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 $\mathbb{C}[X,Y,Z]$:
$$V(Y-X^2,Z-X^3) = \{ (t,t^2,t^3) \mid t \in \mathbb{C}\}\ ?$$ 
$$V(Y-X^2, Z-X^3) = \{(a,b,c) \in \mathbb{C}^3 \mid b-a^2=0, c-a^3=0 \Rightarrow b=a^2, c=a^3\} = \{(t, t^2, t^3) \mid t \in \mathbb{C}\}$$
EDIT 2:
Could we show like that, that: $I(V(Y-X^2,Z-X^3)) = \langle Y-X^2,Z-X^3 \rangle$ ?
$$\begin{align}
I(V(Y-X^2, Z-X^3)) &= I(\{(t, t^2, t^3) \mid t \in \mathbb{C}\}) \\
&= \{f(X,Y,Z) \in \mathbb{C}[X,Y,Z] \mid f(t,t^2,t^3)=0\} \\
&\overset{*}{=} \{(Y-X^2) \cdot g(X,Y,Z)+(Z-X^3) \cdot h(X,Y,Z) \mid g,h \in \mathbb{C}[X,Y,Z]\} \\
&= \langle Y-X^2, Z-X^3 \rangle 
\end{align}$$
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$.
Actually, 
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?

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.
