Is $\mathfrak{m}$ a free/projective/injective $R$-module?

Question

Problem: Let $k$ be a field and $R = k[x_1, x_2, x_3, \ldots] = k[x_i]_{i \in \mathbb{N}}$. Let $\mathfrak{m} = (x_1, x_2, x_3, \ldots)$.

I'm trying to figure out the following:

(i) Is $\mathfrak{m}$ a free $R$-module?

(ii) Is $\mathfrak{m}$ a projective $R$-module?

(iii) An injective $R$-module?

I think for (i) the answer is yes. I would just take as a basis $B = \left\{x_1, x_2, \ldots \right\}$. By def., any $f \in \mathfrak{m}$ is an element of the form $ f = q_1 x_1 + q_2 x_2 + \ldots$ where $q_i \in R$. However, is this supposed to be a finite sum? If so, it would show that $B$ generates $\mathfrak{m}$.

Then (ii) is also true, since free $R$-modules are also projective $R$-modules.

But for (iii) I don't really know.

Help is appreciated.

Answer 1

Freeness

The set you have chosen does generate that ideal, but it is clearly not a basis. You can write $x_1x_2$ as a multiple of two different elements.

Actually this gives us a bit of a clue: show that an ideal of an integral domain that is free can’t have more than one basis element. Then think about what that means for your ideal.

Injectivity

It cannot be injective, because injective submodules are always summands, and this is an integral domain: integral domains only have trivial summands.

Projectivity

Those were easy, but I had a harder time deciding projectivity. In the end my searches arrived at Proposition 1.1 in

Vasconcelos, Wolmer V. "Finiteness in projective ideals." Journal of Algebra 25.2 (1973): 269-278. (Science Direct link)

It reads

Proposition 1.1
Let $I$ be a projective ideal in a commutative ring. Then if $I$ is not contained in any minimal prime ideal, it is finitely generated.

I was quite surprised something like this existed in such generality, and which is also fairly elementary. If you already believe the maximal ideal isn't finitely generated, then it would do the trick to disprove projectivity of the ideal.

There may also be a simpler way to do this, but I haven't noticed it yet.

Answer 2

To see that $\mathfrak{m}$ is not projective, note that $R/\mathfrak{m}$ has a free resolution given by the Koszul complex of the (infinite) regular sequence $(x_n)$. All the maps in this complex are $0$ mod $\mathfrak{m}$, so tensoring with $R/\mathfrak{m}$ we find that $\operatorname{Tor}_i(R/\mathfrak{m},R/\mathfrak{m})$ is nontrivial for all $i$. In particular, $R/\mathfrak{m}$ has infinite projective dimension. If $\mathfrak{m}$ were projective, the short exact sequence $0\to\mathfrak{m}\to R\to R/\mathfrak{m}\to 0$ would imply that $R/\mathfrak{m}$ has projective dimension at most $1$, so $\mathfrak{m}$ cannot be projective (and in fact its projective dimension must be infinite).

Or, for a more elementary proof, let $S=k[x_1,x_2]$ and let $I=(x_1,x_2)\subset S$. If $\mathfrak{m}$ were projective as an $R$-module, it would also be projective as an $S$-module (since $R$ is free as an $S$-module so a direct summand of a free $R$-module is also a direct summand of a free $S$-module). Note that as an $S$-module, $\mathfrak{m}$ is just a direct sum of $I$ and a bunch of copies of $S$ (one for each positive-degree monomial in the other variables). So, it suffices to show $I$ is not projective as an $S$-module. We can show this directly: consider the surjective homomorphism $F:S^2\to I$ given by $F(a,b)=ax_1+bx_2$. If $I$ were projective, there would be a homomorphism $G:I\to S^2$ such that $FG=1_I$. Let $G(x_1)=(a,b)$ and $G(x_2)=(c,d)$. Since $G(x_1x_2)=x_1G(x_2)=x_2G(x_1)$, we must have $x_2a=x_1c$ and $x_2b=x_1d$. Thus $a$ and $b$ must be divisible by $x_1$ and $c$ and $d$ must be divisible by $x_2$. But then $F(G(x_1))=ax_1+bx_2$ has only terms of degree $\geq 2$ and so cannot be equal to $x_1$ as required.

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Injectivity is much easier: if $\mathfrak{m}$ were injective then the exact sequence $0\to\mathfrak{m}\to R\to R/\mathfrak{m}\to 0$ would split but that is impossible since $R/\mathfrak{m}$ has torsion and so cannot embed in $R$.