Are $\Bbb{Z}$ and $K[x]$ (or even all PIDs) hereditary?

I’m trying to prove that $$\Bbb{Z}$$ and $$K[x]$$ are hereditary, so I think I should prove this for every principal ideal domain. Does anyone have any ideas?

At first we need to show that the global dimension of the PID is smaller or equal to 1 so that the derived functor Ext is equal to zero for every n>1 but then I don’t know how to work this.

• You should update your question with some ideas that you have (even if they didn't lead anywhere). Jun 28 at 10:25
• Ok sorry I changed it
– Hey
Jun 28 at 10:28
• (Not meant as an explanation, but worth knowing) the hereditary integral domains are exactly the Dedekind domains among which, of course, are PIDs. Jun 28 at 15:34

A further reference are the lecture notes by P. Clark.

A ring is called hereditary if every submodule of a projective module is projective. Then he states the example, see page $$57$$:

Example: A PID is a hereditary ring. Indeed, any nonzero ideal of a PID $$R$$ is isomorphic as an $$R$$-module to $$R$$.

There are a few ways to do this. The important fact is : $$(*)$$ a submodule of a free module $$R^n$$ over a PID $$R$$ is free. Now there are a few equivalent ways to proceed. $$(*)$$ implies that every projective module $$P$$ over $$R$$ is free, since it is a summand of a free module. So, it follows that any submodule of $$P$$ is free, and hence $$R$$ is hereditary.

You can tie this into the definition using $$\mathrm{Ext}^i(M,N)=0$$ for $$i\ge 2$$ easily as follows. $$\mathrm{Ext}^i(M,N)$$ can be computed using a projective resolution in the first variable. This is constructed as: $$F\xrightarrow{f} M\to 0$$ where $$F$$ is free. Now, $$K=\ker(f)$$ is a submodule of $$F$$ and hence free by $$(*)$$. So, we get a $$2$$ term projective (actually free) resolution $$0\to K\to F\to M\to 0.$$ Then, we compute $$\mathrm{Ext}^i(M,N)$$ by the cohomology of the complex resulting from applying $$\mathrm{Hom}(-,N)$$ to the above resolution. Then it is automatic (because the resolution is length $$2$$, i.e. in degrees $$0$$ and $$1$$) that $$\mathrm{Ext}^{\ge 2}(M,N)=0$$ for all $$R$$-modules $$M$$ and $$N$$.

The following conditions are equivalent for a ring $$R$$:

1. every submodule of a projective right $$R$$-module is projective;
2. every quotient of an injective right $$R$$-module is injective.

The proof is an interesting exercise, that can be done without homological algebra. Of course, both conditions are equivalent to the vanishing of $$\operatorname{Ext}_R^n$$ for $$n\ge2$$.

Now we can apply this to a PID $$R$$. A module $$M$$ over $$R$$ is called divisible if, for every $$x\in M$$ and every $$r\in R$$, $$r\ne0$$, there exists $$y\in M$$ such that $$ry=x$$.

Lemma. A quotient of a divisible module is divisible.

The proof is very easy.

Theorem. If $$R$$ is a PID, then an $$R$$-module $$M$$ is injective if and only if it is divisible.

Proof. Suppose $$M$$ is injective and let $$r\in R$$, $$r\ne0$$, and $$x\in M$$. The map $$\alpha\colon R\to R$$, $$a\mapsto ra$$ is a monomorphism, so there exists $$\gamma\colon R\to M$$ such that $$\gamma\alpha=\beta$$, where $$\beta(a)=ax$$. In particular, $$x=\beta(1)=\gamma\alpha(1)=\gamma(r)=r\gamma(1)$$. (Note: here we haven't used PID, just the fact that $$R$$ is a domain.)

Suppose $$M$$ is divisible. Then we can easily apply Baer's criterion to show that for every (nonzero) ideal $$I$$ and every homomorphism $$\alpha\colon I\to M$$, there exists $$\beta\colon R\to M$$ that extends $$\alpha$$, using the fact that $$I=rR$$ for some $$r\in R$$: set $$x=\alpha(r)$$ and take $$y\in M$$ such that $$ry=x$$. Then $$\beta\colon a\mapsto ay$$ is the required extension. QED

A domain of dimension $$0$$ is a field.

For a domain to have dimension $$1$$ means that every non-zero prime ideal is maximal.

For PIDs this is certainly true, e.g. this SE page

• It looks like you are talking about the Krull dimension, which is bounded above by the global dimension for commutative Noetherian domains. This does not seem to help directly. Jun 28 at 12:22