Question

Verify that K[[X]] is a local ring,where K is a field

Answer 1

Answer on Question # 83120, Math / Abstract Algebra

Question 1. Verify that $k[[X]]$ is a local ring, where $k$ is a field.

Solution. Let $k$ be a field and let $X$ be an indeterminate; denote by $k[[X]]$ the set of all formal expressions

$\sum_{n=0}^{\infty}a_{n}X^{n},\,a_{n}\in k.$

More precisely, $k[[X]]$ as a $k$-vector space is the direct product of a denumerable number of copies of $k$ indexed by the set of monomials $X^{n},\,n\geqslant 0$. $k[[X]]$ is made into a ring by defining addition and multiplication exactly as for polynomials except there is no restriction that the result must have all but a finite number of coefficients . In the formula for the product

$\left(\sum_{i}a_{i}X^{i}\right)\left(\sum_{j}a_{j}X^{j}\right)=\sum_{n}\left(\sum_{i+j=n}a_{i}b_{j}\right)X^{n}$

is a sum with only a finite number of nonzero terms in any case, so the product is well defined.

$U(k[[X]])$ is the set of $\sum_{n}a_{n}X^{n}$ with $a_{0}\in k^{*}$. For suppose $\left(\sum_{i}a_{i}X^{i}\right)\left(\sum_{j}a_{j}X^{j}\right)=1$. Using the above formula yields

$a_{0}b_{0}=1,\quad\sum_{i+j=1}a_{i}b_{j}=0\quad\text{for}\,n>0.$

The first equation may be solved for $b_{0}$ if and only if $a_{0}$ is a unit in $k$. In that case, the remaining equations may then be solved recursively $b_{n}=-a_{0}^{-1}(a_{1}b_{n-1}+\cdots+a_{n}b_{0})$ and the resulting formal series $\sum_{n}b_{n}X^{n}$ is easily seen to be the inverse of $\sum_{n}a_{n}X^{n}$. It follows that the complement of $U(k[[X]])$ is the set of $\sum_{n}a_{n}X^{n}$ with $a_{n}=0$, and that is an ideal, the ideal generated by $X$.

Hence, $k[[X]]$ is a local ring with unique maximal ideal, the ideal generated by $X$. $k[[X]]$ is called the ring of formal power series in the indeterminate $X$. $\Box$