Is $k[x,y]=k(y)[x]$, where $k$ is a field?

Question

Somewhere I saw that

To show that $x^2-y^3$ is irreducible in $k[x,y]$ it suffices to show that $x^2-y^3$ is irreducible in $k(y)[x]$.

My question is what is the relation between $k[x,y]$ and $k(y)[x]$ ?

Also there is a confusion that if $k(y)$ is the smallest field containing $y$ and $k$ (by definition) then what will be the inverse of $y?$ Is it $1/y$ ?

Answer 1

To answer the question in the title, the two rings are different because $y$ has an inverse in $k(y)[x]$ but not in $k[x,y]$.

Answer 2

We have the following: $$k[x,y]\cong k[y][x]\subseteq k(y)[x],$$ since $k[y]\subseteq k(y)$ because $k(y)=Frac(k[y])$.

Now, for simplicity let's set $k[y]=R$, then we have to show that $x^2-y^3$ is irreducible in $R[x]$. By Gauss' lemma it's enough to show that $x^2-y^3$ is irreducible in $Frac(R)[x]=k(y)[x]$. This explains the given argument.