Ring homomorphism extension from $k[x]$ to $k(x)$ I've been trying to prove that if $k$ is an algebraically closed field, $V$ is a $k$-vector space, $\dim_k(V)$ is infinite, and $f:V\to V$ is a linear operator such that $f-rI$ is invertible for all $r\in k$, then the ring homomorphsm
\begin{align} p:k[x]&\to\mathrm{End}_k(V)\\ g(x)&\mapsto g(f)  \end{align}
can be extended to a ring homomorphism
\begin{align} \hat{p}:k(x) &\to\mathrm{End}_k(V)\\ \frac{g(x)}{s(x)}&\mapsto g(f)s(f)^{-1}  \end{align}
It is like universal property of localization ring (which I don't use because I learned this is just for commutative rings, but $\mathrm{End}_k(V)$ is not). I could prove $\hat{p}$ is well defined, also $\hat{p}(a+b)=\hat{p}(a)+\hat{p}(b)$. My problem is with $\hat{p}(ab)=\hat{p}(a)\hat{p}(b)$, because without commutativity I get stuck.
Thanks for any help.

So some ingredients:

*

*Show that if $A$ and $B$ and $C$ and $D$ all commute with each other, then $A + B$ commutes with $C + D$ and $AB$ with $CD$. And also $AB$ with $C + D$ and so on.

*with these base cases + induction, you can show that polynomials in $f$ commute with other polynomials in $f$
Possibly you've already done this.


*you want to show that the fractions commute as well. E.g. if $x$ and $y$ are polynomials in $f$ then $xy = yx$ implies $y^{-1}(xy)y^{-1} = y^{-1}(yx)y^{-1}$, which shows that $x$ commutes with $y^{-1}$.


*Similarly, you can show that $x, y, x^{-1}, y^{-1}$ all commute with each other.
