Extending an isomorphism to $F[x]$ In chapter 13 of Dummit/Foote (p.518 to be more precise) the authors mention that if $\phi: F \to F'$ is an isomorphism, then $\phi$ induces a ring isomorphism 
\begin{equation*}
\tilde{\phi}: F[x] \to F'[x]
\end{equation*}
defined by applying $\phi$ to the coefficients of $f(x) \in F[x]$.  
Now it may be that the authors have explained this concept somewhere else in the text and I have just not realized it, but it seems to me that this statement is far from trivial and really deserves to be a theorem.   This statement also raises a few questions for me:


*

*I'm guessing the phrase "applying $\phi$ to the coefficients of $f(x)$" means I write $f(x)$ in the form 
\begin{equation*}
f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0
\end{equation*}
and then apply $\phi$ to the $a_i$.  But what if $f(x)$ is written in the form 
\begin{equation*}
f(x) = p(x)q(x)
\end{equation*}
Then will I get the same thing if I apply $\phi$ to the coefficients of $p(x)$ and $q(x)$ separately? In other words, do we have:
\begin{equation*}
\tilde{\phi}\big(a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0\big) = \tilde{\phi}\big(p(x)\big) \tilde{\phi}\big(q(x)\big)
\end{equation*}
Since $\tilde{\phi}$ is a homomorphism, these should be the same.  This leads to my second question:

*If it does not matter how I factor $f(x)$ before applying $\tilde{\phi}$, then we need to show that $\tilde{\phi}$ is well-defined.  This seems to be very nontrivial, because there are many different factorizations of a given polynomial.  Do you know an outline of a proof for this result?

*What if I factor $f(x) \in F[x]$ into $f(x) = g(x)h(x)$, where the coefficients of $g(x)$ and $h(x)$ live in some bigger field extension of $F$, i.e. $g(x)$ and $h(x)$ are not in $F[x]$.  Then I'm guessing I can't apply $\tilde{\phi}$ to this factorization of $f(x)$, right?  This seems kind of  strange to me. 
Thanks so much for the help!
 A: Let $\phi:R\to R'$ be a ring homomorphism. Then $\phi$ induces a ring homomorphism $(\cdot)^\phi:R[X]\to R'[X]$ on the corresponding polynomial rings by
$$\sum_{k=0}^n a_kX^k\mapsto \sum_{k=0}^n\phi(a_k)X^k.$$
Example: Reducing an integral polynomial modulo some prime number to see whether or not it's irreducible is quite popular.


*

*Hint: Use the so-called convolution product notation:$$\left(\sum_{k=0}^na_kX^k \right)\left(\sum_{k=0}^mb_kX^k\right)= \sum_{k=0}^{n+m}\left(\sum_{i+j=k}a_ib_j X^k\right)$$

*The representation of an $f(X)\in R[X]$ as a linear combination of monic monomials $X^k$ is unique, and the image of $(\cdot)^\phi$ is immediately in its target space. Hence $(\cdot)^\phi$ is well-defined.

*Indeed, but often you will extend $\phi$ to have the extension field as the domain (quite possibly this extension will not be unique, which is where the fun part begins really) and use the same machinery to that extension to transport factors to some other polynomial ring (e.g. to the polynomial ring over some algebraically closed field that contains the initial field).
