Is it true that $F[(X_{\alpha})_{\alpha \in A}] \cong F[(X_{\beta})_{\beta \in B}] \implies |A|=|B|$, when $F$ is a subring of $\Bbb R$?

Question

Let $F$ be a subring of $\Bbb R$ and let $A$ and $B$ be two non-empty sets. Suppose that the polynomial rings $F[(X_{\alpha})_{\alpha \in A\;}] \cong F[(X_{\beta})_{\beta \in B\;}]$ are isomorphic as rings. Does it follow that $A$ and $B$ have the same cardinality?

Some possible variations:

1. We could see what happens if we assume that we have an $F$-modules isomorphism (or an $F$-algebras) between the two polynomial rings. In that case, I think that we have two free $F$-modules, and as $F$ is a commutative ring with unity, it can be shown (e.g. by quotienting by a maximal ideal) that the cardinalities are indeed the same.
2. It would be interesting to see what happens if we assume that $F$is a subfield and $F((X_{\alpha})_{\alpha \in A\;}) \cong F((X_{\beta})_{\beta \in B\;})$, either a) as fields, or b) as $F$-algebras. This is maybe related to one of my previous questions. As for $F$-algebras, I believe that an argument of transcendence degree over $F$ is sufficient. The result is shown to be wrong if we accepted to take subfields of $\Bbb C$. I don't know if my main question is also wrong if we accept to take subrings of $\Bbb C$.

The result is true for $F=\Bbb Z$. The result of my main question and of $2a)$ is also true for $F=\Bbb Q$ : if $\Bbb Q[(X_{\alpha})_{\alpha \in A\;}] \cong \Bbb Q[(X_{\beta})_{\beta \in B\;}]$ as rings, then they are isomorphic fraction fields, which must have the same transcendence degree over $\Bbb Q$ (because a field isomorphism is a $\Bbb Q$-algebra isomorphism), so we are done. This could probably generalized to extensions of $\Bbb Q$ with finite transcendence degree, I think.

Interesting examples can be found here. Obviously we can easily build counter-examples if we can take any ring for $F$, like $F=\Bbb Z[x_1, \dots, x_n, \dots]$, in which case we have $F[x] \cong F[x,y]$. However, I don't think that $F$ embeds in $\Bbb R$ as a subring (it does as an additive subgroup, I think, because $\Bbb R \cong \Bbb Q^{(2^{\aleph_0})}$ as abelian groups). Here is also a related (but different) question. On MO, this similar question is interesting.

Thank you!

Answer 1

• If you mean an isomorphism of rings, then the answer is no. For example, if $\{x_i\}_{i\in \mathbb{N}}$ is a subset of pairwise distinct elements of a transcendence basis of $\mathbb{R}$ over $\mathbb{Q}$, then $F=\mathbb{Z}[x_i]_{i\in\mathbb{N}}$ is a subring of $\mathbb{R}$ such that $$ F \cong F[X_1] \cong F[X_1, X_2] \cong \ldots $$

• If you mean an isomorphism of $F$-algebras, then the answer becomes yes. Indeed, assume you have an isomorphism $f:F[X_\alpha]_{\alpha \in A}\to F[X_\beta]_{\beta \in B}$. Let $K$ be the field of fractions of $F$; then $f$ induces an isomorphism $\overline{f}:K[X_\alpha]_{\alpha \in A}\to K[X_\beta]_{\beta \in B}$. Now, $A$ is finite iff the Krull dimension of $K[X_\alpha]_{\alpha \in A}$ is finite; in that case, $|A|$ is the Krull dimension of $K[X_\alpha]_{\alpha \in A}$. Since the Krull dimension is invariant under isomorphism, we get $|A|=|B|$.
  Assume now that $A$ is infinite. By the above, this implies that $B$ is infinite as well. Then $\overline{f}$ sends each $X_\alpha$ to a polynomial involving finitely many of the $X_\beta$. Thus $|A|≥|B|$; using the inverse isomorphism, you would get the inverse inequality.

• If you mean an isomorphism of rings, but assume further that $F$ is a field, then the answer is yes as well. Indeed, an isomorphism $f$ would send all elements of $F$ to elements of $F$, since $F\setminus\{0\}$ is the set of invertible elements of the polynomial rings. Then the above argument for the second case can still be applied.