free $R$-algebras: when does $R\langle X\rangle\cong\!R\langle Y\rangle$ $\Rightarrow$ $|X|\!=\!|Y|$ hold?

Solution 1:

Assuming that $R$ is commutative, then we can make the following statements.

1. If $R\langle X\rangle\cong R\langle Y\rangle$ as $R$-algebras, then $\vert X\vert = \vert Y\vert$.
2. If $R\langle X\rangle\cong R\langle Y\rangle$ as rings then either $\vert X\vert = \vert Y\vert$ or $\{\vert X\vert,\vert Y\vert\}=\{0,1\}$.

So, if we only assume that they are isomorphic as rings then we still get the desired conclusion with the exception of the counterexample mentioned by Aaron.

Let's start by proving 1. Let $f\colon R\langle X\rangle\to R$ be any $R$-algebra homomorphism and $K={\rm ker}(f)=f^{-1}(0)$. This is a two sided ideal. Also, define $R^{(X)}$ to be the free $R$-module on $X$. Note that the quotient $K/K^2$ is an $R$-algebra. In fact, I'll show that $R^{(X)}\cong K/K^2$. So, if we have an $R$-algebra isomorphism $\theta\colon R\langle Y\rangle\to R\langle X\rangle$ then $$ R^{(X)}\cong K/K^2\cong\theta^{-1}(K)\big/\theta^{-1}(K)^2={\rm ker}(f\circ\theta)\big/{\rm ker}(f\circ\theta)^2\cong R^{(Y)}. $$ As noted in the question, free $R$-modules have bases of the same cardinality, so $\vert X\vert=\vert Y\vert$.

To construct an isomorphism from $R^{(X)}$ to $K/K^2$, first note that $f$ takes $X_i\in X$ to some element $x_i\in R$. Then (by universality), there is an isomorphism $u\colon R\langle X\rangle\to R\langle X\rangle$ taking $X_i$ to $X_i-x_i$. By composing with $u$, we can suppose that $f(X_i)=0$. Then, $K$ is the linear combinations of monomials without constant term, and $K^2$ is the linear span of monomials of degree at least 2. So, as an $R$-module, $K=K_1\oplus K^2$ where $K_1$ is the span of the degree 1 monomials, which is isomorphic to $R^{(X)}$. Then, $K/K^2\cong K_1\cong R^{(X)}$.

Comparing with Aaron's answer, this proof is very similar to his except that he works in the dual space. The $R$-derivations on $R\langle X\rangle$ is naturally isomorphic to the space of $R$-module homomorphisms from $K/K^2$ to $R$. So, by what I have just said above, $$ {\rm Der}_R(R\langle X\rangle,R)\cong {\rm Hom}_R(K/K^2,R)\cong{\rm Hom}_R(R^{(X)},R)\cong R^{\vert X\vert}. $$By passing to the dual space we pass from the direct sum $R^{(X)}$ to the direct product $R^{\vert X\vert}$ which, in the infinite case loses a bit of information about the cardinality of $X$ (without making set-theoretic assumptions beyond ZFC).

Finally, we can deal with case 2 above where it is only assumed that they are isomorphic as rings. Note that if $\vert X\vert > 1$ then $X_1X_2\not=X_2X_1$, so $R\langle X\rangle$ is not commutative, and it is commutative if $\vert X\vert\le1$.

Let's deal with the $\vert X\vert > 1$ case first. Then $R\langle X\rangle$ is noncommutative with $R$ being its center. $R\langle Y\rangle$ is also noncommutative with center $R$. So, a ring isomorphism $f\colon R\langle X\rangle\to R\langle Y\rangle$ restricts to an isomorphism of the centers of the rings and, hence, restricts to an isomorphism of $R$. By composing with an isomorphism $R\langle Y\rangle\to R\langle Y\rangle$ of $R$ (fixing the $Y_i$), we can reduce to the case that $f$ fixes $R$. So, $f$ is an $R$-algebra homomorphism, and we have reduced to case 1.

Finally, suppose that $\vert X\vert\le1$. Then $R\langle X\rangle$ is commutative, so $R\langle Y\rangle$ is commutative and $\vert Y\vert\le 1$. The only situation where $\vert X\vert\not=\vert Y\vert$ is when $\{\vert X\vert,\vert Y\vert\}$ equals $\{0,1\}$. This situation can occur, as mentioned by Aaron in the comment, so is the only possible exception.

