Polynomial rings and quotients of homogeneous ideals

Question

I fear that this is a stupid question, but I want to have a go anyway.

Let $k$ be a field, and let $f(x,y)$ be an irreducible homogeneous quadratic polynomial in $k[x,y]$.

Question: (when) is $k[x,y]/(f(x,y)) \cong (k[x]/f(x,1))[y]$ ?

Probably I am seeing ghosts, but is there some more general (correct) identity that I am totally missing ? Can the assumptions on $f(x,y)$ be relaxed ?

Answer 1

In characteristic not $2$, when ${b^2-4c}$ is not a square of $k$ (which is impliçed by the irreducibility) then $$k[x,y]/(x^2+byx+cy^2) \cong k[y,\frac{-b+\sqrt{b^2-4c}}{2}y]$$ Which is a subring of the polynomial ring $$k[\sqrt{b^2-4c}][y]$$ As the former doesn't contain $\frac{-b+\sqrt{b^2-4c}}{2}$.

$k[y,\frac{-b+\sqrt{b^2-4c}}{2}y]$,$k[\sqrt{b^2-4c}][y]$ are two integral domains and they have the same fraction field $k[\sqrt{b^2-4c}](y)$.

Answer 2

Let $f \in k[x,y]$ be a homogeneous polynomial. There is a natural injective map $$ k[x,y]/(f(x,y)) \hookrightarrow (k[u]/f(u,1))[v] : x \mapsto uv, y \mapsto v. $$ However, if $\deg f > 1$, then $u$ is visibly not in the image of this map.

Note that $A := k[x,y]/(f(x,y))$ has a grading induced by the usual grading on $k[x,y]$ since $f$ is homogeneous. Hence if an element of $A$ satisfies a polynomial equation over $k$, then it must be an element of degree $0$, i.e. already in $k$. In particular the equation $f(T,1) = 0$ has no root in $A$ if $f$ is irreducible over $k$ but it does have a root (namely $u$) in $(k[u]/f(u,1))[v]$. In conclusion if $f$ is irreducible and homogeneous of degree $> 1$, then $k[x,y]/(f(x,y))$ is not isomorphic to $(k[u]/f(u,1))[v]$.