# Yet Another non-Free Finitely Generated Projective

In the post Examples: Projective Modules that are Not Free, we saw nine examples of projective modules that are not free. On in particular was ‘the’ submodule $M = \oplus_{i=1}^\infty \mathbb{Z}$ of $\prod_{i=1}^\infty\mathbb{Z}$. Now, that’s a cool example to be sure, but the way we showed that $M$ was not free was to cite that $\prod_{i=1}^\infty\mathbb{Z}$ is uncountable. Actually, I like the argument a lot, but it’s possible to use the idea of that example and choose $M$ instead to be finite and different from all the examples in the aforementioned post. In fact, we’ll see a large class of examples that can be constructed from the ideas here.

The idea is to take an abelian group $A$ an consider $A$ as a module over its endomorphism ring $E = \mathrm{Hom}(A,A)$, where the endomorphisms are just homomorphisms $A\to A$ of abelian groups. Sometimes, $A$ can be projective over $E$. Actually, for a while it was believed that the projective dimension of $A$ over $E$ could only be $0$ or $1$, but eventually I.V. Bobylev showed in [1] that $A$ could have any projective dimension over $E$, including infinity!
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# Being Noetherian Is Not Local…Or Is It?

A commutative ring $R$ can be non-Noetherian and have all of its localisations at prime ideals Noetherian, such as the infamous $\prod_{i=1}^\infty \mathbb{Z}/2$. So being Noetherian is not a local property. However, there is an interesting variant of ‘local’ that does work, which I learnt from Yves Lequain’s paper [1]. It goes like this:

Theorem. Let $R$ be a ring and fix a left maximal ideal $M$ of $R$. Then $R$ is left Noetherian if and only if every left ideal contained in $M$ is finitely generated.

The nice thing about this statement is that it avoids localisation so it’s easy to state for noncommutative rings.
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# Wild Spectral Sequences Ep. 4: Schanuel’s Lemma

It’s time for another installment of Wild Spectral Sequences! We shall start our investigations with a classic theorem useful in many applications of homological algebra called Schanuel’s lemma, named after Stephen Hoel Schanuel who first proved it.

Consider for a ring $R$ the category of left $R$-modules, and let $A$ be any $R$-module. Schanuel’s lemma states: if $0\to K_1\to P_1\to A\to 0$ and $0\to K_2\to P_2\to A\to 0$ are exact sequences of $R$-modules with $P$ projective, then $K_1\oplus P_2\cong K_2\oplus P_1$.

We shall prove this using spectral sequences. I came up with this proof while trying to remember the “usual” proof of Schanuel’s lemma and I thought that this would be a good illustration of how spectral sequences can be used to eliminate the dearth of clarity in the dangerous world of diagram chasing.

Before I start, I’d like to review a pretty cool fact I which I think of as expanding the kernel, which is pretty useful More »