TITLE: Verifying that the ideal $(x^3-y^2)$ is prime
QUESTION [18 upvotes]: How to prove that the ideal $I=(x^3-y^2)$ in $k[x,y]$ is prime? 

I have constructed a map from $k[x,y]$ to $k[t]$, which maps $x$ to $t^2$, and $y$ to $t^3$. Then, I want to show that the kernel is exactly $I$. But the difficulty is that how to prove the kernel is contained in $I$. 
Could you help me? Thanks a lot!

REPLY [5 votes]: Take an element $f(x,y)\in Ker f$ i.e. Now, consider $f(x,y)$ as a polynomial in variable $x$ and coefficients coming from $k[y]$. If you divide $f(x,y)$ by $(x^3-y^2)$ we will get $f(x,y)=g(x,y)(x^3-y^2)+r(x,y)$ where $r(x,y)\in (k[y])[x]$ and degree of $r(x,y)$ is less than three. But then $f(t^2,t^3)=0$ implies $r(t^2,t^3)=0$. But $r(t^2,t^3)$ cannot be zero beacuase  $r(x,y)$ is a  polynomial of degree less than three in variable $x$ with coefficients in $k[y.]$
Explanation of "But $r(t^2,t^3)$ cannot be zero beacuase  $r(x,y)$ is a  polynomial of degree less than three in variable $x$ with coefficients in $k[y]$": 
$r(x,y)=h_1(y).x^2+h_2(y).x+h_3(y)$. So $r(t^2,t^3)=h_1(t^3).({t^2})^2+h_2(t^3).t^2+h_3(t^3)$ . Let the highest power of $h_1(t^3).({t^2})^2$ is $t^{3m+4}$ where m is the degree of $h_1(t)$. Similarly the highest power of$h_2(t^3).t^2$ is $t^{3n+2}$  where n is the degree of $h_2(t)$. and the highest power of $h_3(t^{3})$ is $t^{3k}$ where  k is the degree of $h_3(t)$. 
If $r(t^2,t^3) $ has to be zero then $3m+4=3n+2=3k$ for some $m,n,k \in \mathbb Z$
which is same as saying $3m+1=3n+2=3k$ for some $m,n,k \in \mathbb Z$. Whioch is impossible. So $r(t^2,t^3)=0$ And $f(x,y)\in Ker$