\frac{\mathbb{F}_2[X,Y]}{(Y^2 + Y + 1,X^2 + X + Y )}

Benjamin Hampton 2022-05-02 Answered

F2[X,Y](Y2+Y+1,X2+X+Y) and (F2[Y](Y2+Y+1))[X](X2+X+Y¯) are isomorphic

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Answers (1)

narratz5dz
Answered 2022-05-03 Author has 13 answers
Step 1
Let ϕ:F2[X,Y](F2[Y](Y2+Y+1))[X](X2+X+Y)
be the morphism of rings which send X and X and Y to Y. It is surjective and, since ϕ(Y2+Y+1)=ϕ(X2+X+Y)=0 it induces a surjective ring morphism :
ϕ:F2X,YY2+Y+1,X2+X+Y
(F2YY2+Y+1)XX2+X+Y
Assume that ϕ(P)=0 Then P(X,Y)=0 Therefore:
P(X,Y){(X2+X+Y)
and:
P(X,Y)(X2+X+Y,Y2+Y+1)
so F2[X,Y](Y2+Y+1,X2+X+Y)
So ϕ is an isomorphism.
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