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

Benjamin Hampton
2022-05-02
Answered

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

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narratz5dz

Answered 2022-05-03
Author has **13** answers

Step 1

Let$\varphi :{\mathbb{F}}_{2}[X,Y]\to \left(\frac{{\mathbb{F}}_{2}\left[Y\right]}{({Y}^{2}+Y+1)}\right)\frac{\left[X\right]}{({X}^{2}+X+\stackrel{\u2015}{Y})}$

be the morphism of rings which send X and$\stackrel{\u2015}{X}$ and Y to $\stackrel{\u2015}{Y}$ . It is surjective and, since $\varphi ({Y}^{2}+Y+1)=\varphi ({X}^{2}+X+Y)=0$ it induces a surjective ring morphism :

$\stackrel{\u2015}{\varphi}:{\mathbb{F}}_{2}\frac{X,Y}{{Y}^{2}+Y+1,{X}^{2}+X+Y}$

$\to \left({\mathbb{F}}_{2}\frac{Y}{{Y}^{2}+Y+1}\right)\frac{X}{{X}^{2}+X+\stackrel{\u2015}{Y}}$

Assume that$\stackrel{\u2015}{\varphi}\left(P\right)=0$ Then $P(\stackrel{\u2015}{X},\stackrel{\u2015}{Y})=0$ Therefore:

$P(X,\stackrel{\u2015}{Y})\in \{({X}^{2}+X+\stackrel{\u2015}{Y})$

and:

$P(X,Y)\in ({X}^{2}+X+Y,{Y}^{2}+Y+1)$

so$\frac{{\mathbb{F}}_{2}[X,Y]}{({Y}^{2}+Y+1,{X}^{2}+X+Y)}$

So$\stackrel{\u2015}{\varphi}$ is an isomorphism.

Let

be the morphism of rings which send X and

Assume that

and:

so

So

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