Isomorphism $k[x,y]/(y-x^2)$ onto $k[x]$ Perhaps this is very easy but I would like to get a proof of isomorphism $k[x,y]/(y-x^2) \cong k[x]$.
 A: Consider the ring of polynomials in two variables over a field $k$ which we call $k[x,y]$. Then what you can do now is you can consider $k[x,y]$ as $\big(k[x]\big)[y]$; you just simply collect powers of $y$ in a polynomial in $k[x,y]$.
Now you have the evaluation map $\varphi$ that sends $y$ to $x^2$ that is constant on the coefficients (in this case the coefficients are in $k[x]$) , so that $\varphi$ extends to a ring homomorphism from $k[x,y]$ to $k[x,x^2]$. Furthermore, we see that $k[x]$ is not different from $k[x,x^2]$ because $k[x]$ and $k[x,x^2]$ are respectively by definition the smallest rings containing $k$ and the set $\{x\}$, and $k$ and the set $\{x,x^2\}$. Therefore since you know now
$$\varphi: k[x,y] \longrightarrow\!\!\!\!\!\!\!\!\to k[x]$$
the first isomorphism tells you that $k[x,y]/\ker \varphi \cong k[x]$, where $\ker \varphi = (y - x^2).$
Exercise: Prove that $\ker \varphi = (y - x^2)$.
A: Can't this also be seen geometrically?  This may or may not be of interest to the OP, but I think it's worth posting. The only downside is that I think I need $k$ algebraically closed and of characteristic $0$. 
Recall that coordinate rings of affine varieties are isomorphic iff the varieties they cut out are. Next, notice that $k[x,y]/ \langle y-x^2 \rangle$ is the coordinate ring of the usual parabola centered at the origin in $k^2$ and $k[x] \simeq k[x]/\langle 0 \rangle \simeq k[x,y]/\langle y \rangle$ is  the coordinate ring of a copy of $k$. We can view this copy of $k$ as embedded into the plane, in which case the varieties are visually isomorphic, but the correspondence is given explicitly by $x \mapsto (x,x^2)$ or equivalently $(x,y^2) \mapsto x$, as others have noticed in different language. 
