Are all (commutative) rngs ideals of (commutative) rings? [closed]

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.

closed as too localized by Benjamin Steinberg, Yemon Choi, Harry Gindi, Bill Johnson, Mark SapirDec 24 '11 at 2:02

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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 $(s,n)(s',n')=(ss'+ns'+sn',nn')$.
• 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
• @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