Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A question that's been bothering me:

R is a ring with unity. Also consider $M_n(R)$ the matrix ring.

If all ideals $J$ of $M_n(R)$ are finitely generated, does every ideal $I$ of $R$ need to be finitely generated?

share|improve this question
1  
Is that statement true for all $n$? –  wj32 Oct 31 '12 at 8:15
    
Yes, for all $n$. –  drew52 Oct 31 '12 at 15:20
add comment

2 Answers

One can extend the natural homomorphism $R\rightarrow R/I$ for an ideal $I$ to the matrix ring giving a ring homomorphism $f:M_n(R)\rightarrow M_n(R/I)$, the rings $R$ and $R/I$ being diagonally embedded into the respective matrix ring. By assumption the kernel of $f$ is finitely generated, and it contains $I$. In particular the elements of $I$ can be written as sums of products of entries of finitely many matrices in the kernel of $f$. By definition these entries are elements of $I$, hence$I$ is finitely generated.

share|improve this answer
    
Can you say more about how the homomorphism $f$ is created by diagonally embedding $R$ and $R/I$? –  drew52 Oct 31 '12 at 20:36
1  
Map an element $r\in R$ to the diagnal matrix with the element $r$ in the diagonal at all places. The same for $R/I$. The homomorphism $f$ is mapping a matrix $A$ to the matrix obtained by taking the entries of $A$ modulo $I$. –  Hagen Nov 2 '12 at 8:36
add comment

The answer to your question is yes, as already indicated in Hagen's answer. Here's another proof that uses Morita equivalence. This is certainly overkill for such a basic question, but hopefully you can learn something through an alternate proof.

For any $n$, the rings $R$ and $M_n(R)$ are Morita equivalent. One Morita invariant property of a ring is the partially ordered set of its (two-sided) ideals (you can read about this, for instance, in Lam's Lectures on Modules and Rings, Proposition 18.44). And every ideal of a ring is finitely generated if and only if it satisfies the ascending chain condition on its ideals (the proof is the same as in the commutative case).

Thus every ideal in $R$ is finitely generated if and only if $R$ satisfies the ACC on ideals, if and only if $M_n(R)$ satisfies the ACC on ideals, if and only if every ideal of $M_n(R)$ is finitely generated.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.