Ideals-algebraic set

Notice that in $\mathbb{C}[X,Y,Z]$:
$$V(Y-X^2,Z-X^3) = \{ (t,t^2,t^3) \mid t \in \mathbb{C}\}$$

In addition, show that:

$$I(V(Y-X^2,Z-X^3)) = \langle Y-X^2,Z-X^3 \rangle$$

Finally, prove that the ideal $\langle Y-X^2,Z-X^3 \rangle$ is a prime ideal of $\mathbb{C}[X,Y,Z]$. Conclude that the algebraic set $V(Y-X^2,Z-X^3)$ is irreducible.

Could you give me some hints to solve the above exercise?

EDIT 1: Could we show like that, that in $\mathbb{C}[X,Y,Z]$:
$$V(Y-X^2,Z-X^3) = \{ (t,t^2,t^3) \mid t \in \mathbb{C}\}\ ?$$

$$V(Y-X^2, Z-X^3) = \{(a,b,c) \in \mathbb{C}^3 \mid b-a^2=0, c-a^3=0 \Rightarrow b=a^2, c=a^3\} = \{(t, t^2, t^3) \mid t \in \mathbb{C}\}$$

Could we show like that, that: $I(V(Y-X^2,Z-X^3)) = \langle Y-X^2,Z-X^3 \rangle$ ?

I(V(Y-X^2, Z-X^3)) &= I(\{(t, t^2, t^3) \mid t \in \mathbb{C}\}) \\
&= \{f(X,Y,Z) \in \mathbb{C}[X,Y,Z] \mid f(t,t^2,t^3)=0\} \\
&\overset{*}{=} \{(Y-X^2) \cdot g(X,Y,Z)+(Z-X^3) \cdot h(X,Y,Z) \mid g,h \in \mathbb{C}[X,Y,Z]\} \\
&= \langle Y-X^2, Z-X^3 \rangle

At $(*)$ can we say it like that because we know from $V(Y-X^2, Z-X^3)=\{(t,t^2,t^3) \mid t \in \mathbb{C}\}$ that $(t,t^2,t^3)$ is a solution of $Y-X^2=0$ and $Z-X^3=0$ ?

EDIT 3:Let the homomorphism $\phi: \mathbb{C}[x,y,z] \to \mathbb{C}[x]$, with $\phi(x)=x, \phi(y)=x^2, \phi(z)=x^3, \phi(j)=j, \forall j \in \mathbb{C}$

We consider the polynomial $p(x,y,z)=\sum_{k,l,m=0}^n a_{klm} x^k (y-x^2)^l(z-x^3)^m$.

Then, $ \phi(p)=\sum_{k,l,m=0}^n \phi(a_{klm}) (\phi(x))^k (\phi(y)-(\phi(x))^2)^l (\phi(z)-(\phi(x))^3)^m=a_{n00}x^n $

$ ker(\phi)$ contains all the elements from $\mathbb{C}[x,y,z]$, die auf 0, durch die \phi, abgebildet werden.

So, it contains the polynomials with $a_{n00}=0$, so the non-zero elements of the polynomial must contain $y-x^2$ or $z-x^3$, so $ ker \phi=\langle y-x^2,z-x^3 \rangle$.


Let $ p(x,y,z) \in \langle y-x^2,z-x^3 \rangle$.

Then, $p(x,y,z)=p_1(x,y,z) \cdot (y-x^2) +p_2(x,y,z) \cdot (z-x^3), p_1(x,y,z), p_2(x,y,z) \in \mathbb{C}[x,y,z] $

We have that $ \phi(p(x,y,z))=\phi(p_1(x,y,z) \cdot (y-x^2) +p_2(x,y,z) \cdot (z-x^3))=0</math>, also <math> p(x,y,z) \in ker(\phi) \Rightarrow \langle y-x^2,z-x^3 \rangle \subseteq ker(\phi)$

Let $p(x,y,z) \in ker(\phi)$. Then, $\phi(p(x,y,z))=0$. The terms of the polynomial are of the form $a_{ij}(y-x^2)^i (z-x^3)^j$,where at least one of $i,j>0$.

So, $p(x,y,z)$ is a sum of terms of $ \langle y-x^2,z-x^3 \rangle $. So, $ p(x,y,z) \in \langle y-x^2,z-x^3 \rangle $, and so $ ker(\phi) \subseteq \langle y-x^2,z-x^3 \rangle $.

So, we conclude that $ker(\phi)=\langle y-x^2,z-x^3 \rangle$.

Since $\phi(p(x,y,z))=a_{n00}x^n \in \mathbb{C}[x], \forall p(x,y,z) \in \mathbb{C}[x,y,z]$,we have that $im(\phi)= \mathbb{C}[x]$.

So, we have that: $ \mathbb{C}[x,y,z]/ \langle y-x^2,z-x^3\rangle \cong \mathbb{C}[x]$.

Is it right so far? Do we have to show that $\mathbb{C}[x]$ is an integral domain?

Solutions Collecting From Web of "Ideals-algebraic set"

Edit 1 is correct.

