Let $R$ be a ring. Let $A$ be a commutative subring of $R$. Show that $A$ is contained in a maximal commutative subring of $R$.

I'd like to use Zorn's Lemma.

I'd like to start by proving that the family $F$ of commutative subrings of R that contain A is inductive.

In other words, every chain in $F$ (ordered by inclusion) has a maximal element.

But what is the connection between commutativity and this inductive property?


2 Answers 2


Edit: Due to the comments of user119598 and 120579, the answer provided below may be incorrect (in some cases). If by "maximal commutative subring" the subring is not required to be a proper subset of $R$, then the original argument (below the dashed line) is indeed correct. However it seems standard to require that a maximal subring is required to be a proper subset, and for the rest of this edit I will be using this definition.

If $R$ is not commutative, then the original argument is again correct. In this case, the union of a chain of commutative subrings is commutative and hence can not be all of $R$.

If $R$ is commutative, then the argument is not correct as we can not guarantee the union is a proper subset as pointed out in the comments below. Actually, in this case the statement may be false. A maximal commutative subring is just a maximal subring of the commutative ring $R$. The abstract of the paper Most Commutative Rings Have Maximal Subrings by Azarang and Karamzadeh seems to indicate that there are commutative rings for which a maximal subring does not exist. Unfortunately my university does not have access to the journal and I have been unable to get my hands on the paper.

To show that every chain has a maximal element, given a chain you should be able to find (e.g. construct) the maximal element. Hint: If $R$ has a 1, this is similar to how one shows that every ideal is contained in a maximal ideal.

Once you have this candidate, you must show it satisfies the properties required: i.e. you must show it is commutative.

  • $\begingroup$ I'll take a guess and let my candidate be the union of the elements of my chain. $\endgroup$ Nov 25, 2013 at 19:49
  • $\begingroup$ I solved it using this result: proofwiki.org/wiki/Increasing_Union_of_Subrings_is_Subring Thanks. $\endgroup$ Nov 25, 2013 at 20:12
  • $\begingroup$ How do you know that the "candidate" is not equal with the whole ring? $\endgroup$
    – user119598
    Jan 10, 2014 at 23:38
  • $\begingroup$ @user119598 Good catch! Indeed, the family $F$ should be defined as proper subrings of $R$. $\endgroup$
    – RghtHndSd
    Jan 10, 2014 at 23:50
  • $\begingroup$ Dear @rghthndsd, even so there is no guarantee that the union is not equal with $R$. $\endgroup$
    – user120579
    Jan 11, 2014 at 11:50

A general example of an infinite commutative ring R with arbitrary infinite cardinal number $\alpha$ that has no maximal subring (and therefore each proper subring of R can not embedd in a maximal one) : first note that if A is a ring and M is an A-module then the A-subalgebra of the idealization R=A(+)M has the form S=A(+)N where N is an A- submodule of M, in particular S is maximal if and only if N is maximal. Now let A be the ring of integer and M be a field of zero characteristic with cardinality $\alpha$, then R has the same cardinality and it is well-known that M has no maximal A-submodule, hence R has no maximal subrings (note, subring means a unitrary subrings, i.e., has a same identity as R which is not zero, clearly in this case each subring of R contains A and hence is an A-algebra).


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