Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm working on the following.

Let $R=R_0+R_1+ \cdots $ be a graded ring and $u$ a unit of $R_0$. Then the map $T_u$ defined by $T_u(x_0+x_1+ \cdots +x_n) = x_0+x_1u+ \cdots + x_n u^n$ is an automorphism of $R$ (this is clear). If $R_0$ contains an infinite field $k$, then an ideal $I$ of $R$ is homogenous iff $T_\alpha(I) = I$ for every $\alpha \in K^{\times}$.

I see that it fails for non-infinite fields, but I can't see what property to use of infinite fields to make this work. I have been thinking of maybe viewing I as a vector space over $k$, or using prime avoidance of some sort but it doesn't seem to do the trick. Any help would be most welcome.

share|cite|improve this question
This is exercise 13.1 from Matsumura, Commutative Ring Theory. – user26857 Jan 31 '13 at 23:04
up vote 5 down vote accepted

The usual way to prove this is to use Vandermonde determinants. To deal with all possible cases, one needs to have invertible Vandermonde determinants of all sizes, and for that one needs an infinite field.

By the way, this has nothing to do with the fact that $R$ is a ring or $I$ an ideal. It is in fact true that if $V=\bigoplus_{n\geq0}V_n$ is a graded vector space on which you have all those endomorphisms too, and $W\subseteq V$ is a subspace which is invariant, then $W$ is a graded subspace, that is, it is generated by the homogeneous elements it contains.

share|cite|improve this answer
Hmm, I don't think I have ever seen Vandermonde determinants used in arguments like this, but I'll try it. Thanks! – Dedalus Sep 29 '12 at 15:41
Try for a couple of days: if it does not work, let me know and I'll add the complete argument :-) – Mariano Suárez-Alvarez Sep 29 '12 at 15:47
I will! Thanks. – Dedalus Sep 29 '12 at 15:51
Think I got it: Let $f \in I$, and express f as a sum of homogenous terms, $f=f_1 + \cdots f_n$ say. It will be enough to show that each $f_i$ is in I. Let us choose n linearly independent values $\alpha_i \ in K$. Then we can multiply the "vector form" of f with V, and see this as a system of equations. We can solve it to obtain an expression of each $f_i$ in terms of elements in I. It's a sketch, but I think it should be the correct idea :) Thank you! – Dedalus Sep 29 '12 at 16:14
Indeed, that's it. – Mariano Suárez-Alvarez Sep 29 '12 at 16:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.