Proof of Cayley-Hamilton via fields

Let $R$ be a commutative ring, and $M$ a finitely-generated free $R$-module. Let $\phi:M\rightarrow M$ be an $R$-linear map, and $P_\phi(X)$ the characteristic polynomial of $\phi$. Show that $P_\phi(\phi)=0$ for any $R$ and $\phi$ (that is, prove Cayley-Hamilton.)

We may assume Cayley-Hamilton for the case where $R$ is an algebraically closed field, i.e. $P_\phi(\phi)=0$ in that case.

Let $f:R\rightarrow R'$ be a map of rings, and let $\phi':M\otimes_RR'\rightarrow M\otimes_RR'$ be the map induced by $\phi$. We may assume that if $P_\phi(\phi)=0$, then $P_{\phi'}(\phi')=0$, and if $f$ is injective, the converse holds.

Now, the hint is to write $R$ as a quotient of a polynomial ring over $\mathbb{Z},S=\mathbb{Z}[X_1,X_2,\ldots]$ and then embed $S$ in an algebraically closed field. I don't quite understand how to use it.. can anyone fill in the details?

• @JJ Beck: A proof is at pag. $120$ of "Commutative Algebra: With a View Toward Algebraic Geometry" – WLOG Feb 7 '14 at 20:36
• Use math.stackexchange.com/questions/661550 and you are done. Better don't read the solution of this exercise, the hint tells you almost everything. – Martin Brandenburg Feb 7 '14 at 21:46
• @MartinBrandenburg Could you at least explain what it means here to "embed in an algebraically closed field"? – JJ Beck Feb 9 '14 at 5:24
• Fix a (comm, unitary) ring $R$. For each $a\in R$, we associate with an indeterminate $X_a$, then consider the "polynomial ring" $R'=\mathbb Z[\{X_a\}_{a\in R}]$, which is the union of $\mathbb Z[\{X_a\}_{a\in F}]$ where $F$ is any finite subset of $R$. Note that $X_a\mapsto a$ induces a ring epimorphism $R'\to R$. That's so called the presentation. Then we can embed $R'$ to a field, say $\mathbb C(\{X_a\}_{a\in R})$. The trick, which could be used to translate identities from fields to rings, is introduced in, say, M.Artin's Algebra. – Yai0Phah Feb 16 '14 at 7:28