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This is taken from an exercise problem in Dummit and Foote's:

Describe briefly the ring structure of $\mathbb Z[x,y]/(x^2,y^2,2)$ and show that $\alpha^2=1$ or $0$ for every $\alpha$ in the ring.

I proceeded as follows:

Let $\rm S:=\mathbb Z[x,y]/\rm {(x^2,y^2,2)}$.

$x^2\in (x^2,y^2,2):=I\implies x^{r+2}\in I\quad\forall r\in \mathbb Z^+$ and similarly for higher powers of $y$.

Let $\rm \overline{f(x,y)}=\rm \overline{a_n(x)y^n+a_{n-1}(x)y^{n-1}+\cdots+a_1(x)y+a_0(x)}\in S$

$\rm \overline{f(x,y)}=\rm {f(x,y)+I}=\rm \color{blue}{a_0(x)}+\rm \color{green}{a_1(x)y}+I=\rm \color{blue}{a+bx}+\rm \color{green}{(c+dx)y}+I\in\rm \{\overline{p+qx+ry+sxy}:p,q,r,s\in \mathbb Z_2\}=:\rm T$

This is because $\rm I$ absorbs higher powers of $\rm x$ and $\rm y$.

It follows that $\rm S\subset \rm T$.

The other direction $\rm T\subset \rm S$ is obvious. It follows that $\rm T=\rm S$.

Let $\rm {\alpha= p+qx+ry+sxy+I\in S\alpha^2=(p+qx)^2+(ry+sxy)^2+2(p+qx)(rs+sxy)+I=p^2+I}$.

$\rm {\therefore\quad \alpha^2=1\iff p=1\pmod 2}$ and $\rm {\alpha^2=0\iff p=0 \pmod 2}$.

Is this correct? Thanks.

I tried to do it using the third isomorphism theorem also as follows:

$\rm {S\simeq\frac{\mathbb Z[x,y]/{(2,x)}}{(x^2,y^2,2)/{(2,x)}}}$. I know that the numerator is isomorphic to $\mathbb Z_2[y]$ and this is because the homomorphism (from $\rm \mathbb Z[x,y]$ to $\rm \mathbb Z_2[y]$)) $\rm {x\mapsto 0, z\mapsto z\pmod 2 \quad \forall z\in \mathbb Z, y\mapsto y}$ has kernel $\rm {(2,x).(2,x)\subset \ker}$ is obvious. For the other direction, let $\rm {g(x,y)=b_m(x)y^m+b_{m-1}(x)y^{m-1}+\cdots+b_1(x)y+b_0(x)\in \ker}$, it follows by applying the homomorphism on both side that $\rm {0=(b_m(0)\pmod 2)y^m+(b_{m-1}(0)\pmod 2)y^{m-1}+\cdots+(b_1(0)\pmod 2)y+b_0(0)\pmod 2}$

It follows that $b_r(0)=0$ for every $r\in \{0,1,...,m\}$. Hence $x|f(x,y)$ by remainder theorem. It follows that $f(x,y)\in (x)\subset (2,x)\implies \ker\subset (2,x)$.

Using the same map as above, it follows that $\rm {(x^2,y^2,2)/(2,x)}$ is isomorphic to $\rm {(y^2)}$ (an ideal of $\rm \mathbb Z_2[y]$). I am stuck here in concluding formally that ${\rm S\simeq \frac{\mathbb Z_2[y]}{(y^2)}}$.

Koro
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    See the solution to Problem set $3$ here. Your case goes the same way. – Dietrich Burde Apr 19 '22 at 09:32
  • @DietrichBurde: Thank you so much for the link. I've been looking for some solved problems of this type :). – Koro Apr 19 '22 at 09:49
  • @DietrichBurde Can you please explain went wrong in the following argument: $/(+)≅//(+)/$ and it can also be shown that if is a surjective homomorphism from onto then $/(ker)/(ker+)/(ker)≅/()$. In OP, I define a map $$ from $[,]$ to $_2[]$ as follows $()=(mod2)$ for every $∈$ and $()=0,()=$ and I believe that $\ker\tau=(2,)$ so by the theorem stated earlier, I should get $_2[]/(^2)$, which is different from what is obtained by the way in the linked answer. Thanks. – Koro Apr 20 '22 at 18:23
  • @DietrichBurde: I think I have understood now the error I had made in using the third isomorphism theorem and have posted an answer below. – Koro Apr 21 '22 at 12:13

2 Answers2

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Looks like $\Bbb Z_2/\langle x^2,y^2\rangle$, whose elements are $a+bx+cy+dxy$ where $a,b,c,d\in\Bbb Z_2$.

Wuestenfux
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  • Thanks for the response. Is the assertion that $S$ is isomorphic to $\mathbb Z_2[y]/(y^2)$ correct? – Koro Apr 20 '22 at 07:35
  • The latter ring has elements $a+by$ where $a,b\in\Bbb Z_2$. – Wuestenfux Apr 20 '22 at 07:48
  • Yes,I understood how $a+bx+cy+dxy$ came into play. I want to know what went wrong in my usage of the third isomorphism theorem. $R/(I+J)\cong R/I/(I+J)/I$ and it can also be shown that if $\phi$ is a surjective homomorphism from $R$ onto $S$ then $\frac{R/({\ker \phi})}{(\ker \phi+I)/(\ker\phi)}\cong \frac{S}{\phi(I)}$. In OP, I define a map $\tau$ from $Z[x,y]$ to $Z_2[y]$ as follows $\tau(z)=z\pmod 2$ for every $z\in Z$ and $\tau(x)=0,\tau(y)=y$ and I believe that $\ker \tau=(2,x)$ so by the theorem stated earlier, I should get $Z_2[y]/(y^2)$. Can you please help me understand that? Thanks. – Koro Apr 20 '22 at 08:07
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    I think I have understood now the error I had made in using the third isomorphism theorem and have posted an answer below. – Koro Apr 21 '22 at 12:13
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The error in the usage of third isomorphism in OP is that $\rm(2,x)$ is not a subset of $\rm(x^2,y^2,2)$ so $\rm(x^2,y^2,2)/(2,x)$ makes no sense.

Setting $\rm R=\mathbb Z[x,y]$ for brevity, $\rm R/(x^2,y^2,2)=R/((x^2,y^2)+(2))\cong \frac{R/(2)}{((x^2,y^2)+(2))/(2)}$.

The homomorphism $\rm \phi:\mathbb Z[x,y]\to \mathbb Z_2[x,y]$ takes polynomials in $\rm \mathbb Z[x,y]$ and reduces their coefficients modulo $\rm 2$ and is surjective and has kernel $\rm (2)$. It follows by this version of the third isomorphism theorem that $\rm \frac{R/(2)}{((x^2,y^2)+(2))/(2)}\cong \frac{\mathbb Z_2[x,y]}{(x^2,y^2)}$, which is in line with the other answer posted and $\rm T$ in OP as well.

Koro
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