Expert answers to "F2[X,Y](Y2+Y+1,X2+X+Y)and(F2[Y](Y2+Y+1))[X](X2+X+Y¯)are isomorphic" - Plainmath

College
Algebra
Abstract algebra

Benjamin Hampton

Answered question

2022-05-02

$\frac{F_{2} [X, Y]}{(Y^{2} + Y + 1, X^{2} + X + Y)}$ and $\frac{(F_{2} [Y]}{(Y^{2} + Y + 1)} \frac{) [X]}{(X^{2} + X + \bar{Y})}$ are isomorphic

Answer & Explanation

narratz5dz

Beginner2022-05-03Added 13 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)

so

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

So

\overset{―}{ϕ}

is an isomorphism.

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