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I'm trying to prove this ideal:

$$(x^2+y^2+z^2+x+y+z,\ x^5+y^5+z^5+2(x+y+z),\ x^7+y^7+z^7+3(x+y+z))\subset \mathbb C[x,y,z]$$

can't be maximal.

In order to do so, I'm using the Nullstellensatz theorem and showing this ideal is not of this form: $(x-a_1,y-a_2,z-a_3)$, where $a_i\in \mathbb C$.

The problem I don't how to do this.

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    $\begingroup$ What's $3(x,y,z)$? Is it a typo for $3(x+y+z)$? $\endgroup$ – egreg Jun 10 '14 at 22:21
  • $\begingroup$ Note : Even in English, this theorem is usually called with its German name, the Nullstellensatz. (Pronounce it "Nullchtellenzats" with an english pronounciation, that's as close you'll get...) $\endgroup$ – Patrick Da Silva Jun 10 '14 at 22:23
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    $\begingroup$ Let's call your ideal $I$. As a first step, can you specifically identify a maximal ideal that contains $I$? $\endgroup$ – John M Jun 10 '14 at 22:23
  • $\begingroup$ @egreg I'm sorry, I'm going to edit it $\endgroup$ – user85493 Jun 10 '14 at 22:38
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    $\begingroup$ So now can you show strict containment $(x,y,z) \supsetneq I$? So for example, if you could somehow prove that $x \notin I$? $\endgroup$ – John M Jun 10 '14 at 22:53
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Let $J=(x,y,z)$. We have $I+(x^2,y^2,z^2)=(x+y+z,x^2,y^2,z^2)\subsetneq J$ and then $I$ is not maximal because clearly $I\subseteq J$.

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  • $\begingroup$ Why we don't have $(x+y+z,x^2,y^2,z^2)=J$? $\endgroup$ – user85493 Jun 15 '14 at 10:07
  • $\begingroup$ For example, the elements of that ideal which are homogeneous of degree $1$ together with zero form a $1$-dimensional vector space. On the othe thand, the elements of $J$ which are homogeneous and of degreee $1$, together with zero, for a $3$-dimensional vector space. $\endgroup$ – Mariano Suárez-Álvarez Jun 15 '14 at 10:31
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To show your ideal is not maximal, I'll show that the set of common zeros of your ideal contains more than one point. Taking advantage of the symmetry of your ideal, we will look for solutions that even satisfy the extra condition $x+y+z=0.$ Then (using Newton's Identities and remember $e_1=x+y+z=0$) the first equation is $$x^2+y^2+z^2+x+y+z = p_2 + e_1 = p_2 = e_1^2 - 2e_2.$$

The next one is $$ x^5+y^5+z^5 + 2(x+y+z) = p_5 + 2e_1 = -5e_3e_2 $$ and the third is

$$ x^7+y^7+z^7 + 3(x+y+z) = p_7 + 3e_1 = c e_3 e_2^2$$ where $c$ is a non-zero constant.

This system is satisfied if we further impose $e_2=0.$ Hence, any common solutions of $e_1=x+y+z=0$ and $e_2 = xy+yz+xz=0$ are in the set of common zeros of your ideal. It is easy to find solutions to this that aren't the trivial $(0,0,0)$ solution. For example, pick $x=1,$ so $y+z=-1$ and $y+yz+z =0,$ which easily solves for $y= \dfrac{-1}{2} - i \dfrac{\sqrt{3}}{2} \ , \ z = \dfrac{-1}{2} + i \dfrac{\sqrt{3}}{2}.$

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  • $\begingroup$ Interesting approach. $\endgroup$ – John M Jun 11 '14 at 2:31
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Perhaps this works?

For any $f(x,y,z) \in \mathbb{C}[x,y,z]$, it seems clear that $$f(x,y,z)(x+y+z) \not\equiv x (\bmod \langle x^2,y^2,z^2 \rangle).$$ Therefore, $$x \notin \langle x^2,y^2,z^2,x+y+z \rangle.$$

So we have $$\langle x,y,z \rangle \supsetneq \langle x^2, y^2, z^2, x+y+z \rangle \supseteq I,$$ where $I$ is the ideal given by the problem statement.

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