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I was messing with the ring $k[x_1,\dots,x_n,\dots]$ of polynomials in numerable many variables in order to solve an exercise of Atiyah, and the following question came to me and made me curious:

Is there a commutative unitary ring $A$ isomorphic to $A[X]$ but not isomorphic to $A[x_1,\dots,x_n,\dots]$ the ring of polynomials with coefficients in $A$ and numerable many variables?

Alex Ravsky
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Jimmy Page
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  • Sorry, I misread the question in my previous answer so I deleted it. – Marco Vergura May 11 '13 at 09:39
  • Let $φ \colon A[X] → A$ be an isomorphism of rings. Restricting $φ$ to $A$, this induces an injective ring homomorphism $ϕ \colon A → A$. My guess is $ψ \colon A[x_1, …, x_n, …] → A$ defined by $x_i ↦ ϕ^{i-1} \big(φ(X)\big)$ yields an isomorphism … – k.stm May 11 '13 at 09:43
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    @K. Stm: Are you sure? And how do you map the constants? – Martin Brandenburg May 11 '13 at 10:31
  • @MartinBrandenburg I didn’t think this through. (But I also don’t think it’s true either anymore.) – k.stm May 11 '13 at 10:44
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    More generally, one may ask the following, in any category with coproducts: If $A \cong A + P$ (one may say that $A$ is $P$-stable), does it follow that $A \cong A + \coprod_{n \geq 1} P$? A special case of this would be: If $P \cong P + P$, does it follow $P \cong \coprod_{n \geq 0}P$? This fails in many examples. There were similar discussions on MO. The usual proof sketches (construct homomorphisms inductively, then take their colimit) don't work, since we don't get compatibility. I strongly believe that there are (ugly) rings $A$ with $A \cong A[x]$ and $A \not\cong A[x_1,x_2,\dotsc]$. – Martin Brandenburg May 11 '13 at 11:42
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    @Amudhan: Was the answer a proof or a counterexample? – Martin Brandenburg May 27 '13 at 21:18
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    In other words: the easiest way to get a ring s.t. $A\cong A[x]$ is to take $A=B[x_1,...,x_n,...]$ — and the question asks if this is the only example. Interesting indeed. – Grigory M May 28 '13 at 19:56
  • A thought: if $A$ is Noetherian and $A\cong A[x]$, then since $A[x_1,x_2...x_n...]$ is not Noetherian, $A$ cannot be isomorphic to it... –  May 30 '13 at 14:08
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    After reading David's excellent answer I wonder: Can we produce, for every regular cardinal $\kappa$ a commutative ring $A$ with $A \cong A[\lambda]$ for all $\lambda<\kappa$, but $A \ncong A[\kappa]$? Perhaps one first finds a commutative semigroup with this property? – Martin Brandenburg May 30 '13 at 18:57
  • @MartinBrandenburg for any cardinal $\kappa>\aleph_0$, won't $\mathbb{Z}[\kappa_{-}]$ work? ($\kappa$ is the successor cardinal of $\kappa_{-}$) –  Jun 01 '13 at 17:07
  • Possibly related: https://mathoverflow.net/questions/225896/r-is-isomorphic-to-rx-y-but-not-to-rx – Watson Nov 05 '16 at 10:42

1 Answers1

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Let $\Gamma$ be the set of sequences $a_1 a_2 a_3 \cdots $ of nonnegative integers which are eventually constant. A typical element of $\Gamma$ looks like $3921726666666\cdots$. Note that $\Gamma$ is a commutative semigroup under addition. Let $A$ be the semigroup algebra $\mathbb{Z}[\Gamma]$. I claim that $A \cong A[t]$ but $A \not \cong A[t_1, t_2, t_3, \ldots]$.

The semigroup $\Gamma$ is isomorphic to $\mathbb{Z}_{\geq 0} \times \Gamma$, by the map $a_1 a_2 a_3 \cdots \mapsto (a_1, a_2 a_3 \cdots)$. So $\mathbb{Z}[\Gamma] \cong \mathbb{Z}[\Gamma \times \mathbb{Z}_{\geq 0}]$ or, in other words, $A \cong A[t]$.

Let $z$ denote the sequence $11111\cdots$ and let $x_i$ denote the sequence $000\cdots01000\cdots$ with the lone $1$ in the $i$th position. Note that $\Gamma$ embeds in the free abelian group generated by $z$ and the $x_i$, so $A$ embeds in the Laurent polynomial ring $\mathbb{Z}[z,x_1^{\pm},x_2^{\pm},x_3^{\pm},\ldots]$. In particular, $A$ is an integral domain. In this notation, the isomorphism $A \to A[t_1, t_2, \ldots, t_k]$ sends $z \mapsto z t_1 t_2 \cdots t_k$; it sends $x_i \mapsto t_i$ for $i \leq k$ and $x_i \mapsto x_{i-k}$ for $i >k$.

Now, suppose that $\phi$ is a map $A \to A[t_1, t_2, \ldots ]$ with $\phi(z)$ nonzero. We show that $\phi(A) \subseteq A[t_1, t_2, \ldots, t_N]$ for some $N$. In particular, $\phi$ is not an isomorphism.

Choose $N$ large enough that $\phi(z) \in A[t_1, t_2, \ldots, t_N]$. Since $\phi(x_i)$ divides $\phi(z)$, and $A$ is a domain, we must also have $\phi(x_i) \in A[t_1, \ldots, t_N]$. For any $\gamma$ in $\Gamma$, write $\exp(\gamma)$ for the corresponding element of $A$. For any $\gamma \in \Gamma$, we have $\prod_{i=1}^M x_i^{b_i} \exp(\gamma) = \prod_{i=1}^M x_i^{c_i} \cdot z^d$ for some sufficiently large $M$ and some $d$. Therefore, $\phi(\exp(\gamma))$ can be written as a ratio of $\prod_{i=1}^M \phi(x_i)^{c_i} \cdot \phi(z)^d$ and $\prod_{i=1}^M \phi(x_i)^{b_i}$, both of which are in $A[t_1, \ldots, t_N]$ (using again that $A$ is a domain). So $\phi(\exp(\gamma)) \in A[t_1, \ldots, t_N]$. Since the $\exp(\gamma)$ are a $\mathbb{Z}$ basis for $A$, this shows that $\phi(A) \subseteq A[t_1, \ldots, t_N]$.

David E Speyer
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    So $A \approx \Bbb Z[X_1,X_2,\ldots,Y,Y/X_1,Y/X_1X_2, \ldots] \subset \Bbb Q(X_1,X_2,\ldots,Y)$ ; there is an isomorphism $A[X_0] \to A$ by $X_0 \mapsto X_1 \mapsto X_2 \ldots, Y \mapsto Y/X_1$, and any map $A \to A[T_1,T_2,\ldots]$ is contained in a $A[T_1,\ldots,T_N]$. This is great. – mercio May 30 '13 at 15:41
  • Yup! I don't think you need to say $\approx$, your description is equivalent to mine, and maybe clearer. (Except that you should say any map with $\phi(Y) \neq 0$.) – David E Speyer May 30 '13 at 16:32
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    Great example! Note that $A$ can be thought of as a ring of formal power series in $T_1,T_2,\dotsc$, spanned by those monomials $T_1^{a_1} T_2^{a_2} \dotsc$ for which $a_1,a_2,\dotsc$ are eventually constant. So as a ring it is generated by the $T_i$ and the $T_i \cdot T_{i+1} \cdot \dotsc$ (this agrees with mercio's presentation). I had also tried various subrings of formal power series in infinitely many variables, but didn't see this nice example! – Martin Brandenburg May 30 '13 at 18:47