Why are the polynomial rings $k[x]$ and $k[x,y]/\langle y-x^2 \rangle$ isomorphic?

Question

This is related to an exercise in Hartshorne's book which people have previously asked about, but I feel that this particular point hasn't been answered in a very simple way.

Let $k$ be an algebraically closed field. Let $k[x]$ be the ring of polynomials of one variable over $k$ and let $k[x,y]$ be the ring of polynomials of two variables over $k$. How can we see the isomorphism $$ k[x] \simeq k[x,y]/\langle y-x^2 \rangle, $$ where $\langle y-x^2 \rangle$ is the ideal in $k[x,y]$ generated by $y-x^2$?

Answer 1

Define

$$\phi: k[x,y]\to k[x],\;x\to x,\;\;y\to x^2,\text{ and expand accordingly}$$

or if you prefer: $\phi f(x,y):=f(x,x^2)$. Now check stuff.

Answer 2

Hint: Let $D$ be a domain and $a \in D$. Then $D[y] \to D$ given by $f(y) \mapsto f(a)$ is a surjective ring homomorphism with kernel $\langle y-a \rangle$ because $f(y)=(y-a)q(y)+f(a)$. Apply this to $D=k[x]$.