7
$\begingroup$

Assume the definition where a ring has a unity. I assume that under this definition the trivial ring $\{0\}$ is a ring. Here the multiplicative identity and the additive identity are the same.

I have read the definition of a subring that it is a subset that is itself a ring under the same operations. My question is whether or not the trivial ring $\{0\}$ is a subring of any ring $R$. The reason for my confusion is that in $\{0\}$ $0$ is also the multiplicative identity, while the ring $R$ might have different identities. It should be clear that the trivial ring is a subring, but I am not sure. According to the definition on Wikipedia it sounds like the subring need to contain both the additive identity and multiplicative identity which $\{0\}$ doesn't.

$\endgroup$
  • 3
    $\begingroup$ IMO the definition of a subring usually includes that the identity element of the superring should be an element of the subring. That's the one I used when I teach a course where structures such as rings are introduced. Some authors may use a different convention. Obviously including those who don't insist that a ring has a unity in the first place. $\endgroup$ – Jyrki Lahtonen Nov 27 '17 at 13:18
12
$\begingroup$

There are several definitions running around, confusing people like you when you read from different sources. For some of them the answer is "Yes", for some the answer is "no".

Some definitions of unital rings wouldn't consider $\{0\}$ a ring at all, since they require that the multiplicative identity is distinct from the additive one. Disregarding that, a subring is a subring by virtue of the inclusion homomorphism, and some sources require that ring homomorphisms preserve the multiplicative identity. In that case, $\{0\}$ is all by itself without any homomorphisms of any kind to any other ring, so it's not included in any other ring either. However, that's usually unporoblematic since these two definitions often come together (so if your definition of homomorphisms would make $\{0\}$ sit alone, your definition of ring would make it not a ring at all).

So, if we let $\{0\}$ be a ring, and we don't require homomorphisms to treat the multiplicative identity with any more care than any other element, then yes, you can include $\{0\}$ into bigger rings and have it be a subring. The same way you can include $\Bbb Z$ as the first component of $\Bbb Z\times \Bbb Z$, for instance.

When I first learned about rings, I prefered the more relaxed requirements, because I thought they made life easier. These days I prefer the stricter requirements because I feel that they make life easier (knowing that the image of $1$ is $1$, and not just any idempotent element really helps some arguments, for instance).

Note that if your definition of rings requires a multiplicative identity to exist in any ring, then you will most likely be in the stricter domain (requiring $1$ to exist means it is nice if we require homomorphisms to respect it). The more relaxed domain typically doesn't require a multiplicative identity to exist in rings (although, of course, some rings happen to have one).

$\endgroup$
  • $\begingroup$ I hadn't thought about the inclusion homomorphism. Thanks for the answer. $\endgroup$ – John Doe Nov 27 '17 at 13:42
  • 2
    $\begingroup$ In your second paragraph, the trivial mapping $X \to \{0\}$ given by $x \mapsto 0$ is a ring homomorphism (including conservation of the multiplicative identity), so the "from" part of without any homomorphisms of any kind to or from any other ring is wrong, and this is a terminal object in the category. $\endgroup$ – Paŭlo Ebermann Nov 27 '17 at 19:25
2
$\begingroup$

No, $\{0\}$ is not a subring of any nontrivial ring.

The language of ring theory is $\{0,1,+, -, \cdot\}$, where $0,1$ are constant symbols and $+, \cdot$ are $2$-ary function symbols and $-$ is a $1$-ary function symbol, mapping each element to its additive inverse.

Now let $R$ be a nontrivial ring. Then $0,1 \in R$ and $0 \neq 1$. If $S$ is a subring of $R$, we really mean that the set $S$ is a $\{0,1,+, -,\cdot\}$-substructure of $R$, i.e. it contains the correct interpretations of our constant symbols (here $0,1 \in S$) and is closed under $R$'s addition, multiplication and additive inverses. In particular, $\{0,1 \} \subseteq S$.

$\endgroup$
  • $\begingroup$ I find it helpful to regard sub-groups/rings/fields/graphs/... as substructures, in the model theoretic sense. While it may require some additional effort to understand this concept at first, it greatly simplifies the overarching concept (and teaches you to care about the particular language we silently associate to a given first order structure). $\endgroup$ – Stefan Mesken Nov 27 '17 at 13:25
  • 2
    $\begingroup$ How is $\{0,1\}$ always a subring? For example, with $R = \mathbb{Z}$ if $1$ is in the subring, doesn't $2$ have to be there too? $\endgroup$ – John Doe Nov 27 '17 at 13:46
  • 2
    $\begingroup$ @JohnDoe Yes, of course. I don't know what I was thinking. $\endgroup$ – Stefan Mesken Nov 27 '17 at 13:55
0
$\begingroup$

The trivial ring is the only ring in which the multiplicative identity and the additive identity are identical. Therefore, this ring can only ever be a subring of itself.

$\endgroup$
0
$\begingroup$

A sub$\langle$structure$\rangle$ $B$ of a $\langle$structure$\rangle$ $A$ is always subset of $A$ that is itself a $\langle$structure$\rangle$.

Now consider some ring $R$. In $R$ we either have $1\neq0$ or $1=0$. The equality or nonequality of $1$ and $0$ in a ring is part of the structure of the ring (even if $1\notin R$).

A subset of the ring $R$ must have the same structure as $R$. So $$\begin{cases} \text{if }1\neq 0 \implies \text{$\{0\}$ is not a subring of $R$}\\ \text{if }1= 0 \implies \text{$\{0\}$ is a subring of $R$ (because $R=\{0\}$ of course)}. \end{cases} $$

$\endgroup$
  • 1
    $\begingroup$ I don't get this answer. "The equality or nonequality of 1 and 0 in a ring is part of the structure of the ring (even if 1∉R). " What does this mean? Why is this part of the structure? Which other properties are part of the structure? $\endgroup$ – quid Nov 27 '17 at 18:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.