Abstract Nonsense

Injective Modules (Pt. III)

Point of Post: This is a continuation of this post.

April 28, 2012

Submodules of Free Modules Need Not be Free Unless Ring is a PID (pt.I)

Point of Post: In this post we show that submodules of free modules need not be free, and give some intuition why this should be true. We then show that, in fact, submodules of $R$-modules hold (for a unital commutative ring $R$) precisely when $R$ is a PID.

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Motivation

Now that we have discussed free modules we’d like to discuss some of the properties that hold in these free modules. In particular, while tempting it is, in general, wrong to take theorems about the “nicest free modules” (i.e. vector spaces/modules over division rings) and transfer these to general theorems about free modules. This post shall be devoted to resolving one such “inaccurate transfer”. Namely, the “theorem” that every submodules of a free module is free. For example, everyone knows that submodules of vector spaces, or more generally submodules of modules over division rings, are free. Why? Well, the obvious reason–they’re still vector spaces/modules over division rings which we know are always free! Ok, now try to make an analogous proof for a general free left $R$-module $F$ and some submodule $N\leqslant F$. Since we know that, in general, all modules over a given ring are not free (in fact, we shall eventually (and this is a real eventually) show that the only rings for which this statement is true are division rings) we can’t rely on this argument. So, one starts to try to create a basis for $F$ in terms of a basis for $N$. Hmm, a little thought shows that there is no natural way of doing this. We could “hope” that we can nab a basis for $F$ which contains elements of $N$, but we can’t really see a way of necessarily doing this (one may be inclined to, hopefully to some mild embarrassment, try the old linear algebra trick of  “extend a basis from $N$ to $F$“, which of course has a little logical flaw). Ok, fine, so at this point we’ve seen that there is really no natural way of showing that a submodule of a free module isn’t free, but perhaps this is because we aren’t being clever enough, the nonexistence of a proof is not a disproof. As I said though, this is, in fact, a non-theorem. Intuitively, there is an example that typifies the reason why this “theorem” sometimes fails. We know that if we are given a ring $R$ and a left ideal $\mathfrak{a}$ of $R$ then $\mathfrak{a}$ is a submodule of the free $R$-module $R$ and so if our “theorem” was true this would imply that $\mathfrak{a}\cong R^{\oplus \lambda}$ for some cardinal $\lambda$. That’s a pretty steep condition, no? We see that our good friends fields and division rings side-step this issue all together being precisely the rings for which the only ideals of $R$ are $\{0\}$ and $R$, or $R^0$ and $R^1$ respectively. But, if we take, for example some finite ring $R$ which is not a division ring then there exists some proper non-trivial ideal $\mathfrak{a}$. Clearly then we can’t have that $\mathfrak{a}\cong R^{\oplus\lambda}$ for some $\lambda$ by merely appealing to a cardinality argument, and so $\mathfrak{a}$ cannot be a free $R$-module. So, this immediately gives us a whole class of examples. For instance, $\mathbb{Z}_6$ is a free $\mathbb{Z}_6$-module and $2\mathbb{Z}_6$ is an ideal, but is surely not a free $\mathbb{Z}_6$ modules since any such (finite) module would have order $6^n$ for some $n\in\mathbb{N}\cup\{0\}$ and $|2\mathbb{Z}_6|=3$. This is a typifying example precisely because, in the category of commutative unital rings, this is the only possible obstruction in the sense that if a ring does not have this obstruction (i.e. every ideal is a free module) then one can generally state that submodules of free modules over that ring are free.

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November 21, 2011

Polynomial Rings in Relation to Euclidean Domains, PIDs, and UFDs (Pt. II)

Point of Post: This is a continuation of this post.

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October 24, 2011

Polynomial Rings in Relation to Euclidean Domains, PIDs, and UFDs (Pt. I)

Point of Post:

In this post we discuss how polynomial rings act when their underlying rings are PIDs, Euclidean domains, and UFDs. The high points being the proof that $k[x]$ is a Euclidean domain for fields $k$ and $R[x]$ is a UFD when $R$ is a UFD.

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Motivation

We have mentioned several times that polynomial rings occupy a central role in modern mathematics. Consequently, studying them should occupy no small amount of our time. In this post we shall prove some previously made assertions about when polynomial rings are Euclidean domains, PIDs, and UFDs and some of the consequences of thes results.