Edit 2 has a flaw at $(*)$: no, you can’t say that; you have to prove that $f(t,t^2,t^3)=0$ for all $t\in\mathbb C$ implies $f\in (Y-X^2,Z-X^3)$. (This is closely related to prove that $(Y-X^2,Z-X^3)$ is a prime ideal, and in the reasoning I’ve posted below $f(t,t^2,t^3)=0$ reduces to $q(t)=0$ for all $t\in\mathbb C$, so $q=0$.)
You could also use the Hilbert’s Nullstellensatz and this gives you $I(V(Y-X^2,Z-X^3))=\sqrt{(Y-X^2,Z-X^3)}$, but then you have to prove that the ideal $(Y-X^2,Z-X^3)$ is radical in order to get what you want. In fact it is prime, and in some sense all gravitates around this as you can immediately notice and this is what you try to prove in

Edit 3 I can’t see why you write a polynomial like this: $$p(X,Y,Z)=\sum_{k,l,m=0}^n a_{klm} X^k (Y-X^2)^l(Z-X^3)^m.$$ (For instance, how to write $XY$ in that form?)
A better strategy is to use the division algorithm and first write $$p(X,Y,Z)=(Y-X^2)f(X,Y,Z)+r(X,Z)$$ with $\deg_Yr<1$, that is, $Y$ doesn’t occur in $r$, and then $$r(X,Z)=(Z-X^3)g(X,Z)+q(X)$$ with $\deg_Zq<1$ and thus $Z$ doesn’t occur in $q$, so $q$ is a polynomial in $X$ only.
Now, if $p\in\ker\phi$ we get $p(X,X^2,X^3)=0$, so $q(X)=0$ and you get $$p(X,Y,Z)=(Y-X^2)f(X,Y,Z)+(Z-X^3)g(X,Z).$$ This shows that $p\in(Y-X^2,Z-X^3)$.
As a consequence you get $$\mathbb C[X,Y,Z]/(Y-X^2,Z-X^3)\simeq\mathbb C[X]$$ which is an integral domain (no need to prove this in my opinion), and therefore $(Y-X^2,Z-X^3)$ is a prime ideal.

  1. $V(y-x^2, z-x^3) = \{(a,b,c) \in \mathbb{C}^3 \mid b-a^2=0, c-a^3=0\} = \{ (a, b, c) \in \mathbb{C}^3 | b=a^2, c=a^3\} = \{(t, t^2, t^3) \mid t \in \mathbb{C}\}$.

  2. $I(V(y-x^2,z-x^3)) = \{ f(x, y, z) \in \mathbb{C}[x, y, z] | f(t, t^2, t^3) = 0 \}$ (using 1). Obviously $(y – x^2, z – x^3) \subseteq I(V(y-x^2,z-x^3))$. Now we will show that if $f(x, y, z) \in \mathbb{C}[x, y, z] $ is such that $f(t, t^2, t^3) = 0$ then $f(x, y, z) \in (y – x^2, z – x^3)$. Consider the polynomial $y – x^2$ as a polynomial in $y$ over the ring $\mathbb{C}[x, z]$. Since it is a monic polynomial in $y$, we can use division algorithm. We get $f(x, y, z) = f_1(x, y, z)(y – x^2) + f_2(x, z)$. Since the degree of $y – x^2$ is 1 as a polynomial in $y$, $f_2(x, z)$ is a polynomial in $x$ and $z$ only (over $\mathbb{C}$)[Hence the notation is justified]. Similarly, considering the polynomial $z – x^3$ as a polynomial in $z$ over the ring $\mathbb{C}[x]$, we get $f_2(x, y, z) = f_3(x, y, z)(z – x^3) + f_4(x)$. Using the same reasoning as above, $f_4(x)$ is a polynomial in $x$ alone (over $\mathbb{C}$). So, $f(x, y, z) = f_1(x, y, z)(y – x^2) + f_3(x, y, z)(z – x^3) + f_4(x)$. Now $f(t, t^2, t^3) = 0 \Rightarrow f_4(t) = 0 \Rightarrow f(x, y, z) \in (y – x^2, z- x^3)$.

  3. To prove the ideal $(y – x^2, z – x^3)$ is a prime ideal in the ring $\mathbb{C}[x, y, z]$, let’s define a $\mathbb{C}$-algebra homomorphism $\phi : \mathbb{C}[x, y, z] \rightarrow \mathbb{C}[t]$ by $x \mapsto t, y \mapsto t^2, z \mapsto t^3.$ This is a surjective ring homomorphism. Using the same arguments as in (2), we can prove that ker$\phi = (y – x^2, z – x^3)$. Thus by first isomorphism theorem of rings, the ideal $(y – x^2, z – x^3)$ is a prime ideal.

  4. The algebraic set $Y := V(y – x^2, z – x^3)$ is irreducible is equivalent of saying that the ideal $(y – x^2, z – x^3)$ is prime. Suppose $Y$ is not irreducible. Then there is two non-empty proper closed subsets $Y_1, Y_2$ of $Y$ with $Y = Y_1 \cup Y_2$. Thus $I(Y) = I(Y_1 \cup Y_2) = I(Y_1) \cap I(Y_2).$ Since $I(Y)$ is prime, this means either $I(Y_1) = I(Y)$ or $I(Y_2) = I(Y) \Rightarrow Y = Y_1$ or $Y = Y_2$.