I was reviewing the definition of the characteristic of a field where it was defined to be:

"The characteristic of a field $F$, denoted ch$(F)$, is defined to be the smallest positive integer $p$ such that $p\cdot 1_F = 0$ if such a p exists and is defined to be $0$ otherwise."

The further proposition said:" If ch$(F) = p$ then for any $\alpha \in F. p\cdot \alpha=\alpha+\alpha+...(p\, times)+\alpha=0$.

My question was that:

  1. how was the addition associated multiplication?

  2. Further, how did the prime elements associated with the prime number?

Where I found a sentence at the beginning of the chapter: "For $n$ a positive integer, let $n\cdot 1_F = 1_F+...+ 1_F$ ($n$ times). "

So, question:

  1. if we didn't have this relation, does that mean the above proposition would be incorrect and we can no longer associate the addition with multiplication?
  • 8
    $\begingroup$ This is essentially treating the additive group of the field, which is an abelian group, in a natural way as a $\mathbb Z$-module, where multiplication by positive integers is defined by repeated addition, and negation by taking the additive inverse in the field. If you don't already have this structure, you just need to define it. Multiplication by an integer makes sense in any abelian group written additively. $\endgroup$ – Mark Bennet Nov 19 '17 at 20:43

This is not directly linked to field multiplication. By the properties of the integers, given any element $\alpha\in F$, there exists a unique group homomorphism $(\mathbb{Z},+)\to (F,+)$ mapping $1$ to $\alpha$.

This homomorphism maps any positive integer $k$ to $\underbrace{\alpha+\dots+\alpha}_{k\text{ times}}$ and a negative integer $k$ to $-(\underbrace{\alpha+\dots+\alpha}_{-k\text{ times}})$. This is general group theory. It is customary to write $k\alpha$ for the element so described.

Let $\chi_F$ be the group homomorphism we get for $\alpha=1_F$. Like every homomorphism with domain $\mathbb{Z}$, the kernel of $\chi_F$ is a subgroup, so it is of the form $p\mathbb{Z}$, for a unique $p\ge0$. In particular, if $p>0$, we have $p1_F=\chi_F(p)=\underbrace{1_F+\dots+1_F}_{p\text{ times}}=0$ and $p$ is the least positive integer having this property.

If the kernel of $\chi_F$ is $\{0\}=0\mathbb{Z}$, no such integer exists.

What happens in the case of fields (more generally of rings) is that $\chi_F$ is also a ring homomorphism: $\chi_F(mn)=\chi_F(m)\chi_F(n)$, which can be proved by induction on $n$ when $n\ge0$ and by the rule of signs for $n<0$.

This is the key for proving that when the characteristic of $F$ is $p>0$, then $p$ is prime. Indeed, if $p=mn$ with $m,n>0$, then $$ 0=\chi_F(p)=\chi_F(mn)=\chi_F(m)\chi_F(n) $$ so either $\chi_F(m)=0$ or $\chi_F(n)=0$, which implies $m\in p\mathbb{Z}$ or $n\in p\mathbb{Z}$, that is, $m=p$ or $n=p$.


It's important to distinguish between the ring multiplication, which is the operation in the ring, and

the shorthand notation $kr := r+r+\dots +r \ $ ($k$ times)

$\\ $

I like to use $ \ k \cdot r$ for ring multiplication and $kr$ for the shorthand notation.

This basically defines the characteristic of a field to be the order of $1_F$ under addition, if it is finite,

and $0$ if the order of $1_F$ infinite.

You then have $$p\alpha = \alpha + \alpha + \cdots \alpha \ (p \text{ times}) = \alpha \cdot 1_F \ + \alpha \cdot 1_F \ + \cdots + \ \alpha \cdot 1_F $$

$ \\ $

$$ = \alpha \ \cdot \left( 1_F + 1_F + \cdots + 1_F\right) = \alpha \ \cdot ( \ p 1_F \ ) = \alpha \ \cdot 0 = 0$$

$ \\ $

For integers, a number $p$ is prime if and only if $$p | ab \implies p|a \text { or } p|b$$

That was the property that was generalized in defining prime elements. The "elementary school" definition of a prime, that its only divisors are itself and 1, was generalized to the notion of an irreducible element. They are equivalent in PID's, and hence in the integers.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy