Maximal ideals of the ring of formal polynomials over a ring $R$ Consider the polynomial ring $R[x]$ over a ring $R$. Then $I$ is a maximal ideal of $R[x]$ if and only if $R[x]/I$ is a field. When $F$ is a field, we know that $I$ is of the form $(f)$ where $f(x)$ is irreducible over $F$. 
I'd like to know what maximal ideals of  the ring of formal power series $R[[x]]$ look like. A quick search on Wikipedia returns the following:

The maximal ideals of $R[[x]]$ all arise from those in $R$ in the following manner: an ideal $M$ of $R[[x]]$ is maximal if and only if $M ∩ R$ is a maximal ideal of R and $M$ is generated as an ideal by $x$ and $M ∩ R$.

I'm trying a specific example when $R=F$ is a field. Then by the statement above, $M$ is a maximal ideal of $F[[x]]$ if and only if $M\cap F=(0)$ and $M=(x)$. It follows that $M=(x)$ is the unique maximal ideal of $F[[x]]$. Here is my question:

How can I prove that $(x)$ is the unique maximal ideal of $F[[x]]$ without using the quoted statement above?


Some thoughts: 

*

*It is not hard to check that every non-zero polynomial $f\in F[[x]]\setminus (x)$ is a unit. But I don't see if this might help.

*If one wants to use the following statement, which is also in that article:

If $R$ is a local ring, then so is $R[[x]]$.
  
  I would like to know a proof of it, since I know nothing but the definition of local rings (which have exactly one maximal ideal). 

[ADDED: I should have mentioned that the ring $R$ here is assumed to be commutative and has nontrivial unity.]
 A: 
How can I prove that $(x)$ is the unique maximal ideal of $F[[x]]$ without using the quoted statement above?

The fact you mentioned about $F[[x]]\setminus(x)\subseteq U(R)$ is enough to show $F[[x]]$ is local. In fact, that means $F[[x]]\setminus(x)= U(R)$, and therefore $(x)$ is the set of nonunits. A ring is local iff the set of nonunits is closed under addition.
Here's the collection of equivalences anyone studying local rings should prove, at some point:
TFAE for a ring $R$ (with identity)


*

*$R$ has a unique maximal right ideal

*The nonunits of $R$ are closed under addition

*for every $x\in R$, either $x$ or $1-x$ is a unit.
They're very manageable for any abstract algebra student, but hints can be supplied if you ask.


If $R$ is a local ring, then so is $R[[x]]$.

Suppose $R$ is local and let $I$ is a maximal right ideal of $R[[x]]$. Then the projection of $R[[x]]\to R$ carries $I$ to the unique maximal right ideal $M$ of $R$. Then the inverse image of $M$ under this projection is also a right ideal containing $I$, and therefore it's equal to $I$. Thus all maximal right ideals of $R[[x]]$ are equal to the preimage of $M$, so there is only a single maximal right ideal in $R[[x]]$.
