How can I find the kernel of the following map? Suppose that $\varphi\colon F[x, y] \to F[t]$ is given by $\varphi(a(x, y))=a(t^{2},t)$. I want to show that $\ker(\varphi)=\langle y^{2}-x \rangle$.
I can check that $\varphi(y^{2}-x)=t^{2}-t^{2}=0$ so that $\langle y^{2}-x \rangle \subseteq \ker(\varphi)$. I'm not sure how can I show the reverse inclusion.
$F$ is a field.

I have considered taking any $a(x,y)=\sum_{i,j}a_{ij}x^{i}y^{j} \in \ker(\varphi)$. Then $\varphi(a(x,y))=a(t^{2},t)=\sum_{i,j}^{}a_{ij}t^{2i+j}=0$. I'm not sure if there's anything I can conclude about $a(x,y)$ from this.
 A: Let us abbreviate the ideal generated by $y^2 - x$ as $I$.
We already know that $I$ lies in the kernel of $φ$.
It follows that $φ$ descends to a homomorphism of rings
$$
  φ'
  \colon
  F[x, y] / I \to F[t] \,,
  \quad
  [a] \mapsto φ(a) \,.
$$
The kernel of $φ'$ is given by $\ker(φ) / I$, so we need to show that $φ'$ is injective.
But $φ'$ is surjective because $φ$ is surjective (a preimage for $a(t)$ is given by $a(y)$).
The desired injectivity of $φ'$ is therefore equivalent to its bijectivity.
We consider the homomorphism
$$
  ψ \colon F[t] \to F[x, y] \to F[x, y] / I \,,
  \quad
  a(t) \mapsto a(y) \mapsto [ a(y) ] \,,
$$
and claim that $ψ$ is inverse to $φ'$.
To prove this, we need to show the two equations $ψ ∘ φ' = \mathrm{id}$ and $φ' ∘ ψ = \mathrm{id}$.
Both $φ'$ and $ψ$ are $F$-linear ring homomorphisms (i.e., homomorphisms of $F$-algebras).
It therefore suffices to check these two equations on the elements $[x]$ and $[y]$ of $F[x, y] / I$, and the element $t$ of $F[t]$.
We finish by calculating
$$
  φ'(ψ(t)) = φ'([y]) = φ(y) = t
$$
as well as
$$
  ψ(φ'([x])) = ψ(φ(x)) = ψ(t^2) = ψ(t)^2 = [y]^2 = [y^2] = [x]
$$
and
$$
  ψ(φ'([y])) = ψ(φ(y)) = ψ(t) = [y] \,.
$$
