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I wish to know if $$\dfrac{\mathbb Z_2[x_1,x_2,x_3,x_4]}{\langle x_1x_3,x_2x_4,x_1+x_3,x_2+x_3+x_4\rangle}\cong\dfrac{\mathbb Z_2[a,b]}{\langle a^3,b^2,a^2-ab\rangle}.$$

Context - I was computing the cohomology ring of the Klein bottle with $\mathbb Z_2$ coefficients using the result of Davis and Januszkiewicz (Theorem 4.14 page 23) and obtained the first ring (LHS). The second ring (RHS) is what is known to be the cohomology ring of the Klein bottle. So they must be isomorphic. But I am unable to show that they are isomorphic.

I got as far as proving that the LHS is $\dfrac{\mathbb Z_2[x,y,z]}{\langle x^2,x+y+z,yz\rangle}$, but not sure how to go from there. Any help will be appreciated.

Thank you.

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Actually consider the ideal $I:=(b^2, a^2-ab)\in \mathbb{Z}_2[a,b]$. Then $$ a^3=(a+b)(a^2-ab)+ab^2\in I $$ Then the RHS ring is in fact $\mathbb{Z}_2/(b^2, a^2-ab)$. The LHS ring (simplified version) is $\mathbb{Z}_2[x,y,z]/(x^2, x+y+z, yz)\simeq \mathbb{Z}_2[x,y]/(x^2, y(x+y))$. Also since the base ring is $\mathbb{Z}_2$ you have $x^2-xy=x^2+xy$, and the two sides are the same.

Edit: You can take the homomorphism $\phi: \mathbb{Z}_2[x,y,z]/(x^2, x+y+z, yz)\to \mathbb{Z}_2[x,y]/(x^2, y^2+yx)$ given by $$ f(x,y,z)+(x^2, x+y+z, yz)\mapsto f(x,y,x+y)+(x^2, y^2+xy) $$ You can easily check that this is well-defined and surjective. For injectivity note that $$ f(x,y, z)+ (x^2, x+y+z, yz)= f(x,y, x+y)+ (x^2, x+y+z, yz) $$ Now if $f(x,y,x+y)\in (x^2, y^2+xy)$, then $$ f(x,y,x+y)=h_1(x,y)x^2+h_2(x,y)y(x+y)= h_1(x,y)x^2+h_2(x,y)y(x+y+z)+h_2(x,y)yz\in (x^2, x+y+z+yz) $$

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  • $\begingroup$ Could you elaborate a little on $\mathbb{Z}_2[x,y,z]/(x^2, x+y+z, yz)\simeq \mathbb{Z}_2[x,y]/(x^2, y(x+y))$? I don't see it easily. Thank you. $\endgroup$ – R_D Jun 25 '16 at 16:56
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    $\begingroup$ @Rise In the left hand side ring of your comment the coset of $z$ is the same as the coset of $x+y$. So it is not needed as a generator, as we can use $x+y$ in its place everywhere. $\endgroup$ – Jyrki Lahtonen Jun 25 '16 at 17:32

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