Is direct limit of local rings a local ring?

Question

Let $\{R_i\}_{i\in A}$ be a directed set of commutative local rings with directed index set $A$, and let $R$ be the direct limit of this set. I want to know if $R$ is a local ring (we know that $R$ is a commutative ring).

Answer 1

Suppose $\{R_i\}$ is a directed system of local rings, let $R:=\varinjlim R_i$ and suppose further $R\neq \{0\}$.

Consider $\mathfrak{m}:=R\setminus R^\times$. We will show that $\mathfrak{m}$ is an ideal and therefore $R$ is local with maximal ideal $\mathfrak{m}$.

If $x,y\in \mathfrak{m}$ then for some large enough index $i$ there is a homomorphism $\phi_i:R_i\rightarrow R$ and elements $\bar{x},\bar{y}\in R_i$ such that $\phi_i(\bar{x})=x$ and $\phi_i(\bar{y})=y$. Since $x,y\notin R^\times$, $\bar{x},\bar{y}$ are not units in $R_i$ and are therefore contained in the maximal ideal of $R_i$ say $\mathfrak{m}_i$.

Now $x-y=\phi_i(\bar{x}-\bar{y})$ is contained in $\mathfrak{m}$. To see this, note if there was an inverse $z\in R$ then $1=z(x-y)=\phi_i(\bar{z}(\bar{x}-\bar{y}))=\phi_i(w)$ for $w\in \mathfrak{m}_i$. But, $\phi_i(1)=1$ so that $\phi_i(1-w)=1-1=0$. Since $w$ is an element of the jacobson radical of $R_i$, $1-w$ is a unit of $R_i$ so $1=0$ in $R$ which we assumed was not the case.

Suppose $r\in R$ and $x\in \mathfrak{m}$ and consider the multiplication $rx$. Taking their preimages in $R_i$ for sufficiently large $i$ gives $\bar{r}\bar{x}\in \mathfrak{m}_i$ which, by a similar argument to the above, yields $\phi_i(\bar{r}\bar{x})=rx\in \mathfrak{m}$ and therefore $\mathfrak{m}$ is the unique maximal ideal of $R$.

Answer 2

A probably cleaner answer:

Let $(R_i,\mathfrak{m}_i)$ be a direct system of local rings. To each of these we get an exact sequence

$$0\rightarrow \mathfrak{m}_i\rightarrow R_i\rightarrow R_i/\mathfrak{m}_i\rightarrow 0$$

and by applying the exact $\varinjlim$ functor we obtain a sequence $$0\rightarrow \varinjlim \mathfrak{m}_i\rightarrow \varinjlim R_i\rightarrow \varinjlim R_i/\mathfrak{m}_i\rightarrow 0.$$ The nonzero object on the rightmost side is a field (direct limit of fields is a field - choose an element, it comes from a field, so has an inverse) hence the ideal $\varinjlim \mathfrak{m}_i$ is a maximal ideal. It's certainly the only maximal ideal, as every other ideal is contained in it, which completes the proof.