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I'll admit that I know absolutely no set theory (yet) but I'll dare to ask the following question:

Let $\kappa, \mu, \nu$ be cardinal numbers where $\mu, \nu$ are infinite cardinals. Suppose $\kappa^\mu = \kappa^\nu$. Does $\mu = \nu$ follow?

As mentioned in MO/67473 this appears to be independent of ZFC for $\kappa = 2$. I'm sorry if the answer is somehow trivially the same for other $\kappa$.

The background of the question is the proof of Proposition 21.2.18 on p. 570 in Vakil's Rising Sea (November 18, 2017 Draft). Given a $k$-algebra $B$ and $\mathfrak{m} \subseteq B$ a maximal ideal with residue field $k = B/\mathfrak{m}$, we wish to show $\mathfrak{m}/\mathfrak{m}^2 \cong \Omega_{B/k} \otimes_B k$ as $k$-vector spaces. To do so, Vakil begins by saying that it suffices to show an isomorphism of dual vector spaces $$ \operatorname{Hom}_k(\Omega_{B/k} \otimes_B k, k) \cong \operatorname{Hom}_k(\mathfrak{m}/\mathfrak{m}^2,k).$$ I'm not sure why this should suffice if the spaces in question have infinite dimension. (Hartshorne proceeds a little differently on p. 174 - he shows that the above induced by a map $\delta:\mathfrak{m}/\mathfrak{m}^2 \to \Omega_{B/k} \otimes_B k$ is surjective, so $\delta$ is injective and surjectivity of $\delta$ is obtained from the conormal exact sequence.)

So this prompted the following question:

Question. Let $V,W$ be $k$-vector spaces with $V^* \cong W^*$. Does $V \cong W$ follow?

For a counterexample, we clearly need $\dim{V}, \dim{W} > \infty$. In this case, a theorem by Kaplansky, Erdös says that $\dim{V^*} = |k|^{\dim{V}}$. Since there is a field of every cardinality by Löwenheim-Skolem, this exactly prompts the opening question.

On the other hand, I'm probably just not seeing an immediate argument in the specific algebraic geometry situation that I'm dealing with - so I would also be happy to get an answer to why Vakil was able to say that an isomorphism of dual spaces suffices.

Qi Zhu
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2 Answers2

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Wouldn't this work? If $\lvert k\rvert=2^\kappa$ with $\kappa>\aleph_0$, then $\lvert k\rvert^{\aleph_0}=2^{\kappa\times\aleph_0}=2^\kappa=2^{\kappa^2}=\lvert k\rvert^\kappa$, and therefore $\left( k^{\oplus\aleph_0}\right)^*\cong\left( k^{\oplus \kappa}\right)^*$. You may take $k=\Bbb Q(X_i\,:\, i\in\mathcal P(\Bbb R))$.

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Cardinal arithmetic is very forgiving when you change the base of the exponentiation.

If $\kappa\leq\min\{\mu,\nu\}$, then $\kappa^\mu=2^\mu$ and $\kappa^\nu=2^\nu$. So the same result about $\kappa=2$ will trivially hold at least up to $\min\{\mu,\nu\}$.

But in general this can be trivialised by simply taking a very very large $\kappa$. For example, $\kappa=2^{\mu+\nu}$ will satisfy that $\kappa^\mu=(2^{\mu+\nu})^\mu=2^{\max\{\mu+\nu,\mu\}}=2^{\mu+\nu}$ and the same for $\nu$.

Asaf Karagila
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