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let $k$ be algebraic close field , why following rings are isomorphic ?

$k[x, y] /\left(y-x^{2}\right) \cong k\left[x, x^{2}\right] \cong k[x]$

i need theorem for general case for example when $k[x, y] /\left(f(x,y)\right) \cong k[x]$ ? for what $f(x,y)$ this is true ?

amir bahadory
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  • The general case is likely to be quite difficult. Taking specs, we see that this is the case if and only if $C = V(f(x,y))$ is isomorphic to $\mathbb{A}^1$. Hence we get a way of showing when this is not the case: if $C$ not smooth or if $C$ is smooth and $g(C) = (d-1)(d-2)/2 \neq 0$ where $d$ is the degree of $f$. – Mummy the turkey Jan 24 '21 at 18:07

1 Answers1

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There is an isomorphism $f:k[x]\to k[x,y]/(y-x^2)$ sending $x\mapsto x$. The inverse homomorphism is $g:k[x,y]/(y-x^2)\to k[x]$ sending $x\mapsto x$ and $y\mapsto x^2$.

Note that this does not require $k$ to be algebraically closed.

Kenta S
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