We have an answer to "F2[X,Y](Y2+Y+1,X2+X+Y) and (F2[Y]}{(Y2+Y+1)}{)X(X2+X+Y―) are isomorphic"

Rex Maxwell

Answered question

2022-03-30

\frac{F_{2} [X, Y]}{(Y^{2} + Y + 1, X^{2} + X + Y)}

and

\frac{(F_{2} [Y]} {(Y^{2} + Y + 1)} \frac{{)}{X}}{(X^{2} + X + \overset{―}{Y})}

are isomorphic

Yaritza Phillips

Beginner2022-03-31Added 12 answers

Step 1
Let

ϕ : F_{2} [X, Y] \to (\frac{F_{2} [Y]}{(Y^{2} + Y + 1)}) \frac{[X]}{(X^{2} + X + \overset{―}{Y})}

be the morphism of rings which send X and

\overset{―}{X}

and Y to

\overset{―}{Y}

. It is surjective and, since

ϕ (Y^{2} + Y + 1) = ϕ (X^{2} + X + Y) = 0

it induces a surjective ring morphism :

\overset{―}{ϕ} : F_{2} \frac{X, Y}{Y^{2} + Y + 1, X^{2} + X + Y}

\to (F_{2} \frac{Y}{Y^{2} + Y + 1}) \frac{X}{X^{2} + X + \overset{―}{Y}}

Assume that

\overset{―}{ϕ} (P) = 0

Then

P (\overset{―}{X}, \overset{―}{Y}) = 0

Therefore:

P (X, \overset{―}{Y}) \in {(X^{2} + X + \overset{―}{Y})

and:

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

\frac{F_{2} [X, Y]}{(Y^{2} + Y + 1, X^{2} + X + Y)}

\overset{―}{ϕ}

is an isomorphism.

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