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Let $R$ be an rng. (There may be no unity)

Then, does there always exist a ring(with unity) $A$ such that $R$ is a subrng of $A$?

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    $\begingroup$ possible duplicate of Is it possible to extend a commutative ring to have a unity? $\endgroup$ – TomGrubb May 27 '15 at 18:37
  • $\begingroup$ @bburGsamohT: the question you linked refers to commutative rings; this one is more general, referring as it does to general (not necessarily commutative) rngs. Cheers! $\endgroup$ – Robert Lewis May 27 '15 at 22:37
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    $\begingroup$ @RobertLewis well, that difference is completely trivial considering that it makes no difference in the given answers. If the two questions answers were traded, then this should be closed despite the wording. However, the answers here are a bit better written, and this subsumes that one, so the other should be closed as a dupe now. $\endgroup$ – rschwieb May 28 '15 at 3:09
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One can always find a ring with unity containing a ring (without unity). There are many different ways to do that ; let's look at one of them. Let $R$ denote the ring without unity and consider $A=\mathbb{Z}\times R$ with the canonical addition and the following multiplication

$$(m,a)\cdot (n,b)=(mn, na+mb+ab)$$

$A$ with these two operations is a ring and $\left(1,0_R\right)$ is a unity for that ring. And $R\to A$ $a\to (0,a)$ is a canonical injection of $R$ in $A$

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    $\begingroup$ To give some explanation (for OP and future others) for why this definition of multiplication makes sense: if we informally considered this as a sum of two rings, then we'd want to evaluate $(m+a)(n+b)$ the obvious thing to do is $mn + m.b + n.a + ab$. $mn\in\Bbb Z$ but the other three terms are in $R$ hence the reason for collecting them the way we do. $\endgroup$ – Cameron Williams May 27 '15 at 18:49
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    $\begingroup$ It's OK if the starting ring $R$ happens to contain a unity, say $e$.It won't conflict via the embedding with the element $(1,0_R)$ since the image of $e$ is $(0,e)$ and the latter is not an identity for the full ring on $\mathbb{Z} \times R.$. $\endgroup$ – coffeemath May 27 '15 at 18:50
  • $\begingroup$ @Cameron Williams you are perfectly right. In fact it is a general construction based on the direct sum $\mathbb{Z}\oplus R$ which in our case is isomorphic to the direct product. $\endgroup$ – marwalix May 27 '15 at 18:52
  • $\begingroup$ In regard to coffemath's comment, we see that $(0, e)$ is in fact a non-unit idempotent in $\Bbb Z \times R$: $(0, e) \cdot (0, e) = (0, e^2) = (0, e)$, since $e$ is the unit of $R$ in this case. As regard to generality, it appears to me that if $S$ is unital, the set $S \times R$ with componentwise addition and multiplication $(s_1, r_1) \cdot (s_2, r_2) = (s_1 s_2, s_1 r_2 + s_2 r_1 + r_1 r_2)$ will satisfy all the axioms for a unital ring, with $1_{S \times R} = (1, 0)$. Cheers! $\endgroup$ – Robert Lewis May 27 '15 at 19:01
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For future readers, the above construction can be greatly generalized.

Let $R$ be a rng (may have no unity) and $(M,+,•)$ be an $R$-module.

Define $S=\mathbb{Z}\times R$ and define operations as $(n,a)+(m,b)=(n+m,a+b)$ and $(n,a)(m,b)=(nm,ma+nb+ab)$.

Then, $R$ is a subrng of $S$ with respect to an embedding $r\mapsto (0,r)$.

Now, define $(n,a)\ast x = a•x + nx$ where $x\in M$.

Then $\ast$ indeed induces $M$ to be an $S$-module and $•$ is the restriction of $ast$ on $R\times M$.

To sum up, the following is true:

Let $R$ be an rng and $(M,+,•)$ be an $R$-module.

Then, there exists a ring $S$ and an operation $\ast:S\times M\rightarrow M$ such that $(M,+,\ast)$ is an $S$-module and $\ast\upharpoonright (R\times M)=•$.

This means that, defining module on rngs (may be without unity) is equivalent to defining module on rings (with unity).

So, by defining modules on rings with unity, one can handle both two cases.

Moreover, here is one application of the original one : "every rng is a subrng of a ring with unit"

Let $R$ be an rng and consider a polynomial rng $R[X]$.

Since $R$ may not have a unity, you cannot formally define $X$ as an element of $R[X]$.

However, by extending $R$ to a ring $S$, since $R[X]$ is a subrng of $S[X]$, you can consider $X$ as an actual object!

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