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I was trying to find all prime and maximal ideals of ring $R=\mathbb{Z}[x,y]/\langle 6, (x-2)^2, y^6\rangle$.

By correspondence theorem, we know the prime (maximal) ideals of ring $R$ has 1-1 correspondence with prime (maximal) ideals of ring $\mathbb{Z}[x,y]$ which contains the ideal $I=\langle 6, (x-2)^2, y^6\rangle$.

My questions are

(1) For example, ideal $\langle 3, x-2, y\rangle$ is an ideal of $\mathbb{Z}[x,y]$ and it contains $I$. If it is a prime ideal, then $\mathbb{Z}[x,y]/\langle 3, x-2, y\rangle$ must be an integral domain. Is there an easy way to check whether $\mathbb{Z}[x,y]/\langle 3, x-2, y\rangle$ is an integral domain? could we just do it by definition?

(2) It looks like to me that there are many ideals of $\mathbb{Z}[x,y]$ containing $I$, for example, consider $J=\langle 2, x-2, y, p(x,y)\rangle$, where $p(x,y)$ is any irreducible polynomial (other than $x-2$ and $y$ of course). So is there a way to find (or characterize) all the prime ideals of $\mathbb{Z}[x,y]$ containing $I$?

Thanks for any help!

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If $P$ is a prime ideal and $xy\in P$, then either $x\in P$ or $y\in P$, and so by induction, if $x^n\in P$, then $x\in P$. Therefore, any prime ideal containing $I$ actually contains $J=\langle 6, x-2, y \rangle$, and actually must contain one (and only one) of $J_2=\langle 2, x-2, y \rangle$ or $J_3=\langle 3, x-2, y \rangle$ (it cannot contain both, as an ideal which contains both $2$ and $3$ is trivial, and at least some definitions of prime ideal require the ideal to be proper). However, both of these ideals are maximal, as their quotients are the fields with $2$ or $3$ elements, respectively.

To see the isomorphism, we note the following fact: If $R$ is a ring, then $R[x]/(x-a)\cong R$. This follows from the first isomorphism theorem by taking the map $R[x]\to R$ sending $x$ to $a$. A slight extension of this yields that $\mathbb Z[x,y]/\langle 6, x-2, y \rangle\cong \mathbb Z[x]/\langle 6, x-2\rangle \cong \mathbb Z/\langle 6\rangle$. Actually, this gives us another perspective, as we can phrase the original problem as quotients of $\mathbb Z/\langle 6\rangle$ which are domains/fields.

For your second question, note that if an ideal contains both $y$ and $p(x,y)$ as generators, then you can replace $p(x,y)$ with $p(x,0)$ (which is the remainder upon dividing $p$ by $y$). Doing this first with $y$, then $x-2$, you have replaced $p$ with a constant polynomial.

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  • $\begingroup$ It is perhaps worth noting that $J$ is the "radical" of $I$, written $J=\sqrt{I}$, namely the collection of $z$ such that $z^n\in I$ for some $n$. It is not always the case that you can find the radical by taking the roots of the perfect-power generators, but doing so will always give you something contained in the radical, and since the radical of $I$ is always contained in every prime ideal containing $I$ (in fact, it is the intersection of all such ideals), it is a useful first step to try to pass to the radical before beginning your search. $\endgroup$ – Aaron Mar 10 '16 at 19:59
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For (1), I don't think there are much better ways than to just compute the quotient. In this case ($\langle 3,x-2,y\rangle$) it's not too hard : you mod out by $y$, so the quotient is $\mathbb{Z}[x]/\langle 3,x-2\rangle$ ; then you mod out by $3$, so you get $\mathbb{F}_3[x]/\langle x-2\rangle$ ; but $\mathbb{F}_3$ is a field, and $x-2$ is of degree $1$, so you get $\mathbb{F}_3$, and you ideal is actually maximal.

For (2), beware : $J = \langle 2,x-2,y\rangle$ is already maximal, because $\mathbb{Z}[x]/J = \mathbb{F}_2[x,y]/\langle x-2,y\rangle = \mathbb{F}_2[x]/\langle x\rangle = \mathbb{F}_2$. So adding any $p(x,y)$ does not change anything.

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  • $\begingroup$ If one adds the polynomial $p=x+1$ I can see some changes, don't you think? $\endgroup$ – user26857 Mar 10 '16 at 22:21
  • $\begingroup$ Sure, because then the ideal becomes the unit ideal. What I meant was that you can't get a new prime ideal by adding anything (since it's maximal). $\endgroup$ – Captain Lama Mar 10 '16 at 22:41
  • $\begingroup$ Thanks Captain Lama. So $\langle 2, x-2, y, x+1\rangle$ is actually the whole ring $\mathbb{Z}[x,y]$, because $J$ is already maximal and clearly $x+1$ is not in $J$. Is this correct? $\endgroup$ – mori39 Mar 11 '16 at 2:27
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    $\begingroup$ Yes. You can also say that since $x=0$ in the quotient, then $x-1=1$ so this ideal contains $1$. $\endgroup$ – Captain Lama Mar 11 '16 at 9:44

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