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I am reading some course notes and running into some problem here. Suppose, $R=k[x,y,z]$ where $k$ is a field and let $f\in R$. Consider the automorphisms of $R$ given by $F_{u,v,w}$ where $F_{u,v,w}(x)=ux$, $F_{u,v,w}(y)=vy$ and $F_{u,v,w}(z)=wz$ with $u,v,w \in k$.

Given that $f$ has at least two terms, the author says "because $k$ is algebraically closed", $f$ is not a multiple of $F_{u,v,w}(f)$ for some $F_{u,v,w}$.

I don't understand why algebraic closure is needed for this. For since there are at least two terms, there is at least one indeterminate that appears in different degree in at least two terms, say $x$. Then shouldn't $F_{2,1,1}$ do the trick?

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So how are v and w relevant? –  William D. Nov 5 '11 at 15:41
@William: That's basically part of my question. According to the author, there exists some $u,v,w$ for which this is true when $k$ is algebraically closed, and to me it seems we can choose them this way. –  Guy Stiles Nov 5 '11 at 15:53
Are you sure $F_{u,v,w}$ doesn't send $y$ to $vy$ and $z$ to $wz$? Also, if the field $k$ is $\mathbb{F}_2$, the property is not true so you do need some hypothesis on $k$. –  YBL Nov 5 '11 at 16:24
@YBL: Yeah, sorry, that's what I meant to write, but I copy pasted the first thing and planned on editing it for the next two, but forgot to do it. Yeah, I see your point. It still seems a bit strong to assume algebraic closure, but in the light of your example, I don't see a way out. –  Guy Stiles Nov 5 '11 at 16:30

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