the additional structure (concrete categories). Sometimes we have to work with abelian categories. What's an abelian category? Briefly, it is a category $\Acl$ such that:
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Non-unique Factorisation: Part 2

Posted by Jason Polak on 15. April 2017 · Write a comment · Categories: commutative-algebra · Tags:

We are continuing the series on non-unique factorisation. For a handy table of contents, visit the Post Series directory.

In Part 1 of this series, we introduced for a commutative ring three types of relations:

1. Associaties: $a\sim b$ means that $(a) = (b)$
2. Strong associates: $a\approx b$ means that $a = ub$ for $u\in U(R)$
3. Very strong associates: $a\cong B$ means that $a\sim b$ and either $a = b=0$ or $a = rb$ implies that $r\in U(R)$

Here, $U(R)$ denotes the group of units of $R$. We have already seen in a ring with nontrivial idempotents like $\Z\times \Z$, a nontrivial idempotent $e$ will be satisfy $e\sim e$ and $e\approx e$, but $e\not\cong e$ because $e = ee$ and yet $e$ is not a unit and nonzero.

Therefore, $\cong$ is not an equivalence relation for all commutative rings. But it is symmetric:

Proof. Suppose $a\cong b$. Then $a\sim b$ and so $b\sim a$. If $a$ and $b$ are not both zero, write $a = sb, b = ta$. If $b = ra$ then $a = sra = s^2rb$. Since $a\cong b$, this implies that $s^2r$ is a unit and so $r$ is a unit. Hence $b\cong a$.

Guess what? The relation $\cong$ is also transitive. Since the proof is similarly short I'll leave the proof to the reader. So, $\cong$ is just missing being reflexive for all rings to be an equivalence relation for all rings. If $\cong$ is an equivalence relation for a ring $R$, then we say that $R$ is presimplifiable. We introduced this type of ring last time.
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Non-unique Factorisation: Part 1

Posted by Jason Polak on 11. April 2017 · Write a comment · Categories: commutative-algebra · Tags:

If $F$ is a field then the polynomial ring $F[x]$ is a unique factorisation domain: that is, every nonunit can be written uniquely as a product of irreducible elements up to a unit multiple. So in $\Q[x]$ for example, you can be sure that the polynomial $x^2 – 2 = (x-2)(x+2)$ can't be factored any other way, and thus the only zeros of $x^2 – 2$ really are $\pm 2$.

If $F$ is not a field, then a polynomial might have a bunch of different factorisations. For example, in the ring $\Z/4[x]$ we can write $x^2 = (x)(x) = (x+2)(x+2)$. How can we make sense of factorisations in rings that are not unique factorisation domains? In order to do so, we first should make sure we understand what irreducible means in this context.

In the next several posts we'll look at non-unique factorisation more closely, following a paper of Anderson and Valdes-Leon [1], but keeping the posts self-contained. We'll start by looking at the concept of associates. One can in fact look at several different variations of associates. Here are three:
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Generators of Principal Ideals and Unit Association

Posted by Jason Polak on 16. August 2012 · 4 comments · Categories: commutative-algebra · Tags: , ,

Let us ponder a familiar property of any integral domain $R$. Suppose that $a,b\in R$ are such that $(a) = (b)$, where $(a)$ denotes the ideal generated by $a$. Then $a = rb$ and $b = sa$ for some elements $r,s\in R$. Combining these two facts together give $a = rsa$, or $(1 – rs)a = 0$. If $a\not = 0$, this means that $1 – rs = 0$ or that $r$ and $s$ are units. Thus $a$ is a unit multiple of $b$. Of course if $a = 0$ then $b= 0$ and $a$ is still a unit multiple of $b$.

Now, the question is, what happens if we drop the assumption that $R$ is an integral domain? If $(a) = (b)$, can we still prove that $a$ is a unit multiple of $b$? Let us still assume that $R$ is commutative. Of course, the above proof for integral domains can't work because we used the defining property of integral domains to conclude that $1 – rs = 0$. In this post we shall see a counterexample to show that two generators of the same principal ideal do not necessarily have to be unit multiples of each other.

I first heard of this question in Hershy Kisilevsky's algebraic number theory class. I forgot about it for some time until I came across Question 101989 on MathOverflow that gave a reference for a paper that was purported to have a counterexample showing that one cannot prove this statement for arbitrary commutative rings.
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