Posted by Jason Polak on 22. December 2015 · Write a comment · Categories: homological-algebra · Tags: ,

For modules one has the isomorphism theorem $(A/C)/(B/C) \cong A/B$ for $C\leq B\leq A$. One way to remember it is through analogy with canceling of fractions. Another way to remember and prove it is to put all the modules in a 3×3 commutative diagram
C & \to & B & \to & B/C\\
\downarrow & ~ & \downarrow & ~ & \downarrow\\
C & \to & A & \to & A/C\\
\downarrow & ~ & \downarrow & ~ & \downarrow\\
0 & \to & A/B & \to & (A/C)/(B/C)
where there are also zero arrows on the edges of the diagram, but I have omitted them for ease of typesetting. All the columns are exact, and the first two rows are exact, so the remaining row is exact giving the required isomorphism via the 3×3-lemma.

Posted by Jason Polak on 19. December 2015 · Write a comment · Categories: group-theory, homological-algebra · Tags:

Let $A$ be an abelian group. We call an element $a\in A$ torsion if there exists a natural number $n$ such that $na = 0$. The set of all torsion elements $T(A)$ of $A$ form a subgroup of $A$, and we can think of $T$ as an endofunctor on the category of abelian groups. Here are some examples:

  • $T(\Z) = 0$
  • $T(\Z\oplus \Z/n) = 0\oplus\Z/n\cong\Z/n$
  • $T(\Q) = 0$
  • $T(\Q/\Z) = \Q/\Z$

Finding the set of torsion points of an abelian group isn’t always easy as in these examples, since abelian groups may not always be written out in such an explicit way. A fascinating and nontrivial result of Barry Mazur is that the torsion subgroup of $E(\Q)$ for an elliptic curve $E$ over $\Q$ is one of fifteen possibilities: $\Z/n$ for $1\leq n\leq 10$ or $n=12$ or $\Z/2\times \Z/2n$ for $1\leq n\leq 4$. Such strange numerology!
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Let $R$ be any commutative ring. The content of a polynomial $f\in R[x]$ is by definition the two-sided ideal in $R$ generated by the coefficients of $f$. If $f,g\in R[x]$, then $c(fg)\subseteq c(f)c(g)$, because each coefficient of $fg$ is a linear combination of elements of $c(f)c(g)$. Sometimes, however, this inclusion is strict. For example, if $k$ is a field of characteristic two, and $R = k[u,v]$ then $f = u + vX$ satisfies $c(f^2)\subset c(f)^2$, where the inclusion is strict. Indeed, $f^2 = u^2 + v^2X^2$ so $c(f^2) = (u^2,v^2)$, whereas $c(f)c(f) = (u^2,v^2,uv)$. A ring $R$ in which $c(fg) = c(f)c(g)$ for all $f,g\in R[x]$ is called Gaussian. We have just seen that $k[u,v]$ is not Gaussian, and in fact, we didn’t even have to specify that $k$ is characteristic two. What about a polynomial ring $k[u]$ over a field $k$ in one variable? Since $k$ is a field, $k[u]$ is a principal ideal domain (PID), and PIDs are always Gaussian. These observations can be clarified by looking at the concept of weak dimension.
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The flat dimension of an $R$-module $M$ is the infimum over lengths of flat resolutions of $M$, and the weak dimension (or $\mathrm{Tor}$-dimension) of $R$ is the supremum over all possible flat dimensions of modules. Let’s use $\mathrm{w.dim}(R)$ to denote the weak dimension of $R$. As with the global dimension, the weak dimension of $R$ can be computed as the supremum over the set of flat dimensions of the modules $R/I$ for $I$ running over the set of all left-ideals or right-ideals, either is fine!

So, if every ideal is flat, then $\mathrm{w.dim}(R) \leq 1$. What about the converse? If $\mathrm{w.dim}(R) \leq 1$, is it true that every ideal is flat? Let’s make a side remark in that if we replace weak dimension with global dimension, and flat with projective, then the answer follows from Schanuel’s lemma. However, as far as I know there is no Schanuel’s lemma when ‘projective’ is replaced by ‘flat’.

However, we can get away with using part of the proof of Schanuel’s lemma. Before continuing, the reader may wish to check out the statement and proof of Schanuel’s lemma using a double complex spectral sequence.
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Let $R$ be a ring and $M$ be an $R$-module. The flat dimension of $M$ is the infimum over all lengths of flat resolutions of $M$. Usually, the flat dimension of $M$ is denoted by $\mathrm{fd}_R(M)$. For example, $\mathrm{fd}_{\mathbb{Z}}(\mathbb{Q}) = 0$. Since $\mathbb{Q}$ has projective dimension $1$, the flat dimension and projective dimension of a module can be different. Sometimes they can be the same: $\mathbb{Z}/n$ for $n$ a positive integer has the same flat and projective dimension as $\mathbb{Z}$-modules.

