Verifying that the ideal $(x^3-y^2)$ is prime

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!

• @yang now you can try a more general version of your question. $I=(x^m-y^n)$ is a prime ideal in $k[x,y]$ when $(m,n)=1$.
– user114539
Dec 30, 2013 at 13:53

It suffices to prove that $x^3 - y^2$ is irreducible in $k[x, y]$. Hence it suffices to prove that it is irreducible in $k(y)[x]$. This is clear because it has no root in $k(y)$.

• This is an excellent approach. Since somebody will inevitably wonder why irreducibility in $k(y)[x]$ is adequate, I'll preemptively give them the answer: Gauss's Lemma. Dec 30, 2013 at 12:52
• @User-33433 Actually, Gauss' lemma is used twice. The other instance in which is used is to be able to conclude that irreducible implies prime: In fact, since $k[x]$ is a Euclidean domain, then it is a UFD, so by Gauss' lemma $k[x,y]$ is also a UFD (and "prime" is equivalent to "irreducible" in this case). Dec 30, 2013 at 17:12
• Why does having no root in the field $k(y)$ make the polynomial irreducible? This is not the case for fields in general. Feb 5, 2014 at 18:57
• @MohamedHashi Because the degree of the polynomial is 3. Regards, Feb 5, 2014 at 22:17
• @Saru Having no root isn't sufficient to show irreducibility of a cubic over a general ring. For example, $(2x-y)^3$ has no root in $k[y]$. We also have polynomials like $y(x^3-y^2)$, which has no root even in $k(y)$ but is still reducible over $k[y]$ because the coefficients have a common factor. So we should use $k(y)$ then reduce to $k[y]$ with Gauss's lemma. Nov 15, 2022 at 17:30

You want to prove the following: if $f\in k[x,y]$ and $f(t^2,t^3)=0$ then $f\in(x^3-y^2)$.

Let's divide $f(x,y)$ to $y^2-x^3$ by considering both as polynomials in $y$ with coefficients in $k[x]$. (This can be done since $y^2-x^3$ is a monic polynomial in $y$.) We have $$f(x,y)=(y^2-x^3)g(x,y)+r(x,y)$$ with $\deg_yr<2$, so we can write $r(x,y)=r_1(x)+r_2(x)y$. Since $f(t^2,t^3)=0$ we get $r_1(t^2)+r_2(t^2)t^3=0$. Now note that the degree in $t$ of $r_1(t^2)$ is even, while the degree of $r_2(t^2)t^3$ is odd. Can you conclude from this that $r_1=r_2=0$?

• I did a long calculation considering $f(x,y)$ as a polynomial in $x$ with coefficients from $k[y]$ :( Dec 30, 2013 at 12:32

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 $f(x,y)\in Ker$