Why is it impossible for the mapping $x\mapsto 1/x$ to be a polynomial in an infinite field?

Question

Let $K$ be an infinite field and consider the map $x\mapsto 1/x$ on $K^\times$. I want to show it is not possible for this map to be a polynomial in the input $x$.

My idea for a proof goes like this: assume for contradiction that the map is a polynomial. Then there is some polynomial $p$ such that $$ 1/x = p(x) $$ for all $x\in K^\times$. My plan then would be to get $1=xp(x)$ and then make some argument about having infinitely many linear equations... but I'm not super happy with justifying the details of that. Would people agree this is how one should proceed? Or is there a better way?

Answer 1

The polynomial $xp(x)-1$ has at most $\deg p+1$ roots, so it can’t have all the infinitely many elements of $K^\times$ as roots.

Answer 2

From Jose, $xp(x)$ would have positive degree and thus $x(p(x))-1$ would be a non-constant polynomial with infinitely many roots which is impossible.