The weak dimension of a ring $R$ is defined to be $\mathrm{w.dim}(R) = \sup_{M} \{ \mathrm{fd}_R(M) \}$ where $M$ runs over all left $R$-modules. Due to the symmetric nature of the tensor product, we can also take the supremum over all right $R$-modules, in contrast to the asymmetric nature of global dimension.
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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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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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Let $R$ be a ring. The projective dimension $\mathrm{pd}_R(M)$ of an $R$-module $M$ is the infimum over the lengths of projective resolutions of $M$. The left global dimension of $R$ is the supremum over the projective dimensions of all left $R$-modules. There is a notion of right global dimension where left modules are replaced with right modules. Since we’ll be talking about commutative rings only, we’ll just use global dimension to refer to both kinds, and write $\mathrm{g\ell.dim}(R)$ for the global dimension of $R$.

As an example, if $R$ is a field then $\mathrm{g\ell.dim}(R) = 0$ because every module is free. If $R$ is a principal ideal domain (PID), then $\mathrm{g\ell.dim}(R) = 1$. This is because any module $M$ admits a surjection $F\to M$ where $F$ is a free module. But the kernel of this map is also free since over a PID, a submodule of a free module is free. One of the first results in the theory of global dimension is that $\mathrm{g\ell.dim}(R[x]) = 1 + \mathrm{g\ell.dim}(R)$. So far then we have examples of rings with any finite global dimension.
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Here are nine examples of projective modules that are not free, some of which are finitely generated.

Direct Products

Consider the ring $R= \Z/2\times\Z/2$ and the submodule $\Z/2\times \{0\}$. It is by construction a direct summand of $R$ but certainly not free. And it’s finitely generated! Another example is the submodule $\Z/2\subset \Z/6$, though this is the same kind of thing because $\Z/6\cong\Z/2\times\Z/3$. This was the first example I ever saw of a nonfree projective module.

Infinite Direct Products

One can modify the above construction for infinite direct products of rings, too. For instance, $R = \prod_{i=1}^\infty \Z$ contains $\Z$ as a direct summand. Hence $\oplus_{i=1}^\infty\Z$ is a projective $R$ module, yet cannot be free since nonzero free modules are uncountable.

Ideals in Dedekind Domains

In a Dedekind domain $R$, take an ideal representing a nontrivial element in the class group. It will then be projective. As an example, the class number of $\Z[\sqrt{5}]$ is two, and the ideal $(2,1+\sqrt{5})$ represents the nontrivial element in the class group. It is not free since it is not principal, and it is finitely generated projective since it is invertible.

More generally, for any ring extension of commutative rings $R\subseteq S$, one may define invertible $R$-submodules of $S$ as it is done for Dedekind domains. Then any invertible $R$-submodule of $S$ will be finitely-generated and projective. For more details and a further example, see Lam’s ‘Lectures on Modules and Rings’, Sections 2B-2C.

Rings of Continuous Functions

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Posted by Jason Polak on 19. December 2014 · Write a comment · Categories: homological-algebra, modules · Tags:

There are many ways to define the propery of semisimple for a ring $R$. My favourite is the “left global dimension zero approach”: a ring $R$ is left semisimple if every left $R$-module is projective, which is just the same thing as saying that every left $R$-module is injective. In particular, ideals are direct summands, and an easy application of Zorn’s lemma shows that $R$ can be written as a direct product of minimal left ideals, which is actually a finite sum because $R$ contains $1$.

An attack of Schur’s lemma yields the famous Wedderburn-Artin theorem: a ring $R$ is semisimple if and only if it is the finite direct product of matrix rings over division rings.

Since $R$ can be written as a finite direct product of minimal left ideals, we see that $R$ must be Noetherian and Artinian. Is the converse true?

Of course not! Here is a minimal counterexample: $\Z/4$. This ring cannot be semisimple. Indeed if it were, by the Wedderburn-Artin theorem, it would be a direct product of fields since it is commutative. It is not a field so it is not $\F_4$, and the only other possibility is $\Z/2\times\Z/2$, which it is also not isomorphic to since $\Z/4$ is cyclic.

We don’t have to appeal to the Wedderburn-Artin theorem however: the reduction map $\Z/4\to\Z/2$ makes $\Z/2$ into a $\Z/4$-module. If $\Z/4$ were semisimple, then $\Z/2$ would be a projective $\Z/4$-module, and hence at the very least as abelian groups, $\Z/2$ would be a direct summand of $\Z/4$, which is also nonsensical.

Can you think of a noncommutative example?