3

In this famous MO question, a beautiful proof is given of the fact $V\cong V^\ast\iff V$ is finite dimensional.

I'm trying to go through it and I'm having some trouble. First of all, I know the in the infinite dimensional case, $|V^\ast|=|k|^{\dim V}$ and $|V|=\max (|k|,\dim V)$. Now, the assumption in the first claim says $\dim V$ is at least $|k|$. If this means a loose inequality, then I don't understand how we can conclude $\dim V< \dim V^\ast$, since the only result on cardinal exponentiation I know of which gives a strict inequality is:

If $\kappa$ is infinite with $\mathrm{cf}(\kappa)\leq \lambda$ then $\kappa ^\lambda >\kappa$.

If we do assume a strict inequality, then I still don't know why our inequality follows, but even worse, I don't understand why knowing that each field contains a coutnable subfield finishes the proof. So I'm pretty sure we only need to assume a loose inequality, but I just don't understand the cardinal arithmetic.

The second part of my question is about which parts of the proof fail when generalizing to free non-finitely generated modules over commutative rings. It seems the problem can only arise in proposition 2, since the first one involves cardinality games alone, and the last fact seems to be true for most algebraic structures (having a countable substructure).

1 Answers1

0

For any $V$, we have $|V^*|=|k|^{\dim V}\geq 2^{\dim V}>\dim V$. The point of the assumption $\dim V\geq |k|$ is that it implies (via $|V|=\max(|k|,\dim V)$) that $\dim V=|V|$, so we get $|V^*|>|V|$, which then clearly implies $\dim V^*>\dim V$.

The case of arbitrary commutative rings actually follows more or less immediately from the field case. If $A$ is a (nonzero) commutative ring, choose any maximal ideal $m$, with residue field $A/m=k$. If $F=A^{\oplus S}$ is a free module with basis $S$, then $F^*$ is the infinite product $A^S$. Note that $F/mF$ can naturally be identified with $k^{\oplus S}$, and $F^*/mF^*$ has a natural surjection to $k^S$, and in particular, $\dim_k F^*/mF^*\geq\dim k^S$. If $F$ is isomorphic to $F^*$, then $F/mF$ must be isomorphic to $F^*/mF^*$ as a $k$-vector space. This means $|S|=\dim_k F^*/mF^*\geq \dim k^S$. By the field case, this implies $S$ must be finite.

Eric Wofsey
  • 342,377
  • The second part of your answer shows $F\cong F^\ast\implies$ $F$ f.g. How does the second part settle with Qiaochu Yuan's last comment on this answer? Is all well because free f.g $\not\implies$ $F\cong F^\ast$? –  Nov 18 '15 at 09:14
  • Also, I'm wondering about the proof itself - which parts of it are simply not true for modules? Seems you can still extends functionals and all that.. –  Nov 18 '15 at 09:16
  • The second part of my answer shows that if $F$ is free and $F\cong F^$, then $F$ is finitely generated. If $F$ isn't free, it doesn't tell you anything. I think the entire original proof should actually just work directly for any commutative ring, but I haven't checked every detail. Again, assuming it does work, it only tells you about free modules, not arbitrary modules that might satisfy $F\cong F^$. – Eric Wofsey Nov 18 '15 at 09:25
  • Oh, actually, a possibly serious problem with applying the proof to a general ring is that you don't know that $W^$ has to be free. So you can't just take a basis of $W^$ and extend it to a linearly independent subset of $V^*$ for the last step of the proof. – Eric Wofsey Nov 18 '15 at 09:41
  • So we can weaken our assumptions to $M$ is free and $M^{\ast}$ is free? Also, in the proof, why does $|S|\geq \dim k^S$ yield the result? I think $k^S$ is not a dual space. Can't we just say $\dim (F^\ast /mF^\ast)\geq \dim (F/mF)$ and then use $V\cong V^\ast \cong V^{\ast\ast}$? Then if we take $V=F^\ast /mF^\ast$ we get $\dim V \geq \dim V^\ast$ which implies $V$ is f.d. –  Nov 18 '15 at 10:35
  • I don't understand what you mean in the first question. The vector space $k^S$ is the dual of $k^{\oplus S}$, a vector space of dimension $S$. In this situation, you don't know that $V\cong V^$ (for $V=F/mF$); all you know is that $F\cong F^$. – Eric Wofsey Nov 18 '15 at 19:13