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I'm trying to prove that the characteristic of any field $F$ is either $0$ or a prime number, but I have no idea what to do. Help?

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  • $\begingroup$ Hint: Example: Suppose char$(F)=12$. Let $0_F$ and $1_F$ be the additive and multiplicative identities of $F$. Let $a= 1_F+1_F+1_F$ and $b=1_F+1_F+1_F+1_F.$ Then $a\ne 0_F\ne b$ but by the Distributive Law $ab=\sum_{j=1}^{12}(1_F\cdot 1_F)=0_F$. $\endgroup$ – DanielWainfleet Nov 14 '18 at 11:15
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Hint: Show that if the characteristic of $F$ were a composite number, say $n=ab$, then $F$ would have zero-divisors (since $n=0$...). Then show that a zero-divisor cannot be a unit unless $0=1$, which is not the case for fields.

Alternate Hint: Look at the unique homomorphism $\phi:\mathbb{Z}\to F$, defined by $\phi(0)=0_F$, $\phi(1)=1_F$, $\phi(2)=1_F+1_F$, etc. and note that its image, being a subring of a field, must be an integral domain.

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  • $\begingroup$ OK...I'm not very skilled at proofs, so can you see if there are any holes with this? I'm using the 1st Hint... Proof: Consider, for contradiction, that the field is a composite number. (still working) $\endgroup$ – pigishpig Nov 28 '11 at 5:33
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Suppose $char(F) = m \neq 0$.
By definition, $m$ is the smallest integer so that $\underbrace{1 + ... + 1}_{m-times} = 0_F$

Suppose m is not prime and say $m = \alpha.\beta$, where $0 < \alpha,\beta < m$.
Therefore we can write: $$ \eqalign{ 0_F &= \underbrace{1 + ... + 1}_{m-times}\\ &= (\underbrace{\underbrace{1 + ... + 1}_{\alpha-times}) \ + \ ... \ + \ (\underbrace{1 + ... + 1}_{\alpha-times})}_{\beta-times} \tag{1}\label{1} } $$

By closure property, $\underbrace{1 + ... + 1}_{\alpha-times} \in F$. Let $\underbrace{1 + ... + 1}_{\alpha-times} = \zeta \tag{2}\label{2}$ [It's important to not call it $\alpha$. See Note]

There are 2 cases:

Case $\zeta = 0_F$:
Since $\alpha < m$ and $\underbrace{1 + ... + 1}_{\alpha-times} = 0_F, \ \ \ \therefore char(F) = \alpha$.
We arrive at a contradiction.

Case $\zeta \neq 0_F$:
$\exists \zeta^{-1} \in F$ $$ \eqalign{ \therefore 0_F &= 0_F.\zeta^{-1}\\ &= (\underbrace{\zeta + ... + \zeta}_{\beta-times}).\zeta^{-1} \ [by \ (\ref{1}) \ and \ (\ref{2})]\\ &= \underbrace{1 + ... + 1}_{\beta-times} } $$

Since $\beta < m, char(F) = \beta$ and we again arrive at a contradiction.

Hence $m$ is a prime. QED

Note: We should not write $\underbrace{1 + ... + 1}_{\gamma-times} = \gamma$, as $\gamma$ is just an integer and not necessarily a member of the field $F$. Also this may lead to an erroneous conclusion, e.g. say since $char(F) = m$, if we write $\underbrace{1 + ... + 1}_{m-times} = m = 0$, we can say $m = m.1 = 0$ and since $1 \neq 0, \therefore m = 0$.

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  • $\begingroup$ why did you write $0=0C^{-1}$ ? $\endgroup$ – graphtheory92 Apr 17 '19 at 3:56
  • $\begingroup$ because product of additive identity i.e. $0$ with any field element is $0$ $\endgroup$ – Chayan Ghosh Apr 17 '19 at 6:02

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