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If Char$F$ $\neq 0$, then Char$F$ must be prime number.

MY try: If Char$F$$ = nk $ for integers $n$ and $k$, then by definition, $nk = 0 \implies n = 0$ or $k = 0$ which implies Char$F=0$ which is a contradiction.

Is this correct?

ILoveMath
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4 Answers4

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We need to claim $F$ is also an integral domain

Complete proof

As $F$ is a field then $F$ is obviously an integral domain. Assume a contradiction that $\text{Char}(F)$ is not a prime then $\text{Char}(F)=a.b$ where $a,b$ is different from $0$.

So $0=\text{Char}(F).1=(ab).1=a.b$ which implies that $a=0$, or $b=0$ since $F$ is a integral domain. (a contradiction)

Hence, $\text{Char}(F)$ is a prime number.

Matt E.
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An Khuong Doan
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If char $\mathbb{F}$ is $n$, and $n$ is composite. then $n=n_1.n_2$ for some $n_1 \gt 1, n_2 \gt 1$. By the definition of characteristic $n$ is the smallest number such that $n.1 = 0$. But $n=n_1n_2$ implies $n.1=(n_1.1)(n_2.1)$ . Hence $(n_1.1)(n_2.1) =0$. which implies either $n_1.1 = 0$ or $n_2.1 = 0$. contradiction to the minimality of $n$.

GA316
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A sketch would be considering the embedding map $f:\mathbb{Z}\to F$, that is $f(n )=1+1+...+1$ and noticing that $f$ is a (field) domain. The first isomorphism theorem then tells us that $\mathbb{Z}/ker(f)\cong im(f)$. Can you finish the proof? What do we know about $ker(f)$?

LASV
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Char F is the least positive integer m such that m . 1 = 0 (here 1 and 0 are the unit and zero of the field, and m is an integer - and m . 1 is the same as 1+1+...1 m times). It can be proved (by induction) that this "multiplication" obeys several "obvious" properties, such as (mn) . 1 = (m . 1) (n . 1). Using this property it is easy to argue by contradiction. Suppose p=char F is not prime; write p=mn and prove that either (m . 1) or (n . 1) must be zero.