MHB Is the direct sum of isomorphic commutative rings always equal in size?

  • Thread starter Thread starter Euge
  • Start date Start date
Click For Summary
The discussion centers on a problem regarding the isomorphism of direct sums of nonzero commutative rings, specifically questioning whether the direct sums R^m and R^n being isomorphic implies that m equals n. No participants provided answers to the problem posed this week. The original poster shared their solution after the lack of responses. The topic highlights the relationship between the sizes of direct sums in the context of commutative rings. The discussion remains unresolved as no alternative solutions were presented.
Euge
Gold Member
MHB
POTW Director
Messages
2,072
Reaction score
245
Here is this week's POTW:

-----
If $R$ is a nonzero commutative ring such that the direct sums $R^m$ and $R^n$ are isomorphic, show that $m = n$.

-----

Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!
 
Physics news on Phys.org
No one answered this week’s problem. You can read my solution below.

Let $\mathfrak{m}$ be a maximal ideal of $R$, and consider the residue field $k := R/\mathfrak{m}$. If $f : R^m \to R^n$ is an isomorphism, then it induces an isomorphism $1 \otimes f : k \otimes_R R^m \to k \otimes_R R^n$. Now $k \otimes_R R^m$ and $k\otimes_R R^n$ are $k$-vector spaces of dimensions $m$ and $n$, respectively; since those vector spaces are isomorphic, $m = n$.
 

Similar threads

Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
1
Views
3K
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K