Show the ring is isomorphic to $k[x]$ 
Let $k$ be a field. Then, $k[x,y]/(y-x^2)\simeq k[x]$

My attempt is by defining $\phi:k[x,y]\to k[x]$ by $y\mapsto x^2$ and $a\mapsto a$ for $a\in k$ and $x\mapsto x$. Then, $\phi(f(x,y)+g(x,y)) = f(x,x^2)+g(x,x^2) = \phi(f(x,y))+\phi(g(x,y))$ and $\phi(f(x,y)g(x,y)) = f(x,x^2)g(x,x^2) = \phi(f(x,y))\phi(x,y)$ and $\phi(1) = 1$ so it
's a ring homomorphism. Since $\text{ker}(\phi) = (y-x^2)$, by first isomorphism theorem, $k[x,y]/(y-x^2)\simeq k[x]$
Could you check if this proof is ok? I'm a beginner of algebra
 A: Here's the part about the kernel.
First recall that if $A$ is a domain, $A[y]$ is the polynomial ring in the indeterminate $y$, and $f \in A[y]$ is a monic polynomial, then you can do Euclidean division by $f$ in $A[y]$.
Apply this to $A = k[x]$, and $f = y - x^{2}$. If $g \in \ker(\phi)$, divide $g$ by $f$. Since $f$ has degree $1$ in $y$, the remainder will be a constant $r \in A$, that is, an element of $k[x]$.
So you have
$$
g = q f + r,
$$
for some $q \in k[x, y], r \in k[x]$. Now apply $\phi$ to this, and infer that $r = 0$.
A: You have already been told how to complete your proof, but I would like to add an alternative solution which gives you the same isomorphism you found (in an easier way, I would say) and shows that the same result holds for any commutative ring.
Let us recall the following well known fact:

If $A$ is a commutative ring, then $A[X]/(X-a) \cong A$ for any $a\in A$.

You may prove this by considering the homomorphism $\phi: A[X] \to A$, $\phi(f)=f(a)$ and applying the fundamental isomorphism theorem (do you notice the similarity to your solution?).
Now, in the same context, we know that $A[X, Y]$ is defined as $A[X][Y]$ (the polynomials in $Y$ with coefficients in $A[X]$). So, if we pick $X^2\in A[X]$, the fact from above tells us that $A[X, Y]/(Y-X^2) \cong A[X]$, which is precisely what you wanted to show for fields.
Note that my isomorphism is the same as yours, but I didn't have to think too much about how to apply Euclidean division, because I applied it in the "standard" case, which is a ring with only one indeterminate.
