The prime numbers are numbers that are divisible by exactly two numbers. Fields are rings that have exactly two ideals. Moreover, the number of ideals of the product of two rings is the product of the number of ideals of both rings. A natural question one may hence ask is this:

Is every ring a product of fields?

In general, the answer is no, for the following reason: Obviously, the characteristic of a product of rings is the least common multiple of the characteristics of the factors. Hence, rings like $\mathbb Z / 4\mathbb Z$ are not achievable in this way. Moreover, the characteristic of any ring that is the products of fields will have the form $p_1 \cdots p_n$ (where $p_1, \ldots, p_n$ are distinct primes). Moreover, for noncommutative rings, we'll have to allow skew-fields. Hence, the question above becomes the following two questions:

  • What is the class of rings that is the product of (skew-)fields?
  • Is there a decomposition of arbitrary rings into (skew-)fields (that necessarily is not the product)?
  • 2
    $\begingroup$ If you're thinking of fields as "rings with no non-trivial ideals", then you may (though not necessarily) want the non-commutative generalisation to be simple rings - that is, rings with no non-trivial two-sided ideals - rather than skew fields, which are the non-commutative rings with no non-trivial one-sided ideals. $\endgroup$ Jul 23 '18 at 9:10

I'm not aware of any far-reaching characterization of rings in terms of fields (or divison rings.)

Perhaps the most fundamental structure theorem to know in noncommutative algebra is that a semisimple Artinian ring is a finite product of matrix rings over division rings.

From this you could say that the reduced (=no nonzero nilpotent elements) semisimple Artinian rings are exactly finite products of division rings.

Then one could ask about general products of division rings. The only thing along these lines that I know is that every strongly regular ring (=von Neumann regular and reduced) is a subdirect product of division rings. (Subdirect products are a generalization of ordinary products.)

Until this time, we have not considered any very exotic rings. semisimple Artinian and strongly regular rings are among the "nice" rings out there. We haven't gone very far into the wilderness at all.

Departing from division rings, it is also interesting to ask more generally about local rings. I think I'm remembering correctly that commutative semiperfect rings are exactly the rings which decompose into finite products of local rings. Analogously above, I know there is a lot of work showing which rings are matrix rings over local rings of one type or another. And again, it might be interesting to ask which rings are subdirect products of local rings (commutative or otherwise.)


One general theorem in that direction is the decomposition of Artinian rings:

Every commutative Artinian ring is isomorphic to a finite direct product of Artinian local rings


Every simple Artinian ring is a matrix ring over a division ring

and so

Every semisimple Artinian ring is a finite product of matrix rings over division rings


There are much better reasons than just the characteristic.

  • A product of fields has always zero Jacobson radical; in particular a local ring cannot be a product of fields unless it is a field.
  • A product of (at least two) fields is not an integral domain.
  • A product of fields is commutative.

However, any commutative ring $R$ with zero Jacobson radical is a subdirect product of fields, which means that there is a family $(F_i)_{i\in I}$ of fields and an injective homomorphism $$ f\colon R\to\prod_{i\in I}F_i $$ such that $p_i\circ f$ is surjective for every $i\in I$. With $p_i$ I denote the projection from the product to the $i$-th factor.

This is essentially trivial, though: let $(M_i)_{i\in I}$ be the family of the maximal ideals of $R$ and define $F_i=R/M_i$. Then $$ f(r)=(r+M_i)_{i\in I} $$

  • $\begingroup$ Where, of course, the kernel of f is the Jacobson radical. Very nice. And the loss of the integral domain comes from (0,1)(1,0)=(0,0). $\endgroup$ Jul 23 '18 at 13:19
  • $\begingroup$ I'm just noticing that the converse holds as well: The maximal ideals of a product of fields are precisely those ideals that are all of the field except in one place, and then their intersection is zero. Hence we characterized subdirect products of fields as commutative rings with zero Jacobson radical. $\endgroup$ Jul 23 '18 at 13:34
  • $\begingroup$ Finally, it is in fact a direct product, because abelianness and that ideals are subgroups and the fact that the projection on each factor is surjective imply that the product is in fact direct (by a group theoretic statement). Hence, the classification is complete. $\endgroup$ Jul 23 '18 at 13:37
  • $\begingroup$ @AlgebraicsAnonymous It is not, generally, the whole product. Try with $R=\mathbb{Z}$. $\endgroup$
    – egreg
    Jul 23 '18 at 13:38
  • $\begingroup$ What do you mean? We already excluded integral domains. $\endgroup$ Jul 23 '18 at 13:39

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.