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October 24, 2011

UFDs (Pt. II)

Point of Post: This is a continuation of this post.

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October 22, 2011

UFDs (Pt. I)

Point of Post: In this post we discuss the most general “nice integral domain”, unique factorization domains.

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Motivation

So, up until this we point we have discussed the very nice integral domains which admit degree functions (i.e. Euclidean domains) and the, less nice still fantastic, integral domains whose ideals are all principal (i.e. PIDs). In this post we shall take another step down the niceness-ladder and discuss a proper subset of PIDs (and thus a “doubly proper” subset of Euclidean domains)–UFDs. Amusingly enough though, UFDs are perhaps the most useful of the three, because they pop up much more often than either Euclidean domains or PIDs. Perhaps the first indication of why this is so is captured by the nice interaction UFDs have with the construction $R\mapsto R[x]$. Indeed, we have seen $R[x]$ is a PID (let alone a Euclidean domain) if and only if $R$ is a field! So, in the vast majority of cases if one starts with a Euclidean domain or a PID $R$ and passes to the polynomial ring $R[x]$ one is not going to end up with a Euclidean domain or a PID. The classic example is that $\mathbb{Z}$ is about as nice of a Euclidean domain as you could want, yet $\mathbb{Z}[x]$ is not even a PID ($(p,x)$ for a prime $p$ is not principle)! That said, as we shall see the construction $R\mapsto R[x]$ does preserve UFDs in the sense that $R$ being a UFD implies that $R[x]$ is a UFD. So, for example, $\mathbb{Z}[x]$ while not a PID or a Euclidean domain is a UFD. Now, we have indicated before that polynomial rings play a pivotal role in not only algebra but algebra’s application to other fields of mathematics (e.g. algebraic geometry). With this in mind any properties that interact well with the construction $R\mapsto R[x]$ are definitely well-worth our time.

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So, I have yet to say actually what a UFD is. If you pulled some kid off the street and asked him what was one of the most startlingly useful and insightful ring theoretic properties of $\mathbb{Z}$ they might be apt to say “Oh! The fundamental theorem of arithmetic–every integer factors uniquely (up to differences in sign) into a product of primes!” (smart kid, huh?). Well, that kid would be right. Anyone who has done even the slightest bit of number theory is well-aware that the existence and uniqueness of integer factorization into primes is an invaluable tool. Often times easier in just a “makes things” simpler–how often are you trying to decide if some divisibility argument is true, and you just produce the prime factorization of things involved, and it becomes trivial. Well, this is what UFDs attempt to capture. We shall see that in any integral domain there are notions analogous to (integer) primality and UFDs are (roughly) those for which there is a notion of unique factorization into these prime elements. Indeed, this probably would have been clear if I had mentioned that UFD stands for Unique Factorization Domain.

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October 22, 2011

PIDs (Pt. I)

Point of Post: In this post we discuss PIDs, and some of the pursuant theorem.

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Motivation

In this post we discuss a more general class of rings than Euclidean domains. The type of ring we are talking about is a natural one to consider. In particular, we have seen from experience that the ideal theory of a ring is a key-determinant in the rings complexity. For example, the nicest rings (fields) are precisely those with the nicest ideal theories (just the full ring and trivial ideal). Consequently, since our current goals (as laid out in the last post) is create a series of integral domains which are “nice” (so that, when encountered in practice, all our great theory applies to them) it makes sense to create some integral domains with nice ideal theory. So, we saw that having few ideals can make an ideal theory, but it is not the only way. In particular, besides having a small number of ideals its clear that we can make the ideal theory of a ring “nice” by requiring that each ideal is “nice”. Well, it’s pretty clear what the nicest type of ideals are, principal (singly generated). Thus, in this post we shall discuss rings which are integral domains for which every ideal is principal, or as we shall call them, PIDs.  These clearly contain Euclidean domains since, as we proved before, every ideal in a Euclidean domain is principle. In fact, for most basic uses we shall see that Euclidean Domains are just “practical” PIDs in the sense that the theoretical niceties of Euclidean domains are just the niceties from being a PID, but the applications of the theory is much more practical for Euclidean domains (e.g. actually finding the element which generates the ideal, finding greatest common divisors, etc.).

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October 14, 2011