Abstract Algebra – Is Every Ring a Homomorphic Image of Some Ideal?


Since every group is the homomorphic image of a free group, and every module is the homomorphic image of a free module, do we have an analogous result for the rings? To take into account of nonunital rings, I wonder if we can always start from an ideal, or at least a subring, of a free algebra, and get to a given ring by quotiening out some ideal of the ideal/subring? Thanks.

Here "free algebra" is meant to refer to the noncommutative ring of multivariable polynomials over $\mathbb{Z}$, with the concatenation as the multiplication and empty set as the multiplicative identity. (But perhaps can also allow to be over a field or even a PID?)

Also, here, by "homomorphic image", I mean a "ring homomorphic image", not just a module homomorphic image.

Best Answer

This is definitely true. In fact, this is true for universal algebras of arbitrary signature, hence for e.g. $A$-algebra (where $A$ is an arbitrary ring, could be a field or PID or whatever), (noncommutative) rngs, semirings, and Lie algebras as well. Those "free" objects in universal algebras are called "Term algebra".

This is ususally proved with a strong flavor of syntax, like using "words" in group theory. There is also a proof with category theory: Let $\mathcal C$ be the category of certain universal algebras, and $F:\mathcal C\rightarrow Set$ be the forgetful functor. If $F$ has a left adjoint $G$, then $$\text{Hom}_{\mathcal C}(GX, Y)\simeq \text{Hom}_{Set}(X, FY)$$

can be interpreted as $GX$ is the free object generated by $X$. The existence of $G$ can be established by the adjoint functor theorem. Details can be found (in the case of groups) in

Barr, M. (1972). The Existence of Free Groups. The American Mathematical Monthly, 79(4), 364–367. https://doi.org/10.1080/00029890.1972.11993050