On ring isomorphism between cartesian products of rings

Question

Let $R$ be a (non-zero) commutative ring with unity. For $n \in \mathbb N$ denote by $R^n$ the ring under usual co-ordinate wise addition and multiplication. If $R$ is moreover artinian and $R^m \cong R^n$ as rings, then is it true that $m=n $ ?

I know that the similar claim is true if we look at $R^n$ as $R$-module. For my question of the ring case it is easy to see that $R^m \cong R^n$ as rings implies $m=n$ is true if $R$ has finitely many idempotents or say if $R$ has finitely many maximal ideals and is false if $R=\prod_{ \mathbb N}S=S^\mathbb N$ for any ring $S$.

Please help. Thanks in advance.

Answer 1

In the following , all rings are commutative with unity which is non-zero.

The claim is true in general for Noetherian rings ( hence also for Artinian rings as Artinian rings are Noetherian )

First let us recall a fact : If $R$ is Noetherian ring then any surjective ring endomorphism of $R$ is isomorphism . This is a very well known fact and a proof follows by considering the ascending chain of ideals $\{\ker f^n \}_{n\ge 1}$ if $f:R \to R$ is the given surjective ring homomorphism .

Now since ideals of a direct product of rings $R \times S$ looks like $I \times J$ where $I,J$ are ideals of $R , S$ respectively , so if $R,S$ are Noetehrian rings , then $R \times S$ also is a Noetherian ring. Now suppose $R$ is a Noetherian ring and let , if possible , $R^m \cong R^n$ as rings , for some integers $m>n\ge 1$ . Now consider the surjective ring homomprhism $f: R^m \to R^n$ defined as $f(r_1,...,r_m)=(r_1,...,r_n) , \forall (r_1,...,r_m) \in R^m$ ( this is surjective as $m >n$ ) . Now let $g : R^n \to R^m$ be a given isomorphism . Then $g \circ f : R^m \to R^m$ is a surjective ring homomorphism , and then since $R^m$ is Noetherian ring , we get that $g\circ f$ is an isomorphism . Hence $f$ is an injective , so $\ker f =\{0\}$ . But now by definition of $f$ , $\ker f =\{(0,...,r_{n+1},...,r_m)\in R^m | r_{n+1},...,r_m \in R\} \cong R^{m-n}$ as $R$-modules , so $R^{m-n} \cong 0$ as $R$-modules ; but as $m-n \ge 1$ and $R$ has unity which is non-zero , so $R^{m-n}$ cannot be singleton , contradiction ! Thus if $R$ is Noetherian and $R^m \cong R^n$ as rings then $m=n$

Answer 2

Hint: if $A$ is a commutative artinian ring, then $A/J(A)$ is a finite product of fields, where $J(A)$ denotes the Jacobson radical; if a ring is a product of fields, the number of factors is uniquely determined.

The result is generally not true if the ring is not artinian: take an infinite product of copies of a field, which is isomorphic to its square.

Answer 3

This answer uses some topology .

All rings are commutative with unity in the following .

We will show that the result is true for any ring $R$ satisfying ACC on radical ideals .

Let $R$ be a ring satisfying ACC on radical ideals and such that $R^m \cong R^n$ as rings . Then $Spec(R^m)$ and $Spec(R^n)$ are homeomorphic topological spaces ( see Atiyah ; Macdonald's , Commutative Algebra book for this) . Now as $R$ satisfies ACC on radical ideals , so $Spec(R)$ is a Noetherian topological space ( see https://en.wikipedia.org/wiki/Noetherian_topological_space , also Atiyah ; Macdonald's book) ; hence $Spec(R)$ has finitely many connected components (see http://www-personal.umich.edu/~gcheong/schemes/Notherian-fin-conn-components.pdf for example) . Let $Spec(R)$ have $t$ ( $>0$) many connected components . Now since the spectrum of a finite direct product of rings is homeomorphic to the disjoint union of the individual spectra , $Spec (R^n)$ has $tn$ many connected components. And since homeomorphism between topological spaces preserves the no. of connected components , so $tm=tn$ ,and hence $m=n$