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Let $R$ be a ring, $M$ be a finitely generated $R$-module. Let us define $e(M)$ to be the minimal of number of generators of $M$. Clearly, for finitely generated $R$-modules $M,N$ we have

$\mathrm{max}(e(M),e(N)) \leq e(M \oplus N) \leq e(M) + e(N).$

Question A. Are there better bounds, or is every value in between possible for $e(M \oplus N)$?

For example there are rings $R \neq 0$ such that $R \oplus R \cong R$ as $R$-modules, so that $e(R \oplus R)=1$. But for a commutative ring $R \neq 0$, we have $e(R \oplus R)=2$. The following question is more interesting for me:

Question B. Let $R$ be commutative ring, having only trivial idempotents (see the comment by navigetor23). Does it follow that $e(M \oplus N) = e(M) + e(N)$?

If this turns out to be false, which additional conditions should be imposed on $R$? For example, it should suffice that $R$ is a PID, right? But this is quite strong.

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If you take $M$ and $N$ two ideals of $R$ such that $I+J=R$ and $I\cap J=(0)$, then $\mu(I)=\mu(J)=1$ and $\mu(R)=1$. –  user26857 Nov 7 '12 at 9:17
Thank you. This happens iff $R$ is a non-trivial direct product iff there are non-trivial idempotents. I will edit the question. –  Martin Brandenburg Nov 7 '12 at 11:45
Local rings should be also included as "good" examples. –  user26857 Nov 7 '12 at 12:20
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1 Answer

up vote 2 down vote accepted

Question B has obviously positive answer if $R$ is local. But otherwise it is far from to be true.

Suppose we have (or rather $R$ has) two distinct maximal ideals $\mathfrak m$ and $\mathfrak n$. Let $m\le n$ be positive integers. Let $M=(R/\mathfrak m)^m$ and $N=(R/\mathfrak n)^n$. Then by Chinese Remainder Theorem, $$ R^n\to (R/\mathfrak m)^n\oplus (R/\mathfrak n)^n \to M\oplus N $$ are surjective. Therefore $e(M\oplus N)=n$ and $e(M)+e(N)=m+n$.

I think that playing with similar construction, everything between the max and the sum should happen for $e(M\oplus N)$.

Edit The above example uses in fact local computations of $e$. Here is an example of different taste. Let $R$ be a non-principal Dedekind domaine, let $L$ be finitely generated projective module over $R$ of rank $\ell$. Then it is known that $L$ can be generated by $\ell+1$ elements (let $a\in R$ be non-zero. Then $L/aL$ is locally free over the semi-local ring $R/aR$, so it is free of rank $\ell$. Lift a basis to $L$ and add $a$ to form a system of generators). If $L$ is not free then $e(L)=\ell+1$. As $R$ is non principal, there is a non principal ideal $I$. So $e(I)=2$. But the above discussions, $$e(I\oplus I)\le 3<e(I)+e(I).$$

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Thank you. Why is B true when $R$ is local? –  Martin Brandenburg Nov 7 '12 at 12:57
Just tensor the direct sum by $R/m$. –  user26857 Nov 7 '12 at 14:18
@navigetor23: Ah, Nakayama's lemma implies $e_R(M) = e_{R/m}(M \otimes R/m)$. Thanks. –  Martin Brandenburg Nov 7 '12 at 15:12
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