Polynomial Ideals and Primality (9.1.13)

Dummit and Foote Abstract Algebra, section 9.1, exercise 13:

Letting $F$ be a field, prove $F[x,y]/(y^2-x) \not ≅ F[x,y]/(y^2-x^2)$.

Proof: We show something a bit stronger, that $F[x,y]/(y^n-x) \not ≅ F[x,y]/(x^2-y^2)$ for any $n$, with $F$ only an integral domain.

First we show $y-x,y+x∉(y^2-x^2)=(y-x)(y+x)$ so that there are zero divisors in $F[x,y]/(y^2-x^2)$. Assuming the contrary we have for $f$ some polynomial $y-x=(y-x)(y+x)f⇒(y+x)f=1$, an impossibility. Similarly for $y+x∉(y^2-x^2)$.

Now we show $F[x,y]/(y^n-x)=F[y]/(y^n-x)≅F[y]$ is an integral domain, which will complete the proof. To see the first equality, notice that $\supseteq$ is clear and any polynomial of $F[x,y]/(y^n-x)$ can be represented by the image of a polynomial devoid of variables in $x$ since $\overline{y}^n=\overline{x}$. Letting $φ$ be the natural surjective homomorphism of $F[y]$ onto $F[y]/(y^n-x)$, we can observe that if $\text{ker}~φ≠0$ then $0≠(y^n-x)f∈F[y]$ for some $f∈F[x,y]$, an impossibility by $(y^n-x)f=y^nf-xf$ and observing the result of the $xf$ term by choosing the result of the multiplication monomial with maximal degree of $x$ in $f$, which cannot be affected by any terms in $xf$ (else the $f$ monomials' $y$ and $x$ degrees are the same and can be rewritten as one) nor in $y^nf$.$~\square$