let $(R,m)$ be a local ring and let $I,J$ be an ideals of $R$, let $\mu(I)$ be the minimal number of generators of $I$. I've proved that $\mu(IJ)\leq\mu(I)$ but somehow this claim does not feel right to me.
actually in here the authors go to great length to prove that if $\mu(I^n)\leq n$, then for every $r>n$, $\mu(I^r) \leq n$. while by my proof this is trivial.
my proof is:
$I/Im$ is vector space over $R/m$ and $(IJ+Im)/Im$ is subspace of $I/Im$ so the dimension of $(IJ+Im)/Im$ is less then or equals $\mu(I)$ so $IJ+Im$ can be generated by $\mu(I)$ elements. let $A$ be a set containing $\mu(I)$ generators of $IJ+Im$ so $(A)=IJ+Im$ and by Nakayama's lemma $(A)=IJ$.
is this right? and if not, where is my mistake?
I believe that the problem lies in the implication: 'the dimension of $(IJ+Im)/Im$ is less then or equals $\mu(I)$ so $IJ+Im$ can be generated by $\mu(I)$ elements.'
The reason $Im$ appears here below the division bar is that it is the maximal ideal in $I$. But the thing we need to have below the bar for this statement to hold is (I think) the maximal ideal in $(IJ + Im)$, so $(IJ + Im)m$.
The correct implication would be: 'the dimension of $(IJ+Im)/(IJ + Im)m$ is less then or equal to $\mu(I)$ so $(IJ + m)$ can be generated by less than or equal to $\mu(I)$ generators'. But unfortunately we do not know anything about the dimension of $(IJ+Im)/(IJ + Im)m$ as your proof only gives a bound on the dimension of $(IJ+Im)/Im$.
This latter quantity is however not very useful, I think. I figured this out by thinking about a special case:
$$J = m.$$ Then $(IJ+Im)/Im = \{0\}$, a vectorspace whose dimension is certainly smaller than $\mu(I)$ (so the first part of the proof is correct) but that also contains very little information.
We can certainly not infer from $Im/Im = \{0\}$ that $\mu(Im) = 0$.