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