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Let $k$ be an algebraically closed field and $f$ be the polynomial $x_1x_2+x_2x_3+x_3x_1$ in $k[x_1, x_2, x_3]$. Here $f$ is irreducible.

Then this polynomial ring is not a $PID$, it is only an $UFD$. But I have to show that the ideal generated by $f$ is maximal in $k[x_1, x_2, x_3]$. Could anyone help me with this?

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  • $\begingroup$ It must be, because k is a field and it is the polynomial ring $\endgroup$ – Keith Oct 17 '14 at 6:28
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    $\begingroup$ I always thought that maximal ideals of $k[x_1,x_2,x_3]$ are of the form $(x_1-a_1,x_2-a_2,x_3-a_3)$. Nullstellensatz and all that. $\endgroup$ – Jyrki Lahtonen Oct 17 '14 at 6:37
  • $\begingroup$ Not maximal, but a prime nonetheless. $\endgroup$ – Orest Bucicovschi Oct 17 '14 at 6:52
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The ideal generated by $f$ is not maximal. It is properly contained in the maximal ideal $(x_1,x_2,x_3)$.

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Maybe the problem is showing that the ideal is prime. Since $k[x_1, x_2, x_3]$ is an UFD it's equivalent to $f$ is irreducible. Now $f$ is homogenous of degree $2$, a quadratic form of rank $3$ ( I stay away from $\text{char}\, 2$ but should work there too). Such a form cannot be written as a product of two linear forms since that would imply $\text{rank} \le 2$.

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