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?
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.
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.