# An ideal is homogenous iff it is invariant under a certain automorphism.

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

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.
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