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Let $k$ be any field. We consider the ideal $(x^2, xy, y^2)$ in $k[x,y]$. I need to prove that this ideal can not be generated by two elements.

I tried this by considering the contrapositive, but can't see anything useful. I guess some results from dimension theory will be required. Any help is highly appreciated.

user26857
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  • If you want a hands on calculation you can assume $I$ is generated by two elements $f,g$ with degrees $d_1, d_2$. Then by a few arguments you can filter what $d_1, d_2$ can be. Eventually you run out of possibilities. The process is ugly but works. – Hamed Sep 07 '16 at 05:35
  • I went that way but I cannot prove that $d_1, d_2 \leq 2$. –  Sep 07 '16 at 05:38
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    Well actually $d_1, d_2$ cannot be less than two since there are no linear polynomials in $I$. Next use the fact that $x^2, xy, y^2$ are linearly independent. Then you have to argue that as a result no two polynomials of degree two or higher can generate $I$ which rules out everything. – Hamed Sep 07 '16 at 05:43
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    Only two links out of many: http://math.stackexchange.com/questions/1251419/on-the-minimal-set-of-generators-of-monomial-ideals-in-mathbbcx-y?rq=1 and http://math.stackexchange.com/questions/1622081/ideal-of-mathbbcx-y-not-generated-by-two-elements?rq=1. – user26857 Sep 07 '16 at 06:32

1 Answers1

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Set $I=(x^2, xy, y^2)$ and $m=(x,y).$
If $I=(f,g)$, then k-vector, $I/mI$, space has dimension at most $2$. But $x^2+mI, xy+mI, y^2+mI$ are linearly independent in $I/mI$; since every element of $mI$ has degree at least $3.$

user 1
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