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Suppose we have a field $F$ and let $F[x]$ be the ring of polynomials. Define $F[[x]]$ be the ring of formal power series $\sum_{n\geq 0} a_{n} x^{n}$. Let $S \subset F[x]$ be a multiplicative set consisting of all polynomials with a non-zero constant term.
Show that the inclusion $F[x] \rightarrow F[[x]]$ extends to a homomorphism $F[x][S^{-1}] \rightarrow F[[x]]$.

I really do not know what this question is asking. I am thinking this has something to do with power series being invertible, but I am unsure how to proceed.

user100101212
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    By the universal property of localization, this amounts to showing that the elements of $S$ are invertible in $F[[x]]$. – Ayman Hourieh Nov 11 '19 at 22:59
  • Could you explain (or direct me to resources) the universal property of localization and how it is applied to this problem? Also, what is meant by 'inclusion' and what does it mean for an 'inclusion' to be extended to a homomorphism? I am assuming this has to do with some composition of mappings, but it would help if that was further explained. Thanks for the help. – user100101212 Nov 11 '19 at 23:18
  • I added an answer with more details and pointers. – Ayman Hourieh Nov 12 '19 at 11:43

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Let $A$, $B$ and $C$ be rings such that $A \subset B$. Let $\varphi : A \to C$ be a ring homomorphism. We say $\psi : B \to C$ is an extension of $\varphi$ if the restriction of $\psi$ to $A$ is the same homomorphism as $\varphi$.

If $S$ is a multiplicative subset of $A$, and the image of each element of $S$ is a unit in $C$, then $\varphi$ extends to a homomorphism $A[S^{-1}] \to C$ by the universal property of localization.

To apply the result above to your problem, all what we have to do is show that each element of $S$ is a unit in $F[[x]]$. This has already been proved on this site. See this question for example.

Ayman Hourieh
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