# 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.

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

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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
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