Here's another quite neat way of looking at the proof of 1. If $A$ is an $R$-algebra and $M$ is an $A$-bimodule then a derivation is a homomorphism (as $R$-modules) $D\colon A\to M$ satisfying $D(ab)=(Da)b+a(Db)$. The space of Kähler differentials $\Omega_{A/R}$ is an $A$-bimodule with derivation $d\colon A\to \Omega_{A/R}$ satisfying the following universal property: if $D\colon A\to M$ is any other derivation then there is a unique $A$-bimodule homomorphism $\theta\colon\Omega_{A/R}\to M$ satisfying $D=f\circ d$. The universal property ensures that $d\colon A\to \Omega_{A/R}$ is uniquely defined up to isomorphism. It can be seen that the Kähler differentials $\Omega_{R\langle X\rangle/R}$ is just the free $R\langle X\rangle$ bimodule generated by terms $dX_i$ for $X_i\in X$. Then any $R$-algebra homomorphism $f\colon R\langle X\rangle\to R$ allows $R$ to be viewed as an $R\langle X\rangle$-bimodule, and we can form the (bimodule) tensor product $R\otimes_{R\langle X\rangle}\Omega_{R\langle X\rangle/R}$. By the universal property of $R\langle X\rangle$, all homomorphisms to $R$ are equivalent up to an isomorphism of $R\langle X\rangle$ so, up to isomorphism the tensor product is independent of the choice of $f$. In fact, the homomorphism $R\langle X\rangle\to R\otimes\Omega_{R\langle X\rangle/R}$ takes $X_i$ to $f(X_i)\in R$ so, considered as an $R$-module, the tensor product is just the free module over $R$ generated by terms $dX_i$. So, $$ R\otimes_{R\langle X\rangle}\Omega_{R\langle X\rangle/R}\cong R^{(X)} $$ regardless of the choice of $f$. This determines $\vert X\vert$ in terms of the $R$-algebra structure of $R\langle X\rangle$. We can regard the homomorphism $f$ as a generalized point ${\rm Spec}(R)\to{\rm Spec}(R\langle X\rangle)$ (although these definitions only really hold in the commutative case). The tensor product is then restricting the space of Kähler differentials to the point, which gives a free module generated by terms $dX_i$, just like in the commutative case. This is related to $K/K^2$ above because the homomorphism taking $a\in R\langle X\rangle$ to $(a-f(a)1)+K^2\in K/K^2$ is a derivation, giving a natural homomorphism $\Omega_{R\langle X\rangle/R}\to K/K^2$, and tensoring with $R$ turns it into an isomorphism. Using the universal property of Kähler differentials we also relate to Aaron's method, $$ \begin{align} {\rm Der}_R(R\langle X\rangle,R)&\cong{\rm Hom}_{R\langle X\rangle}(\Omega_{R\langle X\rangle/R},R)\cong{\rm Hom}_R(R\otimes\Omega_{R\langle X\rangle/R},R)\\&\cong{\rm Hom}_R(R^{(X)},R)\cong R^{\vert X\vert}. \end{align} $$

Solution 2:

We always have that $R\langle X \rangle \cong R \langle Y \rangle$ implies $|X|=|Y|$, assuming that the map $R\langle X \rangle \cong R \langle Y \rangle$ is a map of $R$-algebras.

Given an $R$-algebra $S$ and an $S$-bimodule $M$, a map $\delta:S\to M$ is an $R$-derivation if it is $R$-linear and $\delta(st)=\delta(s)t+s\delta(t)$. For example, if $S=R[x]$, then $f\mapsto \frac{df}{dx}$ is an $R$ derivation from $S$ to itself. We let $\operatorname{Der}_R(S,M)$ denote the collection of all $R$ derivations. This will be an $R$-module. Note that we need $R$ to be commutative and to lie in the center of $S$, or we run into some minor technical difficulties.

Let us view $R\langle X \rangle$ as an augmented $R$-algebra, that is, we fix a map of $R$ algebras $\epsilon:R\langle X \rangle\to R$. In general, such a map is given by selecting $n$ elements $r_1, \ldots r_n\in R$ and taking the algebra map satisfying $x_i\mapsto r_i$. This allows us to view $R$ as an $R\langle X \rangle$-bimodule. We wish to consider $N=\operatorname{Der}_R(R\langle X \rangle,R)$. We assert the following proposition.

Proposition: There is an isomorphism of $R$-modules $\operatorname{Der}_R(R\langle X \rangle,R)\cong R^{|X|}$.

sketch of proof: If $S$ is generated as an $R$-algebra by a collection of generators $\{x_i\}$, then $\delta\in \operatorname{Der}_R(S,M)$ is determined completely by $\delta(x_i)$. Thus an $R$ derivation $\delta$ of $R\langle X \rangle$ is determined by $\delta(x_i)$ and we have a map $\operatorname{Der}_R(R\langle X \rangle,M)\to M^{|X|}$. This map is an isomorphism (exercise!). Now take $M=R$.

With the proposition in hand, using your result that the rank of a free module is well defined, we easily have that $$R\langle X \rangle \cong R\langle Y \rangle \Leftrightarrow R^{|X|}\cong R^{|Y|} \Leftrightarrow |X|=|Y|.$$

N.B. While we needed to choose an augmentation, and our isomorphism $\operatorname{Der}_R(R\langle X \rangle,R)\cong R^{|X|}$ depended on the augementation, we see that this choice was inconsequential to the final result.