Trying to prove that $\langle y-x^2, z-x^3\rangle$ is prime in $K[x,y,z]$

Question

Let $K$ be a field. I am trying to prove that $I=\langle y-x^2, z-x^3\rangle$ is a prime ideal in $K[x,y,z]$.

Main idea:

Define a homomorphism $\phi:K[z,y,z]\rightarrow K[x]$ as $$\phi(x)=x, \phi(y)=x^2, \phi(z)=x^3.$$

Clearly, $\phi$ is surjective. If we can show that the kernel $\phi $ is $I$, we are done. It is easy to see that $I\subseteq \ker \phi$. But I am struggling to prove that $\ker \phi\subseteq I$. I have been able to prove that any $F\in\ker\phi$ is of the form $$F(x,y,z)=(z-x^3) \frac{H_1(x,y,z)}{G(x)}+(y-x^2)\frac{H_2(x,y)}{G(x)}$$ where $H_1,H_2\in K[x,y,z]$ and $G\in K[x]$. If we can prove that $G=1$, then the proof will be complete.

Before going into the details I will need the following lemma:

Let $R$ be an integral domain and $k$ be its field of fractions. Let $c\in R$ be such that $$c=p_1\frac{a_1}{b_1}+p_2\frac{a_2}{b_2}$$ where

(1) $p_1,p_2\in R$ are primes,

(2) $a_j,b_j\in R$ are such that $(a_j,b_j)=(p_j,b_j)=1$ for $j=1,2$.

Then $$b_1=\pm b_2$$

Proof of Fact:

We have $$b_1b_2c=a_1p_1b_2+a_2p_2b_1$$ So $$b_2(b_1c-a_1p_1)=a_2p_2b_1$$ The LHS of the above equation is divisible by $b_2$, and hence also the RHS. As $b_2$ is coprime to $a_2$ and $p_2$, we must have $b_2\mid b_1$. Similarly we conclude that $b_1\mid b_2$. Hence $b_1=\pm b_2$.

Details of Calculation:

So let us start with $F\in \ker \phi$. Any $F$ is of the form $$F(x,y,z)=\sum_{i,j,k}a x^iy^jz^k\tag{1}$$ where $a\in K$. By the division algorithm in $K(x)[z]$ we get $$z^k=(z-x^3)P(x,z)+\frac{A_k}{B_k}\tag{2}$$ where $A_k,B_k\in K[x]$ and $P(x,z)$ is of the form of $p(x,z)/q(x)$ where $p(x,z)\in K[x,z]$ and $q(x)\in K[x]$.

Substituting the value of $z^k$ from (2) in (1) we get $$ \begin{align} F &=\sum_{i,j,k}a x^iy^j\left[(z-x^3)P(x,z)+\frac{A_k}{B_k}\right]\\ &= \sum_{i,j,k}a x^iy^j (z-x^3)P(x,z)+\sum_{i,j,k}a x^iy^j\frac{A_k}{B_k}\\ &= (z-x^3) \frac{H_1(x,y,z)}{G_1(x)}+\sum_{i,j,k}a x^iy^j\frac{A_k}{B_k}\\ \end{align}\tag{3} $$ where $H_1(x,y,z)\in K[x,y,z]$ and $G_1(x)\in K[x]$. By the division algorithm in $K(x)[y]$ we get $$y^j=(y-x^2)M(x,y)+\frac{C_j}{D_j}\tag{4}$$ where $C_j,D_j\in K[x]$ and $M(x,y)$ is of the form of $m(x,y)/n(x)$ where $m(x,y)\in K[x,y]$ and $n(x)\in K[x]$.

