I want to show that an affine variety $V \subset \mathbb{C}^n$ which is parameterizable in the following sense: There exist Polynomials $g_1, \ldots, g_n \in \mathbb{C}[x]$ such that: $$ V=\{(g_1(t),\ldots, g_n(t) \mid t \in \mathbb{C}\} $$ is irreducible.
This seems to be something clear to a lot of people however I am unable to wrap my head around it. Any help/tips are appreciated
I thought about using the fact that an affine variety is irreducible iff it the corresponding Ideal is a prime ideal, however I was not able to make that work.
I have since stumbled across a solution:
It is sufficient to show that $I(V) \subseteq \mathbb{C}[x_1,\ldots,x_n]$ is prime. As such let $f,h \in \mathbb{C}[x_1, \ldots, x_n]$ such that $f \cdot h \in I(V)$.
This implies that $$ \forall t \in \mathbb{C}: (f \cdot h) (g_1(t), \ldots g_n (t)) = 0 $$ This implies that $$ \forall t \in \mathbb{C}: f(g_1(t), \ldots g_n (t)) = 0 \vee h(g_1(t), \ldots g_n (t)) = 0 $$ Now either $f$ or $h$ has to evaluate to $0$ at infinitely many $t \in \mathbb{C}$. Without loss of generality let $f(g_1(t), \ldots g_n (t)) = 0$ for infinitely many $t \in \mathbb{C}$
Now consider $f(g_1(x), \ldots g_n (x)) \in \mathbb{C}[x]$ a univariate polynomial.
It holds that $f(g_1(x), \ldots g_n (x)) = 0$ for infinitely many $x \in \mathbb{C}$ which implies that (since $f(g_1(x), \ldots g_n (x)) \in \mathbb{C}[x]$ is a univariate polynomial) $$\forall x \in \mathbb{C}: f(g_1(x), \ldots g_n (x)) = 0$$
Notably then $$\forall t \in \mathbb{C}: f(g_1(t), \ldots g_n (t)) = 0$$ which implies that $f \in I(V)$.
By which we have shown that $I(V)$ is a prime ideal, which implies, that $V$ is irreducible.