I have been trying to study the basics of algebraic geometry using nonstandard analysis and I can't wrap my head around this issue.
Let $^*\mathbb{C}$ be the extension of the complex numbers. Now fundamental theorem of algebra says that all polynomials in $\mathbb{C}[x]$ have solutions in $\mathbb{C}$. It also implies that a degree $n$ polynomial will have $n$ complex solutions. By transfer, this should also be true for polynomials in $^*\mathbb{C}[x]$. Now if you have a degree $n$ polynomial $f \in \mathbb{C}[x]$, then it has $n$ zeroes in $^*\mathbb{C}$ and these zeros must all be standard.
This also means that if $x \in {}^*\mathbb{C}$ is nonstandard, then $f(x) \neq 0$. However, I'm reading Nonstandard Generic Points by Guy Wallet, and he uses Nullstellensatz to show that for a prime ideal in $\mathbb{C}[x]$, $P$, there is a point $x \in {}^*\mathbb{C}$ such that $f(x)=0$ iff $f \in P$.
This means there are standard polynomials that are zero on nonstandard numbers. Does this contradict the fundamental theorem of algebra? I appreciate any explanation. Thanks.
There's no contradiction here. The prime ideals of $\mathbb{C}[x]$ are the maximal ideals $(x - a), a \in \mathbb{C}$ and zero (the generic point). For the maximal ideals the desired point is $x = a$, which is standard. And for the zero ideal we can take any nonstandard point, since as you say a standard polynomial vanishes on a nonstandard point iff it's identically zero.