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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.

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    $\begingroup$ As a base case, it's true for commutative unital rings since every such ring is a $\mathbb{Z}$-algebra. $\endgroup$ Apr 19 at 10:57
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    $\begingroup$ What does "free algebra" mean to you? Because in general $R$ is the free $R$-algebra on the empty set, giving a trivial answer to your question. But the answer is still yes if you mean for example "free noncommutative $\Bbb Z$-algebra", since the ideal of $\Bbb Z\langle X_i : i \in I \rangle$ generated by the $X_i$ is a free nonunital-noncommutative-ring on $X$, I believe. $\endgroup$ Apr 19 at 11:38
  • $\begingroup$ @IzaakvanDongen Thank you. I updated the question to cover that. Can you please elaborate on how the quotient is taken? $\endgroup$
    – user760
    Apr 19 at 12:21
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    $\begingroup$ I think first it helps if you understand what a free nonunital-noncommutative-ring on a set $X$ looks like. Do you have any guesses as to what universal property such an object should have, and how you would show such an object exists? Do you see why every nonunital noncommutative ring is a quotient of such an object for some $X$? $\endgroup$ Apr 19 at 12:36

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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

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  • $\begingroup$ Thank you. I think my difficulty in understanding this is, for a ring, there are two algebraic operations, and if we take the forgetful functor to the category of sets, are we guaranteed to have a "left adjoint" without hassles? Perhaps we should forget one algebraic structure at a time? $\endgroup$
    – user760
    Apr 19 at 13:24
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    $\begingroup$ No, we can just forget all of them at the same time (even if the signature is infinite). This is the magic of the adjoint functor theorem. It only requires the forgetful functor to have very basic properties such as preserving the product (and some other more technical conditions but easy to check for universal algebras). Intuitively, I guess from the Yoneda lemma, all information about the structure has been encoded in the hom sets. $\endgroup$ Apr 19 at 13:31
  • $\begingroup$ Thank you for the reply! I'm only exposed to some terminologies from category theory, but never really learned it. Though, if I'm not mistaken, for topological vector spaces, freeness is only defined as the left adjoint to a less thorough forgetful functor, one that only takes the TVS to Top, or the category of Hausdorff uniform spaces. If there's also completeness requirement, then it's even more restricted, where you only arrive at the category of complete Hausdorff uniform spaces. $\endgroup$
    – user760
    Apr 19 at 14:20
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    $\begingroup$ The point of my previous comment is that, sometimes, people don't take the forgetful functor to Set. So I'm not sure what to expect when thinking of "freeness". $\endgroup$
    – user760
    Apr 19 at 14:22
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    $\begingroup$ @user760 I see what you mean. I only know one such (useful) example: Free profinite groups generated by a (pointed) profinite topological space. I guess in the most general setting, any left adjoint functor (not necessarily to a forgetful functor) defines a "freely" generating machine. I guess "free = left adjoint" morally. $\endgroup$ Apr 19 at 14:50

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