Here is the question I want to answer letter $(b)$ in it:
A commutative ring $R$ is local if it has a unique maximal ideal $\mathfrak{m}.$ In this case, we say $(R, \mathfrak{m})$ is a local ring. For example, if $R$ is a field, then $(R,(0))$ is a local ring, since the only proper ideal of a field is $(0).$
$(a)$ Let $(R, \mathfrak{m})$ be a local ring. Show that $R^* = R\setminus \mathfrak{m}.$
$(b)$ Show that, for a field $K,$ $R = K[[x]]$ is a local ring.
Hint: According to part $(a),$ $\mathfrak{m} = R\setminus R^{*}$ and you know what $R^*$ is.
My questions are:
I already know the proof for letter $(a).$ Also, I have proved thoroughly before that:
If $R$ be an integral domain and let $R[[x]]$ be the corresponding ring of formal power series,then $R[[x]]$ is an integral domain. and $R[[x]]^*$ consists of the series $\sum_{n \geq 0}a_{n}x^n (a_{n} \in R)$ such that $a_{0} \in R^*.$
And I have the following hint given to solve my question:
Hint: According to part $(a),$ $\mathfrak{m} = R\setminus R^{*}$ and you know what $R^*$ is.
1-But I do not understand how to use it. Could anyone show me how can I use this hint please?
Also, I understood that I should prove that $K[[x]]$ has a unique maximal ideal.
And according to the hint given here by Arthur:
The set of formal power series over a fieldis a local ring? which is:
" Hint: take an element with non-zero constant term, and construct an explicit inverse, degree by degree (or at least show that it can be done, by finding the first three or so terms of the inverse and point out that you can keep going indefinitely). This shows that $(x)$ is the only maximal ideal."
I should construct an explicit inverse of an element $x$ with a non-zero constant term and it will be the only maximal ideal $<x>$.
2-I do not know what is explicityly the form of this ideal and I do not know how to prove that this is the only maximal ideal, could anyone show me the proof of this please?
Here is my detailed proof for $R[[x]]^*$ consists of the series $\sum_{n \geq 0}a_{n}x^n (a_{n} \in R)$ such that $a_{0} \in R^*.$
Let $R$ an integral domain (commutative division ring with no zero divisors), and let $R[[x]]$ be the corresponding ring of formal power series. i.e. $$\displaystyle R[[x]]=\left\{\sum_{n=0}^{\infty}a_n x^n\;\middle\vert\; a_n\in R\right\}$$ With addition and multiplication as defined for polynomials.
\textbf{First: showing that if $a_0\in R$ is a unit, then $\displaystyle\sum_{n=0}^{\infty}a_n x^n$ is a unit in $R[[x]]$}
Let $a=\sum_{n=0}^\infty a_nx^n\in R[[x]]$, where $a_0$ is a unit. We want to construct some $b=\sum_{n=0}^\infty b_nx^n\in R[[x]]$ such that $ab=1$, or after expanding, $$ab=a_0b_0+(a_1b_0+a_0b_1)x+\cdots=1+0x+0x^2+\cdots \quad \quad (1)$$ We therefore need $b_0=a_0^{-1}$ (recall that $a_0$ is a unit by the given). We want to have $a_1b_0+a_0b_1=0$, so our only choice for $b_1$ is $$b_1=\frac{-a_1b_0}{a_0}=-a_1a_0^{-2}.$$Also, We want $a_2b_0+a_1b_1+a_0b_2=0$, so we must have $$b_2=\frac{-a_2b_0-a_1b_1}{a_0}=-a_2a_0^{-2}+a_1^2a_0^{-3}.$$ So, to find a recursive definition of $b_{n}$ we will use the definition of multiplication in the ring of formal power series, we have that $$\sum_{n = 0}^{\infty}a_n x^n . \sum_{n = 0}^{\infty}b_n x^n = \sum_{n\geq 0} (\sum_{i=0}^{n} a_{i} b_{n-i})x^n = a_{0}b_{0} + (a_{0}b_{1} + a_{1}b_{0})x + \cdots + (\sum_{i=0}^{k}a_i b_{k-i}) x^k + \cdots .$$ Now, we need in our problem here $ab = 1,$ i.e. $$ a_{0}b_{0} + (a_{0}b_{1} + a_{1}b_{0})x + \cdots + (\sum_{i=0}^{k}a_i b_{k-i}) x^k + \cdots = 1, $$ So we need all terms except the constant term to vanish.
Assume that for some natural number $n,$ we know that the coefficients of $b$ are nonzero up to $(n-1),$ then the $n^{th}$ coefficient of $ab$ is zero. So, we can write $$0 = a_{0}b_{n} + a_{1}b_{n-1} + \cdots + a_{n-1}b_{1} + a_{n}b_{0},$$Or equivalently, $$ a_{0}b_{n} = -( a_{1}b_{n-1} + \cdots + a_{n-1}b_{1} + a_{n}b_{0}),$$Hence, $$ b_{n} = \frac{-1}{a_{0}}( a_{1}b_{n-1} + \cdots + a_{n-1}b_{1} + a_{n}b_{0}) = \frac{-1}{a_{0}} (\sum_{i=1}^n a_{i} b_{n-i}). $$And this is the recursion relation describing the coefficients $b_{n}$ of $b$ that will make $b$ an inverse of $a.$
\textbf{Second: showing that if $a = \displaystyle\sum_{n=0}^{\infty}a_n x^n$ is a unit in $R[[x]]$ then $a_0\in R$ is a unit}
Assume that $a = \displaystyle\sum_{n=0}^{\infty}a_n x^n$ is a unit in $R[[x]]$ and we want to show that $a_0\in R$ is a unit.
Since $a$ is a unit, then $\exists b = \displaystyle\sum_{n=0}^{\infty}b_n x^n\in R[[x]]$ such that $ab =1.$ But this means that $(a_0 + a_1 x+\cdots)(b_0 + b_1 x+ \cdots)=1+0x+\cdots$ so $a_0b_0+ (a_1b_0+a_0 b_1)x+\cdots=1+0x+\cdots$ therefore $a_0b_0=1$ and hence $a_{0}$ is a unit as required.
Using your hint, $M=K[[x]]\setminus K[[x]]^*$ is the candidate for the unique maximal ideal. Now, $M$ is the set of formal power series with constant term equal to zero. That means that your elements in $M$ have the form $$a=a_1x+a_2x^2+\cdots=x(a_1+a_2x+\cdots)=xb,$$ so they are in the ideal generated by $x$, which is denoted by $(x)$. Actually, $M=(x)$. This ideal is maximal since the quotient $K[[x]]/(x)=K$ is a field (you can think of the quotient, as the ring $K[[x]]$ with the condition that $x=0$, so you get $K$. Another way to see this, is taking the evaluation map $e_0:K[[x]]\rightarrow K$ that takes a formal power series and evaluate it at $0$. That is a homomorphism with kernel $(x)$). Finally, any ideal (maximal or not) which is not the entire ring, does not have invertible elements, that is, it is contained in $M$. Hence, $M$ is the only maximal ideal.