# Commutative rings with trivial automorphism group

The commutative rings $\mathbb{Z}/p\mathbb{Z}$, $\mathbb{Z}_{(p)}$, $\mathbb{Z}$, and $\mathbb{Q}$, where $p$ is a rational prime, all have trivial automorphism groups. Are there any other (unital) commutative rings which have trivial automorphism group?

## 1 Answer

There are many examples, including the reals $\mathbb R$ and the $p$-adic numbers $\mathbb Q_p$.

For the reals, the ordinary order topology is algebraically defined: $a\le b$ if and only if $b-a$ is a square in the field. This has the consequence that every automorphism $\varphi$ of $\mathbb R$ is continuous. Indeed, if a sequence $\{a_n\}$ has $\lim_na_n=L$, then $\lim_n\varphi(a_n)=\varphi(L)$, and conversely by looking at $\varphi^{-1}$. Since your automorphism $\varphi$ is identity on $\mathbb Q$, it also is identity on the closure of $\mathbb Q$.

Similarly, the $p$-adic topology is algebraically defined, because the maximal ideal $p\mathbb Z_p$ of $\mathbb Z_p$ has a purely algebraic characterization: an element $z$ of $\mathbb Q_p$ is in $p\mathbb Z_p$ if and only if it’s unequal to $1$ and for every $m$ prime to $p$, $1-z$ is an $m$-th power in $\mathbb Q_p$. (Of course you have to check this assertion!) But since the topology of $\mathbb Q_p$ is defined by powers of $p\mathbb Z_p$, again every automorphism of $\mathbb Q_p$ is continuous, and again since it’s identity on $\mathbb Q$, it also is on the whole field.

Note that the argument that every automorphism of the local field $\mathbb Q_p$ is continuous applies perfectly well to finite extensions of $\mathbb Q_p$, and to $\kappa((t))$, the field of formal Laurent series over a finite field $\kappa$. The finiteness of the residue field plays an essential part in the proof of my assertion above.