Is $\mathbb{C}[x,y]/(y)$ always isomorphic to $\mathbb{C}[x]$ I'm supposed to prove the algebras $\mathbb{C}[x,y]/(y)$ and $\mathbb{C}[x,y]/(y-x^2)$ are isomorphic.
Attempt:  $\mathbb{C}[x,y]/(y-x^2)$ is isomorphic to $\mathbb{C}[t]$ and $\mathbb{C}[x,y]/(y)\cong \mathbb{C}[x]$ so just map $t\mapsto x$. But I'm not sure if $\mathbb{C}[x,y]/(y)\cong \mathbb{C}[x]$ or how to show that
A hint was given which I don't think I used at all: treat the rings as functions on algebraic sets and build a function that induces an isomorphism.
 A: Consider the map $\mathbb{C}[x,y]\rightarrow \mathbb{C}[x]$ given by $f(x,y)\mapsto f(x,0)$.
See that this is a ring homomorphism.
Compute its kernel and image.
A: always $\Bbb C[x,y]/(y-x^2) \cong \Bbb C[x,x^2]\cong \Bbb C[x] $ and $\Bbb C[x,y]/(y)\cong (\Bbb C[x])[y]/(y)\cong \Bbb C[x] $
A: Although the answer by cello is probably the easiest, here is one explicitly using the hint.
Consider the algebraic varieties
$$V(y) = \{(x,y) \in {\Bbb C}^2 \mid y = 0 \} = \{ (x,0) \mid x \in {\Bbb C}\}$$
and
$$V(y-x^2) = \{(x,y) \in {\Bbb C}^2 \mid y = x^2 \} = \{ (x,x^2) \mid x \in {\Bbb C}\}.$$
Your two algebras are the coordinate rings of these two varieties and if the varieties are isomorphic, so are their coordinate rings.
Now the projection onto the $x$-axis induces an isomorphism from $V(y-x^2)$ to $V(y)$. Explicitly: $(x,y) = (x,0) \ni V(y) \mapsto (x,x^2) \in V(y-x^2)$ and $(x,y) = (x,x^2) \ni V(y-x^2) \mapsto (x,0) \in V(y)$.
Note that this also gives the isomorphisms of the coordinate rings: from $\Bbb C[x,y]/(y)$ to $\Bbb C[x,y]/(y-x^2)$ by sending (the residue class of) $x$ to (the residue class of $x$) and (the residue class of) $y$ to $0$; its inverse sends $x$ to $x$ and $y$ to $x^2$.
