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Consider the ring $A=\mathbb{C}[x,y]/(y^2-x^4+5x-4)$, and consider the ideal $\mathfrak{m}=(y, x+2)$. Is $\mathfrak{m}$ maximal?

My sketched solution: consider an arbitrary element of $A$ and a representative $p(x,y)\in \mathbb{C}[x,y]$ for this element. Then $p(x,y)$ will be of the form $p(x,y)=y(\ldots)+g(x)$. Now by the Euclidean algorithm, there exist $q,r$ so that $g(x)=q(x)(x+2)+r(x)$ with $deg(r)<1$, i.e. $r(x)=c$ for some $c\in\mathbb{C}$. Summing up, $p(x,y)=y(\ldots)+(x+2)(\ldots)+c$ for some complex number $c$. Therefore there exist a surjective homorphism $\mathbb{C}\rightarrow A/\mathfrak{m}$, so $A/\mathfrak{m}\simeq\text{some quotient of $\mathbb{C}$}$. Clearly it must be $\mathbb{C}$ itself (since it's a field), so $A/\mathfrak{m}\simeq \mathbb{C}$ which implies that $\mathfrak{m}$ is maximal.

Is it a valid answer? Thanks in advance.

Edit: $A=\mathbb{C}[x,y]/(y^2-x^4+5x^2-4)$

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    $\begingroup$ "Clearly, it must be $\mathbb{C}$ itself": How do you prove that $A/\mathfrak{m}$ is not the zero ring? $\endgroup$
    – Stefan
    Aug 25, 2012 at 11:49
  • $\begingroup$ @tomasz: $p(x,y)$ is necessarily of this form: $g$ is just the sum of the terms of p(x,y) not containing y. Of the rest, you can factor out $y$. $\endgroup$
    – Stefan
    Aug 25, 2012 at 11:55
  • $\begingroup$ @StefanWalter: Oh. I thought the solution assumed the $(\ldots)$ to be $y$-free, my mistake. :) $\endgroup$
    – tomasz
    Aug 25, 2012 at 12:02

2 Answers 2

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As Stefan Walter pointed out, you need to show that the quotient is nonzero. To see that, notice that we have $$(y^2-x^4+5x^2-4,y,x+2)=(x^4-5x^2+4,y,x+2)=(y,x+2)$$ So $A/\mathfrak m\cong {\bf C}[x,y]/(x+2,y)\cong {\bf C}[x]/(x+2)\cong {\bf C}$. Since the quotient is a field, $\mathfrak m$ must be maximal.

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  • $\begingroup$ Oh tomasz, i'm sorry. There's a typo: it's not $(x^4-5x+4)$ it's $(x^4-5x^2+4)$ in the whole excercise $\endgroup$
    – random
    Aug 25, 2012 at 12:33
  • $\begingroup$ @random: I see. Will correct the answer. :) $\endgroup$
    – tomasz
    Aug 25, 2012 at 12:37
  • $\begingroup$ Ok, sorry again for the waste of time :) $\endgroup$
    – random
    Aug 25, 2012 at 12:39
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    $\begingroup$ @tomasz Alternatively by the Nullstellensatz you can conclude that that $(y,x+2)$ is maximal in $\Bbb{C}[x,y]$ since $\Bbb{C}$ is algebraically closed... $\endgroup$
    – user38268
    Aug 25, 2012 at 13:41
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Yes, since $\rm\ (y,x\!+\!2)\subset M\:$ we have $\rm\:mod\ M\!:\ x\equiv -2,\,y\equiv 0\:\Rightarrow\: p(x,y)\equiv p(-2,0),\,$ $\rm\forall\,p\in \Bbb C[x,y].$ Hence the natural map $\rm\:h:\Bbb C\to A\:$ is surjective. But, you must also show that $\rm\,h\,$ is injective. Suppose $\rm\:\Bbb c\in ker\, h,\:$ $\rm\,c\in \Bbb C.\:$ Then $\rm\:c = (x\!+\!2)\,f + y\, g + (y^2\!-\!x^4\!+\!5x^2\!-\!4)\,h\:$ for $\rm\:f,g,h\in \Bbb C[x,y].\:$ Therefore, evaluating at $\rm\:x=-2,\, y=0\:$ yields $\rm\:c = 0,\:$ so $\rm\:ker\, h = 0,\:$ so $\rm\:h\:$ is injective.

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