Let $\varphi: \mathbb{Q}[X] \longrightarrow R$ an inclusion of commutative rings. Suppose that the map $$- \circ \varphi: \operatorname{Hom}_{\mathbb{Q}\operatorname{-alg}}(R, R^{\otimes_{\mathbb{Q}} n}) \longrightarrow \operatorname{Hom}_{\mathbb{Q}\operatorname{-alg}}(\mathbb{Q}[X], R^{\otimes_{\mathbb{Q}} n}) \cong R^{\otimes_{\mathbb{Q}} n}$$ is a bijection for all nonnegative integers $n$ (where $R^{\otimes_{\mathbb{Q}} 0} = \mathbb{Q}$). Must the map $\varphi$ be an isomorphism? If so, then, replacing $\mathbb{Q}$ with an arbitrary field $k$ of characteristic zero, is the corresponding statement true?

The question is motivated by questions about plethories, specifically, Classification of plethories over $\mathbb{Q}$

In the original version of the question I did not include the assumption that $\operatorname{char} k = 0$. If you leave out that assumption then $\varphi$ need not be an isomorphism. A nice counterexample is provided below by Julian Rosen.

  • $\begingroup$ The symbol $\cong$ in the second line seems to be wrong. $\endgroup$ – Fernando Muro Dec 25 '13 at 22:27
  • $\begingroup$ I assume OP means for these to be homs for the category of $k$-algebras, where the $\cong$ is correct. $\endgroup$ – Todd Trimble Dec 25 '13 at 22:37
  • $\begingroup$ Sorry. I meant $k$-algebra homs. I edited it. $\endgroup$ – Jesse Elliott Dec 25 '13 at 23:14
  • $\begingroup$ Is $R$ commutative? $\endgroup$ – darij grinberg Dec 26 '13 at 1:16
  • $\begingroup$ Yes. I meant to say $R$ commutative (and associative). $\endgroup$ – Jesse Elliott Dec 26 '13 at 6:07

Let $p$ be a prime. Take $k=\mathbb{F}_p$, $$ R=k[X,X^{1/p},X^{1/p^2},\ldots]. $$ For any $k$-algebra $A$, there is a natural bijection $$ \mathrm{Hom}_{k\text{-alg}}(R,A)\cong\big\{(a_0,a_1,\ldots):a_n\in A, a_0=a_1^p, a_1=a_2^p,\ldots\big\}, $$ where $\varphi:R\to A$ corresponds to $(\varphi(X),\varphi(X^{1/p}),\ldots)$. If every element of $A$ has a unique $p$-th root, then the natural map $$ \mathrm{Hom}_{k\text{-alg}}(R,A)\to \mathrm{Hom}_{k\text{-alg}}(k[X],A) $$ is a bijection. Every element of $R^{\otimes_k n}$ has a unique $p$-th root, so this gives a counterexample to the first question.

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  • $\begingroup$ Nice! How about if $k$ is assumed to have characteristic zero? $\endgroup$ – Jesse Elliott Dec 26 '13 at 3:51

To answer my own question...I just proved (finally) that it's true assuming a positive answer to the question Classification of plethories over $\mathbb{Q}$, that is, assuming that every $\mathbb{Q}$-plethory is linear. Basically, the proof goes by showing that the given condition implies that $R$ has a $\mathbb{Q}$-plethory structure such that the comonad structure $W_R \longrightarrow W_R W_R$ is an isomorphism, and then its purported linearity implies that $R$ is isomorphic to $\mathbb{Q}[X]$ as a $\mathbb{Q}$-plethory.


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