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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 $f_n$...2012-02-27
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    Dear Pierre, thank you for point it out for me. I has just edited.2012-02-27

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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?$2012-02-27
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    Thanks. That is what I want to say, the codimension of $\sum_{i=1}^{n-1}I_iS_{n-i}$ in $I_n$. :)2012-02-27
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    Do you agree with my statement? (Note that there is no $S_i$ in it.)2012-02-27
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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.2012-02-27
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    You're right. $+1$ for your nice answer!2012-02-28
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    @wxu: How do you know that $I_{1}$ has structure of a vector space? What is the difference between the number $k$ of $I_k$ and the number $k$ in $I_{1}$ as $k$-vector space. And, what does $S_{i}$ mean ?2012-02-29
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    There is a similar question here http://mathoverflow.net/questions/15584/minimal-number-of-generators-of-a-homogeneous-ideal-exercise-in-harsthorne , however I could not understand the answer there. Can anyone point it out for me ? Thanks2012-02-29
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    @msnaber $S=k[x_1,\ldots,x_n]$ is a graded ring, $S=k\oplus S_1\oplus S_2\ldots$, $S_i$ denotes the set of all homogeneous polynomials of degree $i$, i.e., $S_i$ is generated by those monomials $x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}$ of degree $i$ as a $k$-vector space. We say an ideal $I$ is homogeneous if $I=\oplus_i I\cap S_i $. So here $I_i=I\cap S_i$. Since $I$ is an homogeous graded ideal, it then is a $k=S_0$ module. etc.2012-02-29
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    @msnaber, I apologize that my answer above assumes that we are considering those generators are all homogeneous. So I have not considered your original question.2012-02-29
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    Dear @wxu, can you tell me more about the codimension of $\sum^{n-1}_{i-1}I_{i}S_{n-i}$? In your original answer, if $I$ do not have $I_{1}$ component, for example let $I$ be $(x^2, y^{2})$, then can we think $I_{1}S_{1}=0$?2012-04-25
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    @msnaber, Yes. Of course, in this case, we can think $I_1S_1=0$. In this case $I_1=0$. I think what is important is that just write $I$ by homogeneous parts then consider the lowest degree (homogeneous) elements, then second lowest degrees, and so on. You then may find the formula. Just consider a concrete example.2012-04-25
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    Dear @wxu, I still have a stupid question. Can a homogenous ideal of polynomial ring generated by generators of different degree? I beg your pardon for my stupid question.2012-04-28
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    @msnaber, Of course can, see the last part of the answer, $(x,y^2)$ can be generated by $x,y^2$ with different degrees. And, $(x,y^2)$ is a homogeneous ideal, you should check this. In fact, in a graded ring $S=\oplus S_n$, depending on what your definition of homogeneous ideal, the following is equivalent: 1,$I$ is generated by homogeneous elements; 2, $I=\oplus I\cap S_n$. If 1,or 2, holds for an ideal $J$, we say $J$ is homogeneous or graded.2012-04-28
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    @wxu : Can you tell me the dimension of a homogenous ideal $I=\circ I_{d}$ as a $k$-vector space? Since $d$ can tend to infinity, then can dimension of $I$ tend to infinity. What happened in my mind is that $I$ is finitely generated, and I got confusion of a generating set for $I$ as an ideal(which is finite) and a basis for $I$ is a vector space(which can tend to infinity). I beg your pardon if this question wastes your time. Thanks2012-05-02
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    @msnaber, You are welcome. The dimension of an ideal $I$ as $k$-vector space is always infinite if $I\neq 0$. Take an example, $I=(x)$ in $\mathbb{Q}[x]$, then $I$ has a basis $x,x^2,x^3,\ldots,x^n,\ldots$ as a $\mathbb{Q}$-vector space. But $I$ as an ideal can be generated by $x$. You will see that $\mathbb{Q}[x]=\mathbb{Q}\oplus (x)$. Aslo see $\mathbb{Q}[x]$ has a basis $1,x,x^2,\ldots$ as a $\mathbb{Q}$-vector space.2012-05-02
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    @msnaber, I should say that we are talking the polynomial ring over a field and with the usual grading.2012-05-02