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Let $I$ be a homogenous ideal in the ring $k[x_{1},\dots,x_{n}]$. My question is:

If $\lbrace f_{1},\dots,f_{r}\rbrace$ is a minimal system of generators of $I$, then are the integers $r$ and $\deg f_i$ determined uniquely by $I$?

More precisely:

If $\lbrace g_{1},\dots,g_{s}\rbrace$ is another minimal set of generators of $I$, then do we necessarily have $r=s$ and $\deg f_{i}=\deg g_{\sigma(i)}$ for some $\sigma\in S_r$.

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I suspect the $n$ of $x_n$ is not the same as the $n$ of $f_n$... –  Pierre-Yves Gaillard Feb 27 '12 at 12:20
    
Dear Pierre, thank you for point it out for me. I has just edited. –  Arsenaler Feb 27 '12 at 13:19
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up vote 3 down vote accepted

I think it is true. And we assume those generators are homogeneous, otherwise, the ideal $(x,x^n+y)=(x,y)$ of $k[x,y]$ gives an example that degrees are not unique determined.

Write $I=I_1+I_2+I_3+\cdots$, where $I_k$ is the part with degree $k$ of $I$. Then $I$ is generated by the homogeneous elements of degree $1,2,3,...$. But view $I_1$ as $k$-vector space, we see that those generators of degree $1$ form a $k$-base of $I_1$, so the numbers of those generators of degree $1$ are unique. Similarly, in degree $2$, $I_2=(I_1S_1,f_{21},f_{22},\ldots)$ (as $k$-vector space). Since generators are minimal, those generators of degree $2$ are not in $I_1S_1$ and form a base of $I_2/I_1S_1$. (Here $S_i$ is the set of homogeneous polynomials of degree $i$.)

We can do this step by step to see that your claim is true.

Hence the number of the minimal generators of $I$ is $\sum_n\operatorname{dim}_kI_n/\sum_{i=1}^{n-1}I_iS_{n-i}$. (By Noetherian property, there are only finite terms in the sum.)


Edit: If we do not require the generators are homogeneous, the statement is not true in general.

In $k[x,y]$, ideal $J=(x+y^3,y^2+y^3,y^4)=(x,y^2)$ is a homogeneous ideal, let us verify that $\{x+y^3,y^2+y^3,y^4\}$ is a minimal generating set.

If $I= (x+y^3,y^4)$, then $I=(x+y^3,y^4,x)=(x,y^3)$, contradiction. If $I=(x+y^3,y^2+y^3)$, then there exist $f,g\in k[x,y]$ such that $x=(x+y^3)f+(y^2+y^3)g$, let $x=1,y=-1$, we will get $1=0$, also contradiction.

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Dear wxu: Can't you just say that the number of degree $n$ generators is the codimension of $$\sum_{i=1}^{n-1}\ I_i\ I_{n-i}$$ in $I_n?$ –  Pierre-Yves Gaillard Feb 27 '12 at 16:02
    
Thanks. That is what I want to say, the codimension of $\sum_{i=1}^{n-1}I_iS_{n-i}$ in $I_n$. :) –  wxu Feb 27 '12 at 16:13
    
Do you agree with my statement? (Note that there is no $S_i$ in it.) –  Pierre-Yves Gaillard Feb 27 '12 at 16:17
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I do not agree. Let $I=(x)\subset k[x,y]$, then $\{x\}$ is a minimal generating set, $I_2=(x^2,xy)$ as $k$-module, $\operatorname{dim}_k I_2/I_1^2=1$, but we want it to be zero. –  wxu Feb 27 '12 at 23:09
    
You're right. $+1$ for your nice answer! –  Pierre-Yves Gaillard Feb 28 '12 at 0:33
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