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Let $F$ be a field. Apparently we know that $\langle x,y \rangle \neq \langle g(x,y) \rangle$ for any $g \in F[x,y]$. Why is this the case?

user26857
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dumbquestiondude
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1 Answers1

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Assume $x\in \langle g(x,y)\rangle$. That means that $x=f(x,y)\cdot g(x,y)$. Since the polynomial $x$ is irreducible in $F[x,y]$, we don't have much options left: either $g(x,y)=c$ or $g(x,y)=cx$ for some constant $c\in F$. In the former case $\langle g(x,y)\rangle=F[x,y]$, in the latter case $\langle g(x,y)\rangle=\langle x\rangle$.

Berci
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  • Perfect. This is more elegant than what I had in mind (choosing a generator of lowest multi-degree and working towards a contradiction). – dumbquestiondude Oct 20 '14 at 21:44