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A is generated by all polynomials $g(x)$ such that $g(0)=g(1)$. In particular it contains $x^2-x$ and $x^3-x^2$, and those are algebra generators. They satisfy $a^3+ab-b^2=0$, so $A \cong F[a,b]/(a^3+ab-b^2)$F[a,b]/(a^3+ab-b^2)$.] If you define your map $f$ by $f(X) = (1,-1)$ instead, you get $a^3+a^2-b^2$ on the nose. |
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A is generated by all polynomials $g(x)$ such that $g(0)=g(1)$. In particular it contains $x^2-x$ and $x^3-x^2$, and those are algebra generators. They satisfy $a^3+ab-b^2=0$, so $A \cong F[a,b]/(a^3+ab-b^2)$. |
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