# A ring isomorphic to its finite polynomial rings but not to its infinite one.

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?

#### Solutions Collecting From Web of "A ring isomorphic to its finite polynomial rings but not to its infinite one."

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 $\phi(A) \subseteq A[t_1, \ldots, t_N]$.