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To avoid repeating it endlessly, assume all rings and rngs are commutative. I do not know if this is necessary.

The question then is exactly the title, but I think a stronger statement is true:

For any rng $S$ there is a ring $R$ and an injective rng-homomorphism $f:S\rightarrow R$ such that for any ring $T$ and any rng homomorphism $g:S\rightarrow T$, there is a ring homomorphism $h:R\rightarrow T$ such that $h$ extends $g$.

In fact I think the construction is pretty clear; let $X=( x_s : s\in S )$ be a set indexed by $S$, and let $R=\mathbb Z[X]/I$, where $I=( x_a+x_b-x_{ab} : a,b\in S) \cup (x_a*x_b-x_{ab} : a,b\in S)$.

It seems clear that if a universal object can exist, this has to be it. But I'm having trouble proving the natural map $f:S\rightarrow R$ (given by $f(a)=s_a$) is actually injective like it ought to be. Is there some classical universal property I'm missing here, or is there a slick way to ignore the details?

Also, I don't think the commutativity is at all necessary for the problem, it's just the situation I'm most used to. I think a similar construction (the free algebra on $S$ and $1$, modulo the same $I$) would do fine for the noncommutative case, and is isomorphic to this in the commutative case.

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As Fernando Muro points out, every ring (unital or otherwise) is an ideal in its "forced" unitization. (I apologize if my terminology is anqtiquated, but then again I don't study Banach algbras) –  Yemon Choi Dec 22 '11 at 0:10
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See also these questions and their answers: mathoverflow.net/questions/22579/… mathoverflow.net/questions/34332/… Especially this answer: mathoverflow.net/questions/34332/… –  Emerton Dec 22 '11 at 9:02
    
See also Wikipedia's entry on adjoint functors en.wikipedia.org/wiki/Adjoint_functors#in_algebra for some further information/background –  Yemon Choi Dec 22 '11 at 12:39
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closed as too localized by Benjamin Steinberg, Yemon Choi, Harry Gindi, Bill Johnson, Mark Sapir Dec 24 '11 at 2:02

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1 Answer

up vote 8 down vote accepted

You're wondering about the existence of a left adjoint to the forgetful functor from rings to rngs. Of course it exists. It sends a rng $S$ to $R=S\oplus \mathbb{Z}$ with multiplication $(s,n)(s',n')=(ss'+ns'+sn',nn')$.

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The "Dorroh extension" above completely answers the question, but I wanted to tack on this interesting paper I ran across recently: Burgess, W. D.(3-OTTW); Stewart, P. N.(3-DLHS) The characteristic ring and the "best'' way to adjoin a one. J. Austral. Math. Soc. Ser. A 47 (1989), no. 3, 483–496. In addition to the Dorroh extension, I think they argue for another economical adjunction of one. I'm not really up to speed on the categorical properties of either extension, but it all looks pretty interesting. –  rschwieb Dec 22 '11 at 2:59
    
If you are working in a setting with a version of Gelfand theorem, then commutative rngs are like non-compact spaces, and rings are the compact ones. Fernando's unitalization corresponds to adding a point, such that it is in any open neighborhood which is the complement of a compact set of your original space. There is another compactification, due to Stone and Cech. On the ring side, this corresponds to replacing the rng $S$ with the ring of $S$-module maps $S \to S$. I think this is the other adjoint, and probably the one rschwieb likes. –  Theo Johnson-Freyd Dec 22 '11 at 4:23
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@Yemon Choi: For what it's worth, sometimes a forgetful functor can have two adjoints -- a left adjoint and a right adjoint -- which typically are quite different from each other. For example, if $R \rightarrow S$ is a ring homomorphism and $U: {}_SMod \rightarrow {}_RMod$ is the forgetful functor, then $F := (S \otimes_R -): {}_RMod \rightarrow {}_SMod$ is the left adjoint and $G := Hom_R(S, -): {}_RMod \rightarrow {}_SMod$ is the right adjoint of $U$. In OP's case, perhaps the Dorroh extension is the left adjoint and the Stone-Cech-ish extension is right adjoint? –  Neil Epstein Dec 22 '11 at 11:22
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I didn't think hard about the actual adjunctions, and meant only to mention that there's a close analogy between some basic theory of rings and some basic theory of spaces. Of course @Yemon is correct that there are many more rng homomorphisms Forget(Z) -> Z+Z than there are ring homomorphisms Z -> D(Z+Z). (There are four of the former.) But actually, this makes me doubt that there is any right adjoint, because there's never more than one ring homomorphism Z -> anything, and often more than one from Forget(Z). Of course, left adjoints also have to be cocontinuous, and I think Forget is not. –  Theo Johnson-Freyd Dec 23 '11 at 6:19
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So I should have thought a moment longer, and never mentioned an adjoint on the other side. What confused me is that the analogy between rngs and spaces isn't perfect, and is not functorial in this set-up. The Stone-Cech compactification of spaces is the left adjoint to forgetting that a compact space is compact, whereas I think the one-point compactification is not an adjoint on either side. In fact, the one-point compactificiation isn't even functorial! –  Theo Johnson-Freyd Dec 23 '11 at 6:24
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