I've been working on the following problem from Hartshorne:
Let $Y\subseteq \mathbb{ A }^3 $ be the set $Y = \{(t,t^2 , t^3) \mid t \in k \}$. Show that $Y$ is an affine variety of dimension $1$.
To start with, I proved that $Y$ is an algebraic set: define polynomials
$$ f(x,y,z) = x^2 - y \ , \ g(x,y,z) = x^3 - z \in k[x,y,z] $$
Then it's not bad to check that
$$ Y = Z(f) \cap Z(g) = Z (\{f,g\})$$
Now I need to show that $Y$ is irreducible. This is equivalent to showing that the ideal of $Y$,
$$ I(Y) = \{ p(x,y,z) \in k[x,y,z] \mid p (t,t^2,t^3) = 0, \forall t \in k \}$$
is prime.
I tried a direct approach: suppose that $pq \in I(Y)$ for $p,q \in k[x,y,z]$, and suppose that $q \notin I(Y)$. Then there exists some $s \in k$ for which $q(s,s^2, s^3) \ne 0$. I don't believe that it necessarily follows that $p \in I(Y)$. Even though $k$ is an integral domain, the fact that we have one triple $(s,s^2,s^3)$ at which $q$ does not vanish does not force $p$ to vanish at every triple $(t, t^2, t^3)$, $t \in k$.
The Nullstellensatz tells us that
\begin{align*}
I(Y) &= I(Z(\{f,g\})) \\
&= \sqrt{(f,g)} \\
\end{align*}
But I wasn't getting anywhere using this fact either. I don't know much about non-principal ideals and how they relate with prime ideals. Any help would be greatly appreciated!
My apologies if this is overkill, but I'm really trying to understand:
So when you say that
$$ k[x,y,z]/ (x^2 - y , x^3 - y) = k[x] $$
You mean that the two are isomorphic right?
We define a homomorphism $\varphi: k[x,y,z] \rightarrow k[t]$ which sends $x \mapsto t$, $y \mapsto t^2$, $z \mapsto t^3$. This is surjective: given $p(t) = a_0 + a_1 t + \ldots + a_n t^n$, then $\varphi ( \tilde{p}) = p$ where $\tilde{p} = a_0 + a_1 x + \ldots + a_n x^n \in k[x,y,z]$.
Observing that
$$ \mbox{Ker} (\varphi) = \{ h \in k[x,y,z] \mid h(t,t^2 , t^3) = 0 \} = I (Y) $$
then we have that
\begin{align*}
A(Y) &= k[x,y,z]/I(Y) \\
&= k[x,y,z]/ \mbox{Ker}(\varphi) \\
&\cong \mbox{Im}(\varphi)\\
&= k[t]
\end{align*}
and so it seems to me that it is $I(Y) = \sqrt{J}$, (as opposed to $J$ itself) which is prime. Of course this gives us that $Y$ is irreducible.
The problem goes on to ask for generators of $I(Y)$.
If we get that $J$ is prime immediately, then (since primes are radical) $I(Y) = J$ (as you mention). If we get that $I(Y)$ is prime, then we don't immediately get that $I(Y) = J$, and we have something left to prove.
If $J=(f,g)$, we have $k[x,y,z]/J=k[x,y,z]/(x^2 - y , x^3 - z) =k[x]$, hence $J$ is prime since the quotient $k[x,y,z]/J=k[x]$ is a domain.
As you proved that $Y=Z(J)$ we have $I(Y)=I(Z(J))=\sqrt J$ (by the Nullstellensatz) $=J$ , since we just saw that $J$ is prime. So $I(Y)=J$, a prime.
But the condition that $Y$ be irreducible is exactly that $I(Y)$ be prime, so $Y$ is indeed irreducible.
Our variety $Y$ is of dimension $1$ because it is the image of $\mathbb A^1_k$ under the non-constant morphism $t\mapsto (t,t^2,t^2)$.