In this post, we will see that the usual definition of the characteristic of a ring obscures its true nature, and explain why another equivalent characterization is conceptually more useful.

Throughout this discussion, every ring has a multiplicative identity $1$, and every ring homomorphism $\varphi: A \to B$ must satisfy $\varphi(1_A) = 1_B$. The ring of integers is denoted by $\mathbb{Z}$.

Several sources define the characteristic of a ring as follows.

Definition. The characteristic of a ring $A$, denoted by $\operatorname{char}(A)$, is the smallest positive integer $n$ such that $1 + 1 + \cdots + 1 = 0$ ($n$ times) in $A$; if no such integer exists the characteristic of $A$ is said to be $0$.

A merit of this definition is that it is simple and purely elementary. However, it is not surprising that many students who first encounter this definition feel confused by ‘the characteristic $0$’. Many students think characteristic $\infty$ is more natural than $0$ according to the above definition. The following is the elementary proof showing that the characteristic of a field must be a prime in [1].

Proposition 1. The characteristic of a field $F$, $\operatorname{char}(F)$, is either $0$ or a prime $p$. Furthermore, if $n \cdot 1_F = 0$ for $n > 0$, then $n$ is divisible by $p$.

Proof. For a positive integer $n$, let $n \cdot 1_F \coloneqq 1_F + \cdots + 1_F$ ($n$ times). It is easy to see that

\[(n \cdot 1_F)(m \cdot 1_F) = mn \cdot 1_F\]

for positive integers $m$ and $n$. If $n = ab$ is composite with $n \cdot 1_F = 0$, then $ab \cdot 1_F = (a \cdot 1_F)(b \cdot 1_F) = 0$, and since $F$ is a field, one of $a \cdot 1_F$ or $b \cdot 1_F$ is $0$. Hence, the smallest such integer is necessarily a prime. It also follows that if $n \cdot 1_F = 0$ for $n > 0$, then $n$ is divisible by $p$.$\qquad\square$

However, as one studies (commutative) algebra more deeply, one comes to realize that another equivalent characterization of the characteristic of a ring is conceptually more useful. It is straightforward to check that the following fact holds, using the assumption that every ring homomorphism preserves the multiplicative identities.

Fact. For a ring $A$, there exists a unique ring homomorphism $\mathbb{Z} \to A$.

Remark. The fact says that $\mathbb{Z}$ is the initial object of the category of rings.

One can easily check the following definition is equivalent to the original usual definition of the characteristic of a ring.

(New) Definition. The characteristic of a ring $A$, denoted by $\operatorname{char}(A)$, is the non-negative generator of the unique ring homomorphism $\mathbb{Z} \to A$.

In this sense, the concept of ‘characteristic $0$’ becomes very natural. If there is no positive integer $n$ such that $1 + 1 + \cdots + 1 = 0$ ($n$ times) in $A$, then the unique homomorphism $\mathbb{Z} \to A$ must be injective, implying that the kernel is $0$. This definition is not only conceptually useful, but also practical in actual problem solving. The following is a new proof of Proposition 1.

New proof of Proposition 1. Consider the unique ring homomorphism $\mathbb{Z} \to F$. Since any subring of $F$ is an integral domain, the kernel of the map must be prime, so the characteristic is $0$ or a prime $p$. If $n \cdot 1_F = 0$ for $n > 0$, then the ideal $(n)$ must be contained in $(p)$, implying that $p \mid n$.$\qquad\square$

As you can see, the proof becomes more systematic and is beautifully described in the language of ideal theory. This approach can be applied to other problems.

Problem 1. Suppose $((A_i)_{i \in I}, (f_{ij})_{i \le j \in I})$ is an inverse system of commutative rings. Suppose for each $i \in I$, the ring $A_i$ has a positive characteristic. Then what can we say about the characteristic of its inverse limit?

Solution. Consider the unique ring homomorphism $\varphi_i : \mathbb{Z} \to A_i$. Then the characteristic $n_i$ of $A_i$ is precisely the kernel $n_i\mathbb{Z}$ of $\varphi_i$. Hence, we get an exact sequence of $\mathbb{Z}$-modules

\[0 \to n_i\mathbb{Z} \to \mathbb{Z} \to A_i.\]

Since the inverse limit is left exact (see Proposition 10.2 of [2]), we get

\[0 \to \varprojlim_I n_i\mathbb{Z} \to \mathbb{Z} \to \varprojlim_I A_i.\]

Thus, the characteristic of $\displaystyle\lim_{\underset{I}{\longleftarrow}} A_i$ is the non-negative generator of $\displaystyle\lim_{\underset{I}{\longleftarrow}} n_i\mathbb{Z}$. Note that $n_j\mathbb{Z} \le n_i\mathbb{Z}$ whenever $i \le j$; i.e., $n_i$ must divide $n_j$. Thus, if $n_i$ is unbounded, then $\displaystyle\lim_{\underset{I}{\longleftarrow}} n_i\mathbb{Z} = 0$. Otherwise, $n_i$ is eventually stabilized to their least common multiple $L$, and it is precisely the characteristic of $\displaystyle\lim_{\underset{I}{\longleftarrow}} A_i$. $\quad\square$

Remark. It might sound surprising that the inverse limit of rings of positive characteristic can have characteristic $0$. However, there is a well-known example. For a fixed prime $p$, consider the inverse system $\mathbb{Z}/p^i\mathbb{Z}$ for $i \ge 1$, where $\mathbb{Z}/p^j\mathbb{Z} \to \mathbb{Z}/p^i\mathbb{Z}$ is the natural quotient by $(p^{j-i})\left(\mathbb{Z}/p^j\mathbb{Z}\right)$ for $j > i$. The inverse limit is the ring of $p$-adic integer $\mathbb{Z}_p$, which is known to have characteristic $0$.

We can use the same strategy for the dual problem.

Problem 2. Suppose $((B_i)_{i \in I}, (g_{ij})_{i \le j \in I})$ is an direct system of commutative rings. Suppose for each $i \in I$, the ring $B_i$ has a positive characteristic. Then what can we say about the characteristic of its direct limit?

Solution. Consider the unique ring homomorphism $\varphi_i : \mathbb{Z} \to B_i$. Then the characteristic $m_i$ of $B_i$ is precisely the kernel $m_i\mathbb{Z}$ of $\varphi_i$. Hence, we get an exact sequence of $\mathbb{Z}$-modules

\[0 \to m_i\mathbb{Z} \to \mathbb{Z} \to B_i.\]

Since the direct limit is exact (see Exercise 2.19 of [2]), we get

\[0 \to \varinjlim_I m_i\mathbb{Z} \to \mathbb{Z} \to \varinjlim_I A_i.\]

Thus, the characteristic of $\displaystyle\lim_{\underset{I}{\longrightarrow}} B_i$ is the non-negative generator of $\displaystyle\lim_{\underset{I}{\longrightarrow}} m_i\mathbb{Z}$. Note that $m_j\mathbb{Z} \ge m_i\mathbb{Z}$ whenever $i \le j$; i.e., $m_j$ must divide $m_i$. Thus, $m_i$ must be eventually stabilized to their greatest common divisor $D$, and it is precisely the characteristic of $\displaystyle\lim_{\underset{I}{\longrightarrow}} B_i$. $\quad\square$


References

  1. Dummit, D. S., & Foote, R. M. (2003). Abstract Algebra (3rd ed.). Wiley.
  2. Atiyah, M. F., & MacDonald, I. G. (2018). Introduction To Commutative Algebra. CRC Press. https://doi.org/10.1201/9780429493638