 # \frac{\mathbb{F}_2[X,Y]}{(Y^2 + Y + 1,X^2 + X + Y )} Benjamin Hampton 2022-05-02 Answered

$\frac{{\mathbb{F}}_{2}\left[X,Y\right]}{\left({Y}^{2}+Y+1,{X}^{2}+X+Y\right)}$ and $\frac{\left({\mathbb{F}}_{2}\left[Y\right]}{\left({Y}^{2}+Y+1\right)}\frac{\right)\left[X\right]}{\left({X}^{2}+X+\overline{Y}\right)}$ are isomorphic

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Step 1
Let $\varphi :{\mathbb{F}}_{2}\left[X,Y\right]\to \left(\frac{{\mathbb{F}}_{2}\left[Y\right]}{\left({Y}^{2}+Y+1\right)}\right)\frac{\left[X\right]}{\left({X}^{2}+X+\stackrel{―}{Y}\right)}$
be the morphism of rings which send X and $\stackrel{―}{X}$ and Y to $\stackrel{―}{Y}$. It is surjective and, since $\varphi \left({Y}^{2}+Y+1\right)=\varphi \left({X}^{2}+X+Y\right)=0$ it induces a surjective ring morphism :
$\stackrel{―}{\varphi }:{\mathbb{F}}_{2}\frac{X,Y}{{Y}^{2}+Y+1,{X}^{2}+X+Y}$
$\to \left({\mathbb{F}}_{2}\frac{Y}{{Y}^{2}+Y+1}\right)\frac{X}{{X}^{2}+X+\stackrel{―}{Y}}$
Assume that $\stackrel{―}{\varphi }\left(P\right)=0$ Then $P\left(\stackrel{―}{X},\stackrel{―}{Y}\right)=0$ Therefore:
$P\left(X,\stackrel{―}{Y}\right)\in \left\{\left({X}^{2}+X+\stackrel{―}{Y}\right)$
and:
$P\left(X,Y\right)\in \left({X}^{2}+X+Y,{Y}^{2}+Y+1\right)$
so $\frac{{\mathbb{F}}_{2}\left[X,Y\right]}{\left({Y}^{2}+Y+1,{X}^{2}+X+Y\right)}$
So $\stackrel{―}{\varphi }$ is an isomorphism.