Substituting the value of $y^j$ from (4) in the second summation in (3) we get

$$ \begin{align} F &= (z-x^3) \frac{H_1(x,y,z)}{G_1(x)}+\sum_{i,j,k}a x^i\left((y-x^2)M(x,y)+\frac{C_j}{D_j}\right)\frac{A_k}{B_k}\\ &= (z-x^3) \frac{H_1(x,y,z)}{G_1(x)}+ \sum_{i,j,k}a x^i(y-x^2)M_j(x,y)\frac{A_k}{B_k}+\sum_{i,j,k}a x^i\frac{C_jA_k}{D_jB_k}\\ &= (z-x^3) \frac{H_1(x,y,z)}{G_1(x)}+(y-x^2)\frac{H_2(x,y)}{G_2(x)}+\frac{E_1}{E_2} \end{align}\tag{5} $$ where $H_2\in K[x,y]$ and $G_2,E_1,E_2\in K[x]$.

As $F\in\ker\phi$ we have $F(x,x^2,x^3)=0$. Substiting $y=x^2$ and $z=x^3$ in (5) we get $$ \begin{align} 0&=0\cdot \frac{H_1(x,y,z)}{G_1(x)}+0\cdot\frac{H_2(x,y)}{G_2(x)}+\frac{E_1}{E_2}\\ &=\frac{E_1}{E_2} \end{align}\tag{6} $$ Thus we get $$F=(z-x^3) \frac{H_1(x,y,z)}{G_1(x)}+(y-x^2)\frac{H_2(x,y)}{G_2(x)}\tag{7}$$ WLOG we can assume that $(G_1, H_1)=1$ and $(G_2, H_2)=1$. Then by the Lemma we conclude that $G_1=\pm G_2$. WLOG we can assume that $G_1=G_2$.

From here onwards I am unable to move ahead. Can you help?

Answer 1

HINT:

Useful to consider polynomials in $y$, $z$, with coefficients in $K[x]$, or polynomials in $y$ with coefficients in $K[x,z]$. Like @Hellen: suggested, given a polynomial $P \in K[x,y,z]$, divide it first by $y-x^2$, and get a remainder a free terms ( a polynomial in $K[x,z]$) $$P(x,y,z) = (y-z^2)Q(x,y,z) + R(x,z)$$ Now, divide $R(x,z)$ by $z-x^3$, and get a reminder, a polynomial in $x$ $$R(x,z) = (z-x^3)Q_1(x,z)+ R_1(x)$$ All together: $$P(x,y,z) = (y-x^2) Q(x,y,z) + (z-x^3)Q_1(x,y) + R_1(x)$$ This works for every $P(x,y,z)$. Now from the above we get $$P(x,x^2, x^3) = R_1(x)$$

Answer 2

The main idea of looking at the map $\phi \colon K[x,y,z] \to K[x]$ with $x \mapsto x$, $y \mapsto x^2$, and $z \mapsto x^3$ is the right idea, but after that you get bogged down in computational details.

In such cases, it is often easier to generalize the problem so that irrelevant details don't get in the way. In this case, $x^2$ and $x^3$ are irrelevant details; they could be any element of $K[x]$ and the argument would still work.

In general, let $R$ be a (commutative) ring (with unit) and $r_1, \dots, r_n \in R$. Then the kernel of the evaluation map $R[x_1, \dots, x_n] \to R$, $f \mapsto f(r_1, \dots, r_n)$ is the ideal $\langle x_1 - r_1, \dots, x_n - r_n\rangle$. This gives an isomorphism $R[x_1,\dots,x_n]/\langle x_1 - r_1, \dots, x_n - r_n\rangle \cong R$.

Ofcourse, you still need to prove this and in essence it is still the argument you are trying to give, but you can simplify even more by first considering the case that $r_1 = \dots = r_n = 0$ and then use the fact that $x_1 \mapsto x_1 - r_1, \dots, x_n \mapsto x_n - r_n$ gives an isomorphism of $R[x_1,\dots,x_n]$.

Now, apply this to $R = K[x]$ and $r_1 = x^2$, $r_2 = x^3$ (and $x_1 = y$, $x_2 = z$).