5

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?

Vincent
  • 10,614
yairb
  • 63
  • You don't define the number $n$, but from context it seems it both equals $\mu(I)$ and the dimension of $I/Im$ as a vector space over $R/Im$. But is it clear in general that these numbers should be equal? It is a serious question, not a hint or so. I am not very familiar with this topic. – Vincent Nov 11 '22 at 12:10
  • Also I feel a bit unconfortable with the notion of vectorspace over $R/Im$, since $Im$ is smaller than $m$ and hence not maximal in $R$ so that $R/Im$ is not a field. Sure instead of vector space you could have written module, but the notion of dimension there is a bit more slippery than in the vectorspace case, I believe – Vincent Nov 11 '22 at 12:13
  • Vincent - thanks. for the second comment it was my mistake, should be $R/m$ instead of $R/Im$ I edited the question. – yairb Nov 11 '22 at 12:54
  • Why $(A)=IJ+Im$? I see only that $(A)+Im=IJ+Im$ because $(A)$ and $IJ$ have the same image in $R/Im$. But I cannot figure out $(A)\supset IJ+Im$. – Acrobatic Nov 11 '22 at 13:12
  • Vincent - for the first comment, n is any natural number. and $\mu(I^n)$ are not necessarily equals to $\mu(I)$. my proof (even if is true) shows that $\mu(I^n)\le mu(I^n)$. for Counterexample see here – yairb Nov 11 '22 at 13:16
  • Acrobatic - $A$ is by definition a generating set of $IJ+Im$. the reason I think there is such a set is because there is some set $\bar{A}$ of leangth at most $\mu(I)$ generating the image of $IJ$ in $R/Im$. the image is $(IJ+Im)/Im$ so there is some set $A$ of length at most $\mu(I)$ generating $IJ+Im$. – yairb Nov 11 '22 at 13:29
  • 1
    Thanks for the reply! But now I am even more confused about the notion of $n$. I was ignoring the $n$ in the comparison between $\mu(I^n)$ and $\mu(I^r)$ because I thought your proof was of the first claim in the post, that $\mu(IJ) \leq \mu(I)$ where no number $n$ plays a role. My confusion was about the sentence 'so the dimension of $(IJ+Im)/Im$ is less then or equals $n$' further down. From context I gathered that this $n$, which shows up here for the first time, is the dimension of the vectorspace $I/Im$, because in the previous line you said that $(IJ+Im)/Im$ is a subspace of $I/Im$ – Vincent Nov 11 '22 at 13:39
  • But if this $n$ is the $n$ from $\mu(I^n) < n$ then it is not clear at all how it follows that $\dim (IJ+Im)/Im$ is less than or equal to $n$. This $n$ is much more mysterious to me – Vincent Nov 11 '22 at 13:41
  • 1
    vincent - thanks again, and i'm sorry, I didn't noticed the use of n in my proof so I misunderstood your question, (I thought that you claim $\mu(I^n)=\mu(I)$, and answer to this). of course you are right. I changed the use of $n$ in the proof to $\mu(I)$. this is equals to the dimension of the vector space $I/Im$. to see this it's enough to prove that ${a_1,...,a_n}$ is set of generators of $I$ iff ${a_1+Im,...,a_n+Im}$ generates $I/Im$. then because $I/Im$ is vector space $dim(I/Im)=\mu(I)$ – yairb Nov 11 '22 at 14:06

1 Answers1

2

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 $\mu(Im) = 0$.

Vincent
  • 10,614