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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.

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This is exercise 13.1 from Matsumura, Commutative Ring Theory. –  user26857 Jan 31 '13 at 23:04
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The usual way to prove this is to use Vandermonde determinants. To deal with all possible cases, one needs to have invertible Vandermode 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.

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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
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