Let us suppose there is a ring $R$ with the multiplicative identity $1$.

  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 $1$?

  2. 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!


7 Answers 7


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 $\mathbb{Z}/n\mathbb{Z}$.


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 $\mathbb{Z}$.

  • 2
    $\begingroup$ Actually, the element $2$ will act precisely as you would expect (ie, $2a = a+a$), and the same goes for all the "integers" in the ring. Where you need to be careful is in statements about reducibility and such. $\endgroup$ Jul 18, 2013 at 8:46
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    $\begingroup$ To give an example of what I mean, $2$ is prime in $\mathbb{Z}$, but not prime in the Gaussian integers. $\endgroup$ Jul 18, 2013 at 8:51
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    $\begingroup$ Yes, that is why I mentioned reducibility as being something that might behave differently. $\endgroup$ Jul 18, 2013 at 8:53

I'll assume all rings to be unital.

Every ring contains a canonical homomorphic copy of the integers; more precisely, there is a unique morphism $\mathbb Z\to R$ for any ring $R$, where "morphism" means a map $f\colon \mathbb Z\to R$ with $f(x+y) = f(x) + f(y)$, $f(xy)=f(x)f(y)$, and $f(1_\mathbb Z) = 1_R$.

However, this map may not be injective; for example, the ring $(\mathbb Z/3)[x]$ contains the integers $\{\dotsc, -1, 0, 1, 2, 3, \dotsc\}$, but we have $-1 = 2,\, 0 = 3, \dots$ in this ring.


Your observation is very nice! Indeed, the ring $R$ is closed under addition (and, more generally) subtraction so:

$1\in R$ (if $R$ is unital, which we shall henceforth assume without further mention)

$2=1+1\in R$

$3=1+1+1\in R$


$n=1+\cdots+1\in R$ (the $1$ is added $n$ times with itself for each positive integer $n$).

Similarly, $-n$, the additive inverse of $n$, is an element of $R$ for all positive integers $n$. Therefore, we have a map $\mathbb{Z}\to R$ (the integer $n$ is mapped to the element denoted by $n$ in $R$). However, here's the catch: this map needn't be injective. So, we can't say in general that $R$ contains all integers. What can happen is that different integers $n$ and $m$ may be equal in $R$.

For example, a ring is said to have characteristic $n$ if $n$ is the smallest positive integer such that $1+\cdots+1$ ($n$ times) is equal to $0$. The ring $\mathbb{Z}/n\mathbb{Z}$ is the simplest example of a ring of characteristic $n$ but there are others. In a ring of characteristic $n$, integers which differ by $n$ will equate. If there is no such smallest integer $n$ such that $1+\cdots+1$ ($n$ times) is equal to $0$, then the ring has characteristic $0$.

Exercise 1: Prove that this map $\mathbb{Z}\to R$ is a ring homomorphism. If $R$ has characteristic $0$, then prove that this map is injective. What is the kernel of this map if $R$ has characteristic $n$?

Exercise 2: Prove that this ring homomorphism $\mathbb{Z}\to R$ is unique.

Exercise 3: If $A$ is a ring such that there exists a unique ring homomorphism $A\to R$ for all rings $R$, then prove that $\mathbb{Z}\cong A$.

In the language of category theory, we say that $\mathbb{Z}$ is the initial object in the category of (unital) rings.

Exercise 4: Let $R$ be a ring. Prove that there exists a unique smallest subring of $R$; we call this the minimal subring of $R$. What can this minimal subring be? (Hint: this minimal subring of $R$ will depend on the characteristic of $R$ and there are only countably many possibilities; use Exercise 1 and the first isomorphism theorem.)

Exercise 5: Let $R$ be an integral domain. Prove that the characteristic of $R$ is a prime number.

Exercise 6: As an application of this theory, prove the following: let $F$ be a finite field. Prove that the cardinality of $F$ is a prime power. (Hint: use Exercise 5 and the fact that $F$ is a vector space over its minimal subring. In this case, the minimal subring of $F$ is also the minimal subfield of $F$.)

I hope this helps!


You have to keep in mind that a ring does not have to consists of integers at all, and thus, while a ring must contain an identity which we denote by $1$, an additive inverse which we denote by $-1$, and thus all elements of the form $1+1+1+1+\cdots +1$ (any finite addition of $1$ to itself), which we may denote by $n$, and similarly all elements which we may denote by $-n$, the rind does not have to contain even a single integer. In fact, the ring may very well be finite. For instance, the ring $\mathbb Z_n$, for each $n\in \mathbb N$, is a finite ring. Usually, we write $\mathbb Z_n=\{0,1,2,\cdots, n-1\}$, but none of these is an integer. These represent equivalence classes of integers, or, if you like, they are just symbols.


In the case of characteristic $0$ (which means that $n\cdot 1 = \underbrace{1 + 1 + \ldots + 1}_{n\text{ times}}$ never is zero for $n > 0$), you are right: The ring contains a copy of $\mathbb{Z}$, and it often is convenient to name its elements with the common terms for the integers.

However, your ring might have characteristic $n > 0$, meaning that $n$ is the smallest integer $>0$ such that $n\cdot 1 = 0$. Here, the ring doesn't contain a copy of $\mathbb{Z}$ (but a copy of $\mathbb{Z}/n\mathbb{Z}$).


The theory of rings has many constants. While $0$ and $1$ are the only constants introduced by the usual definition of a ring, the operations $+$, $-$, and $\cdot$ allow us to construct more constants. It is useful to use "2" to refer to the constant $1+1$, "3" for $2+1$, and so forth.

Incidentally, we can also construct additional operations; e.g. "multiplication by 3", or "raising to the seventh power" are also ring operations. In fact, for every polynomial $f$ in zero or more variables with integer coefficients, or even "plug values into $f$" is also a ring operation. (and every ring operation -- including the constants -- is of this form)

Another convention that is often useful is that when we have a ring homomorphism $\varphi :R \to S$, it can be useful to represent elements of $S$ by specifying elements of $R$. e.g. if $r \in R$, we might also use the notation $r$ for the element $\varphi(r) \in S$ in any setting where it's clear we are talking about elements of $S$.

Some situations where this convention is particularly useful are

  • $\varphi$ is monic, and we want to think of $R$ as a subring of $S$
  • $\varphi$ is surjective, and we want to think of $S$ as a quotient of $R$ (i.e. $S$ is the ring $R$ modulo an equivalence relation)
  • There is only one map $R \to S$, such as when $R = \mathbb{Z}$.

Note that the third bullet above results in the exact same notation as suggested in the previous section. (which is part of the reason why it's useful notation!)


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