TITLE: Polynomial with degree greater than or equal to 2 not injective over algebraically closed field?
QUESTION [5 upvotes]: I know that this is probably a silly question. Given a polynomial in an algebraically closed field, why is the polynomial not injective if its degree is greater than or equal to 2? I understand that the polynomial must factor into its linear roots and that it's roots must be distinct, for otherwise, more than one value would map to $0$. 
However, what if we have a repeated root? For example, if $p(z) = (z-1)^3$ over $\mathbb{C}$, why can't this be injective, since only $p(1) = 0$?

REPLY [4 votes]: The claim is false if the characteristic of the field is non-zero. In any field $k$ of characteristic $p>0$, the Frobenius endomorphism $\phi:k\to k$ defined by $\phi(x)=x^p$ is injective (regardless of whether $k$ is algebraically closed). So let's focus on the characteristic zero case now.
As you've observed, if a polynomial $f\in k[x]$ has at least two distinct roots $\alpha,\beta\in k$, then it is not injective because $f(\alpha)=f(\beta)=0$.
Suppose that $f$ is of degree $\geq 2$ and has only one root, say $f=(x-\alpha)^n$ for some $n\geq 2$. For any non-zero $\gamma\in k$, the polynomial $x^n-\gamma$ has $n$ distinct roots (here's where we need both $k$ to be algebraically closed as well as $\mathrm{char}(k)=0$), and therefore so does the poynomial $(x-\alpha)^n-\gamma$; namely, the roots of $x^n-\gamma$, all translated by $\alpha$. But this means there are distinct elements of $k$ that are sent to $\gamma$ by the polynomial $f$, so that $f$ is not injective.