There is a classical theorem that no field can be expressed as finite union of proper subfields.

In contrast, there is an example of an integral domain that can be expressed as finite union of proper subrings.

Therefore, I wonder whether there are any known results about the existence of Principal ideal domain (or even Euclidean domain) that can be expressed as finite union of proper subrings?

  • $\begingroup$ The example you link to – is it known that it's never a principal ideal domain? $\endgroup$ Jan 17, 2019 at 12:11
  • $\begingroup$ @GerryMyerson: That example is not integrally closed: $D^{1/3}$ is in the fraction field but not in the ring. $\endgroup$ Jan 17, 2019 at 12:46
  • 1
    $\begingroup$ Consider the PID $R=\mathbb{F}_2[t]$. Given a $1$-dimensional $\mathbb{F}_2$-subspace $V \subseteq \mathbb{F}_2^2$, let $R_V=\{f\in R: (f(0),f(1))\in V\}$. Then $R=\bigcup_V R_V$, where $V$ ranges over the $1$-dimensional subspaces of $\mathbb{F}^2_2$. $\endgroup$ Jan 17, 2019 at 13:24
  • $\begingroup$ Actually, $R_V$ is not a ring since $1\notin R_V$ if $(1,1)\notin V$. Sorry. This question is more tricky than it seems. However, if $\mathbb{F}_2^n$ can be expressed as a union of proper subrings for some $n$, then this should give the required example. $\endgroup$ Jan 17, 2019 at 15:36

1 Answer 1


Consider the following subrings of $\mathbb{F}_2^3$: $$ V_1=\{(0,0,0),(1,1,1),(1,0,0),(0,1,1)\}\\ V_2=\{(0,0,0),(1,1,1),(0,1,0),(1,0,1)\}\\ V_3=\{(0,0,0),(1,1,1),(0,0,1),(1,1,0)\} $$ Then $\mathbb{F}_2^3=V_1\cup V_1\cup V_3$.

Let $R$ be a PID admitting $3$ distinct surjective ring maps $f_1,f_2,f_3$ onto $\mathbb{F}_2$ and write $$ R_i=\{a\in R: (f_1(a),f_2(a),f_3(a))\in V_i\} $$ for $i=1,2,3$. Then $R_1,R_2,R_3$ are subrings of $R$ with $R=R_1\cup R_2\cup R_3$. They are proper subrings by the Chinese Remainder Theorem.

To construct such $R$, take a cubic extension $K$ of $\mathbb{Q}$ in which $2$ splits into a product of $3$ primes $P_1,P_2,P_3$, and choose $R$ to be the localization of $\mathcal{O}_K$ at the multiplicative set $R\setminus(P_1\cup P_2\cup P_3)$.

  • $\begingroup$ The existence of $R$ can also be shown using a theorem of Heitmann. See this answer: mathoverflow.net/questions/56513/exotic-principal-ideal-domains/… $\endgroup$ Jan 17, 2019 at 16:10
  • $\begingroup$ Sorry for accepting the answer after such a long time as it‘s not so obvious to me to prove the existence of such $3$ distinct surjective ring homomorphisms from $R$ onto $\mathbb{F}_2$. Besides, for the 'Euclidean Domain' part, may I ask whether it can be expressed as finite union of proper subrings? $\endgroup$ Jan 28, 2019 at 2:57
  • 1
    $\begingroup$ @SteveJacob No problem. I agree that it is not easy to write down an explicit example. Concerning your question about a Euclidean example, every semilocal PID, e.g., as in the end of my answer, is Euclidean. Indeed, if $P_1,\dots,P_t$ are the nonzero prime ideals in $R$, then the sum of their corresponding additive valuations is a Euclidean function on $R$ (so says Wikipedia). $\endgroup$ Jan 28, 2019 at 11:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.