The characteristic of a field is defined to be the smallest positive integer $p$ such that $$p \cdot 1 = 0.$$ But I have learned that field has no zero divisors. How is this possible?

  • 2
    A zero divisor is an element of the field. When you define the characteristic via that equation you consider $1$ an element of the field, whereas $p$ is an integer and the $\cdot$ is not the field operation: it stands for 'multiple' i.e. $p\cdot1$ means $1+1+...+1$ ($p$ times). – rafforaffo Feb 3 '15 at 14:58

The notation $p \cdot 1$ just means $1+1+\cdots+1$ ($p$ times), where $1$ is the unit of the field and $+$ is the addition in the field.

You can also interpret $p$ as this same element in the field, but in this case you'll have $p=0$.

For any field $\mathbb{F}$, we denote $$n \cdot 1 := \underbrace{1 + \cdots + 1}_n.$$ Now, $n \mapsto n \cdot 1$ is by construction a group homomorphism $\phi: (\mathbb{Z}, +) \to (\mathbb{F}, +)$, and if $\mathbb{F}$ has positive characteristic $p$, by definition we have $\phi(p) = 0$, or more informally, "$p = 0$ in $\mathbb{F}$".

Consider, for example, the field $\mathbb{F}_2$ with underlying set $\{0, 1\}$, with elements just an additive and multiplicative identity (respectively) and satisfying $1 + 1 := 0$. Since $0 \neq 1$, the characteristic of this field is $\text{char } \mathbb{F}_2 = 2$.

We can see quickly using a zero divisor argument that the characteristic of any field with positive characteristic must be prime. (In fact, any finite field must have $p^n$ elements for some prime $p$ and positive integer $n$, and for each such number $p^n$ there is precisely one field up to isomorphism with that order.)

By the way, if $n \cdot 1 \neq 0$ for all positive integers $n$, we usually say by convention that the field has characteristic $0$; some older sources say such fields have characteristic $\infty$ instead.

If $R$ is a ring with identity, let's denote it by $1$. Let's also denote integers with boldface type. So $\mathbf{1}$ is the integer number one, for instance, $\mathbf{0}$ is the integer zero, whereas $0$ and $1$ are the neutral elements in $R$.

Since $R$ under addition is an abelian group, we can as usual define $$\def\z#1{\mathbf{#1}} \z{0}r=0,\qquad (\z{n}+\z{1})r=\z{n}r+r $$ that defines $\z{n}r$ for $\z{n}\ge\z{0}$. If $\z{n}<\z{0}$, define as usual $$ \z{n}r=(-\z{n})(-r) $$ It's easy to prove that, if $\varepsilon_R\colon\mathbb{Z}\to R$ is the map $$ \varepsilon_R(\z{n})=\z{n}1 $$ (according to the definition above), then

  1. $\varepsilon_R(\z{m}+\z{n})=\varepsilon_R(\z{m})+\varepsilon_R(\z{n})$
  2. $\varepsilon_R(\z{m}\z{n})=\varepsilon_R(\z{m})\varepsilon_R(\z{n})$
  3. $\varepsilon_R(\z{1})=1$

so $\varepsilon_R$ is a homomorphism of unital rings. Actually it is the unique unital ring homomorphism $\mathbb{Z}\to R$, because $\mathbb{Z}$ is a cyclic group under addition and we're required that a homomorphism maps $\z{1}$ into $1$.

Like all ring homomorphisms, $\varepsilon_R$ has a kernel which is an ideal of $\mathbb{Z}$, so we have $\ker\varepsilon_R=\z{k}\mathbb{Z}$ for a unique $\z{k}\ge\z{0}$.

If $\z{k}=\z{0}$, then $\varepsilon_R$ is injective. Otherwise, we have $$ \z{k}1=0 $$ by definition of kernel, which means that $$ \z{k}r=\underbrace{r+r+\dots+r}_{\z{k}}=(\underbrace{1+1+\dots+1}_{\z{k}})r= (\z{k}1)r=0r=0 $$ for every $r\in R$.

This integer $\z{k}$ is the characteristic of $R$.

If $R$ is an integral domain and $\z{k}>0$, suppose we have $\z{k}=\z{a}\z{b}$, with $\z{1}\le\z{a}<\z{k}$. Then, by definition of the characteristic, we know $\z{a}1\ne0$, but $$ 0=\z{k}1=\varepsilon_R(\z{k})=\varepsilon_R(\z{a})\varepsilon_R(\z{b}) $$ and so, being $R$ an integral domain, we have $\varepsilon_R{\z{b}}=0$ which implies $\z{k}\mid\z{b}$. Therefore $\z{b}=\z{k}$ and $\z{a}=1$. It follows that $\z{k}$ is prime.

Saying $\z{k}1=0$ in a field (which is an integral domain) is not a contradiction, because $\z{k}$ is not an element of the field.

Normally one doesn't denote integers with different type than elements of the rings, which could lead to confusion. Most of the times, one simply writes $n$ instead of $n1$ (when $n$ is an integer), so we can even find $p=0$ when the characteristic of a field is the prime $p$. This should be simply interpreted as $p1=0$. With some experience, the notational problems disappear.

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