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I seem to hold a very loose grasp of the concept of fields - I've encountered this question: Show that every finite field with characteristic $p$ contains $\mathbb{Z}_p$ (i.e. $\mathbb{Z}_p = \{0,...,p-1\}$). Now my question is what exactly is a field of characteristic $p$?

My second question: Show that every field of characteristic $0$ contains $\mathbb{Q}$ as a subfield?

Regards.

Dmoreno
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Nimrodshn
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    Write $1$ for the multiplicative identity of the field $F$. Look at $1,1+1,1+1+1,1+1+1+1,\cdots$. If none of these is $0$, we say $F$ has characteristic $0$. If one of these is $0$, let $k$ be the smallest integer such that $1+\cdots +1=0$. It turns out that $k$ is prime. This $k$ is called the characteristic of $F$. – André Nicolas Aug 08 '14 at 13:33
  • So doesnt this mean that every field of charectaristic p is actoualy Zp ? or isomorphic to Zp? but i cant seem to understand why it would contain Zp. – Nimrodshn Aug 08 '14 at 13:36
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    @Nimrodshn No, not every element of a field needs to be represented as $1+1+1+\dots +1$. For example, in $\mathbb Q$, you can't represent $1/2$ that way. – Thomas Andrews Aug 08 '14 at 13:37
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    The field $F={0,1,a,a+1}$ with addition modulo 2 (ie. for example $a+(a+1) = (a+a)+1=1$) and multipliction goven by $a^2=a+1$ (hence e.g. $(a+1)^2=a^2+2a+1=3a+2=a$) is a field with four elements. It properly contains $\mathbb Z/2\mathbb Z$ – Hagen von Eitzen Aug 08 '14 at 13:39
  • It contains a field isomorphic to $\mathbb{Z}_p$. It can be much bigger, even infinite. – André Nicolas Aug 08 '14 at 13:40

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If $F$ is a field use the ring morphism $h:\Bbb Z\to F$ defined by $n\mapsto n.1_F$ and regarded it' kernel. It's kernel is a subgroup of $(\Bbb Z,+)$ so it is of the form $\ker h=p\Bbb Z$. Now discuss the cases $p=0$ and $p\neq 0$.

Hamou
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  • If $p=0$ then $F$ you can show that $F$ contain $\Bbb Q$. If $p\neq 0$ by the universal propriety of the quotient $h$ induce an injective morphism fro $\Bbb Z/(\ker h)=\Bbb Z/(p\Bbb Z)=\Bbb Z_p$ to $F$ – Hamou Aug 08 '14 at 13:41