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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.

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    $\begingroup$ Suppose $m<n$. Then the composition $R^m\simeq R^n\stackrel{\pi}\to R^m$ (here $\pi$ is the projection on the first $m$ components) is a surjective ring homomorphism, and let $I$ be its kernel. We have $R^m/I\simeq R^m$ and since $R^m$ is noetherian we get $I=(0)$ (see here), a contradiction. $\endgroup$ – user26857 Jul 27 '17 at 18:14
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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 $m=n$

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  • $\begingroup$ @users : Wow! That's awesome and easy too . Thanks . Your $\ker f :=\{(0,...,r_{n+1} , ..., r_m) \in R^m : r_{n+1},...,r_m \in R\}$ ; is right ? definitely it can not be $\{0\}$ if $m>n$ without even having to identify with $R^{m-n}$ ; but yours is okay too. Thanks again $\endgroup$ – user456828 Jul 22 '17 at 16:18
  • $\begingroup$ @misao : You are welcome . And yes, that is exactly my kernel of $f$ . I have now added that explicit expression in my answer $\endgroup$ – user Jul 22 '17 at 16:38
  • $\begingroup$ @users yes, I think this approach works. The same argument will work for rings with the ACC on ideals, a much larger class containing right noetherian rings. (There is no need to stick to commutative rings, it was never important at any point. $\endgroup$ – rschwieb Jul 22 '17 at 21:03
  • $\begingroup$ @rschwieb : yes true , commutativity was not needed ; its actually a bit bad habit of mine to assume commutative beforehand . Thank you for your attention $\endgroup$ – user Jul 23 '17 at 7:16
  • $\begingroup$ @rschwieb : what do you think about the claim when $R$ is an arbitrary $k$-algebra where $k$ is a field ? Can we then conclude $m=n$ ? (the isomorphism is still only assumed to be ring isomorphism) . We only need to look at infinitely generated algebras due to this answer $\endgroup$ – user456828 Jul 24 '17 at 14:59
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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.

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  • $\begingroup$ How do I get to $R/J(R)$ from $R$ ? The statement is certainly true for semisimple artinian rings as you suggest, I can see that .. but I don't really get how to work with $R/J(R) $ ... $\endgroup$ – user456828 Jul 20 '17 at 14:19
  • $\begingroup$ @misao The Jacobson radical is obviously invariant under ring isomorphisms. So, under your assumptions, $R^n/J(R^n)\cong R^m/J(R^m)$; on the other hand, $R^n/J(R^n)\cong(R/J(R))^n$. $\endgroup$ – egreg Jul 20 '17 at 14:21
  • $\begingroup$ ah I see ... since the Jacobson radical gets send to Jacobson radical ... thanks . Do you know what happens if $R$ is Noetherian ? $\endgroup$ – user456828 Jul 20 '17 at 14:28
  • $\begingroup$ @misao I feel there are counterexample, but have none at hand. $\endgroup$ – egreg Jul 20 '17 at 14:31
  • $\begingroup$ I could be misremembering, but I think there is probably a result for semiperfect rings about the uniqueness of the number of idempotents in a decomposition of $1$ into a complete orthogonal set, and that would probably extend the argument given for Artinian rings. Noetherian rings are surprising enough that they might furnish a counterexample (but none occurred to me either.) $\endgroup$ – rschwieb Jul 20 '17 at 16:39
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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 $m=n$

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