Is the trivial ring a subring?

Question

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.

Answer 1

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).

Answer 2

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, $\{0,1 \} \subseteq S$.

Answer 3

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.

Answer 4

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 $$\begin{cases} \text{if }1\neq 0 \implies \text{$\{0\}$ is not a subring of $R$}\\ \text{if }1= 0 \implies \text{$\{0\}$ is a subring of $R$ (because $R=\{0\}$ of course)}. \end{cases} $$

Answer 5

Since {$0$} is a non-empty subset of $R$, this subset is eligible to undergo the subring test (if $a, b$ are members of $R$ then $a-b$ and $ab$ are also members of $R$). In this case the only possibility is $a=b=0$, and we know that both, $a-b=0$ and $ab=0$ are in $R$, therefore {$0$} forms a subring of $R$.