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 $\sigma\in S_r$.

I have asked this question on MS, however I did not get a satisfactory answer. Please help me answer this question. Thank you very much!

  • $\begingroup$ I do not know why the sentences after "more precisely..." is bigger and in bold. Moderators of MO, please edit it for me. Thanks $\endgroup$
    – Arsenaler
    Mar 26, 2012 at 2:32
  • 1
    $\begingroup$ @msnaber: The boldface was caused by your "----" following that paragraph. I added a line break which removed it. If you intended to produce a horizontal line, then this should probably look like what you intended, otherwise, remove the "----". $\endgroup$ Mar 26, 2012 at 7:23
  • $\begingroup$ @SándorKovács : Thank you very much! $\endgroup$
    – Arsenaler
    Mar 26, 2012 at 14:03

2 Answers 2


Yes, they are uniquely determined by $I$. This holds more generally:

Let $R = \oplus_{i \ge 0}R_i$ be a graded ring with $R_0 = K$ a skew field and let $M= \oplus_{i\ge 0}M_i$ be a finitely generated, graded $R$-module. Set $N_k := \sum_{i> 0} R_iM_{k-i}$. If $E$ is a minimal homogeneous set of generators of $M$ over $R$, then, for each integer $k$, the number of elements in $E$ having degree $k$, equals $\dim_K M_k/N_k$.

In particular, this number is independent of $E$. So, if $E,E'$ are two minimal generating sets, they have the same number of elements in each degree (giving a degree-wise bijection) and hence they have the same number of elements altogether.

Proof of the statement: Let the generators be $x_i$ of degree $d_i$. We want to show that $\bar{E}_k := \lbrace x_iN_k \mid d_i = k \rbrace$ is a $K$-basis of $M_k/N_k$.

First note that $M_k = \sum_i R_{k-d_i}x_i= \sum_{d_i \le k}R_{k-d_i}x_i$. Since $\sum_{d_i < k}R_{k-d_i}x_i \subseteq N_k$ and $R_0=K$ it follows that $\bar{E}_k$ is a generating set.

In order to show linear independence, let $a_p \in K$ such that $\sum_p a_px_p = n \in N_k$ (sum over $p$ with $d_p = k)$. By definition $n = \sum_{i+j=k}r_im_j$ with $r_i \in R_i\;(i>0)$ and $m_j \in M_j$. As above we may write $m_j = \sum_l s_{j,l}x_l$ with $\deg(s_{j,l}) = j-d_l$. Putting all together we have

$$\sum_p a_px_p = \sum_{i+j=k}r_i\sum_l s_{j,l}x_l= \sum_l \big (\sum_{i+j=k}r_is_{j,l} \big )x_l=: \sum_l t_l x_l $$ where $\deg t_l=k-d_l \ge 1$, since $\deg r_i \ge 1$. In particular, $\deg x_l = d_l < k$ $= \deg x_p$, so $x_p \neq x_l$ for all $p,l$.

Now, if $\alpha_p \neq 0$ for some $p$, we can express $x_p$ as $R$-linear combination of the $x_i, i \neq p$, contradicting the minimality of $E$. Hence $\alpha_p = 0$ for all $p$.


A simple minded direct approach would be this:

Q: How do we find a set of generators for $I$?


  1. Let $d$ be the smallest integer for which $I_d\neq \emptyset$. Then $I_d$ is a $k$-vector space, so we must choose $r_d=\dim_kI_d$ number of elements of $I_d$ just to generate $I_d$ and we can choose that many to do so.
  2. Next we need to generate $I_{d+1}$. Obviously we already have $x_1\cdot I_d+\dots+x_n\cdot I_d\subseteq I_{d+1}$ and to generate the entire $I_{d+1}$ we minimally need $r_{d+1}=\dim_k I_{d+1}/(x_1\cdot I_d+\dots+x_n\cdot I_d)$ elements.
  3. (this should really be "i.") In every subsequent step, in order to determine the needed generators in an arbitrary $I_{d+i}$, we have "stuff" coming from lower degree parts and whatever they generate determines exactly the number of generators we need: $r_{d+i}=\dim_k I_{d+i}/\{\text{stuff coming from lower degree parts}\}$

This shows that the number of degree $e$ homogenous elements in a minimal set of generators is uniquely determined by $I$. This is equivalent to the desired statement.

Remark: Of course, I realize that this must be essentially the same proof as Ralph's, but perhaps it is a bit more down-to-earth.

  • $\begingroup$ Sándor, nice explanation. I think in 2. you should have used $R_d$ instead of $I_d$ in the linear combinations. $\endgroup$
    – Ralph
    Mar 26, 2012 at 8:02
  • $\begingroup$ Ralph, I think it is OK this way. The $\\{x_i\\}$ stand for $R_1$. $\endgroup$ Mar 26, 2012 at 8:22
  • $\begingroup$ Ralph, I am using the original setup, so the "$x_i$" are the variables in $k[x_1,\dots,x_n]$ and not the generators. $\endgroup$ Mar 26, 2012 at 8:27
  • $\begingroup$ Ah, I see, you're right. $\endgroup$
    – Ralph
    Mar 26, 2012 at 8:35

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