Does every unital ring contain all the integers?
Let us suppose there is a ring $R$ with the multiplicative identity $1$.
We know that $1+r\in R$, where $r$ is any element of the ring $R$. Does this mean $1+1$ is also part of the ring, or does $r$ have to be an element of the ring different from $1$?
Is $1+1$ called $2$ in the ring? Similarly, as $-1$ is also part of the ring, is $-1+ -1$ called $-2$ in the ring? If it is, then I suppose all integers are contained in every unital ring.
These questions are very elementary. However, I read contradictory remarks in some places which tend to confuse me. So I thought it would be best to clear any doubts, however trivial the questions.
Thanks in advance for your help!
Solution 1:
We do indeed call $1+1 = 2$ in any unital ring, and similarly, we have in any unital ring a subring generated by $1$, which will consist precisely of elements of the form $1+1+\cdots +1$ and their negatives (and $0$).
However, this does not necessarily mean that the ring contains all integers (by which I mean a subring isomorphic to the integers). The reason is that it can happen that adding $1$ to itself some number of times gives $0$, such as is the case in the ring $\mathbb{Z}/n\mathbb{Z}$ where adding $n$ copies of $1$ gives $n$ which equals $0$ in that ring.
If there is no way to get $0$ by adding $1$ to itself, we say that the characteristic of the ring is $0$, and the subring generated by $1$ is isomorphic to the integers.
If there is a way to get $0$, and $n$ is the smallest number of $1$'s you need to add to get $0$, we say that the ring has characteristic $n$. In this case the subring generated by $1$ will be isomorphic to $\mathbb{Z}/n\mathbb{Z}$.
Solution 2:
There is a subtle distinction to be made between whether a unital ring "contains integers" or contains a subring "that looks like integers". The elements you are describing are constructed in the way one would construct the integers; namely, you start with $1$ and iteratively proceed to add it to itself. However, if your ring is a ring of matrices, and the element $1$ is the identity matrix, you are building objects that are more than just integers, even though, as a subring, they are structurally identical to integers.
In answer to your question: you can continue to add $1$ to itself in every ring. You may keep on creating new elements, in which case you have a subring that looks like $\mathbb{Z}$, or you may eventually hit $0$, in which case you have a subring that looks like $\mathbb{Z}/n\mathbb{Z}$, the integers modulo $n$. The exact nature of the subring you construct is an important property of the ring itself.
Again, it is important to note that although, as a subring, this may appear to be $\mathbb{Z}$ or $\mathbb{Z}/n\mathbb{Z}$, these elements may interact in interesting ways with the rest of the ring. You should be careful to avoid the trap of convincing yourself that the element $2$ in a ring must act in a very ordinary way, simply because $2$ seems so ordinary in $\mathbb{Z}$. For example, $2$ is prime in $\mathbb{Z}$, but it is not prime in the Gaussian integers, which contain a copy of $\mathbb{Z}$.