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I translate this from a German book: "For every finite field $K$ there exists a prime number $p$ such that $\mathbb Z/p\mathbb Z$ is a subfield of $K$"

But how is this possible? For example the field $K = \{0,1\}$ contains integers but $\mathbb Z/p\mathbb Z$ contains equivalence classes. To be a subfield it would also have to be a subset of $K$.

user104025
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    Strictly speaking, the author should say a subfield isomorphic to $\mathbb{Z}/p\mathbb{Z}$. – Casteels Oct 29 '13 at 13:29
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    Inside finite field you can find $\mathbb{Z}/p\mathbb{Z}$ and inside any field of charecteristic zero you can find $Q$ as sub field which are called prime sub fields. – GA316 Oct 29 '13 at 13:30
  • @GA316 If by $Q$ you mean the rationals, then that is not correct. – Tobias Kildetoft Oct 29 '13 at 13:31
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    @GA316, some care is needed as what you wrote is completely wrong. Perhaps you meant "inside infinite field of characteristic zero we can find (an isomorphic copy of) the rationals $;\Bbb Q;$ . – DonAntonio Oct 29 '13 at 13:33
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    @Casteels, that's speaking too strictly, imo. – DonAntonio Oct 29 '13 at 13:33
  • why it is not correct? see here http://mathworld.wolfram.com/PrimeSubfield.html – GA316 Oct 29 '13 at 13:33
  • How does $K={0,1}$ contain the integers? – dezign Oct 29 '13 at 13:34
  • Because there are tons (ok, infinite number) of infinite fields which don't have characteristic zero and thus do not contain any isomorphic copy of the rationals with them, @GA316 . – DonAntonio Oct 29 '13 at 13:35
  • @DonAntonio yes correct. by infinite field Itry to mean field of char zero. now I have correct it – GA316 Oct 29 '13 at 13:37
  • There's a huge difference between "infinite field" and "field with characteristic zero", as the former follows from the latter but not the other way around. – DonAntonio Oct 29 '13 at 13:38
  • @DonAntonio agreed, agreed, agreed. thanks. – GA316 Oct 29 '13 at 13:39
  • @user104025 question title should be Is $\mathbb{Z}/p\mathbb{Z}$ sub field of every "finite" field? – GA316 Oct 29 '13 at 13:43

2 Answers2

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Every field contains a subfield isomorphic to either $\mathbb{Z}/p\mathbb{Z}$ (for some prime $p$) or $\mathbb{Q}$.

This follows from basic laws of additive exponents.

Let $\mathbb{K}$ be a field with multiplicative identity $1_\mathbb{K}$. Then consider the map $f:\mathbb{Z}\to\mathbb{K}$ defined by $n \mapsto n1_\mathbb{K}=\underbrace{1_\mathbb{K}+1_\mathbb{K}+\cdots+1_\mathbb{K}}_{n-\mbox{times}}$.

Basic laws of (additive) exponents tell us that $f(n+m)=(n+m)1_\mathbb{K} = n1_\mathbb{K}+m1_\mathbb{K}=f(n)+f(m)$ and $f(nm)=(nm)1_\mathbb{K} = (n1_\mathbb{K})(m1_\mathbb{K})=f(n)f(m)$ so $f$ is a ring homomorphism. Thus by the first isomorphism theorem $\mathbb{Z}/\mathrm{Ker}(f) \cong \mathrm{Im}(f)$.

The kernel of $f$ is either $\{0\}$ (this means $\mathbb{K}$ has characteristic $0$) and so $\mathbb{Z} \cong \mathrm{Im}(f)$. Thus $\mathbb{K}$ has a subring isomorphic to the integers and so (since it's a field) must have (multiplicative) inverses for these elements and so has a subfield isomorphic to $\mathbb{Q}$.

Otherwise the kernel of $f$ is a nonzero ideal of $\mathbb{Z}$ and so has the form $p\mathbb{Z}$ where $p$ must be prime (otherwise $\mathbb{K}$ would a contain a subring which has zero divisors). Thus $\mathbb{K}$ has a subring isomorphic to $\mathbb{Z}/p\mathbb{Z}$ (for some prime $p$).

If we begin with the assumption that $\mathbb{K}$ is finite, this rules out characteristic zero (there's not enough room to fit the infinitely large copies of $\mathbb{Z}$ or $\mathbb{Q}$). So finite fields must contain a subfield isomorphic to $\mathbb{Z}/p\mathbb{Z}$ for some prime $p$.

By the way, these subfields (the subfield generated by $1_\mathbb{K}$) are called prime subfields.

For your particular example, $K=\{0,1\} \cong \mathbb{Z}/2\mathbb{Z}$ (this field is of characteristic 2 and is equal to its prime subfield).

Bill Cook
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  • thank you, so the statement would be true if the author had written "isomorphic". – user104025 Oct 29 '13 at 14:10
  • Yes, but it's probably a deliberate omission, and this is common practice in abstract algebra, so be aware. – Casteels Oct 29 '13 at 14:19
  • Sorry for the necropost: Is every finite field isomorphic to $\mathbb Z_p$ for some prime $p$? I guess we could work on the multiplication/addition tables of $Z_p$ and see if we can teak it without losing "fieldness" – gary Dec 30 '17 at 20:35
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    @gary no. Every finite field contains an isomorphic copy of Z mod p (for some prime p), but there are finite fields of order p^k for every prime p and positive integer k. (Their multiplication tables are a bit complicated and aren't isomorphic to Z mod p^k unless k=1.) – Bill Cook Dec 31 '17 at 00:47
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$\mathbb Z / 2\mathbb Z$ contains $K=\{0,1\}$. It means $\mathbb Z / 2\mathbb Z$ and $K=\{0,1\}$ are isomorphism.

Hoseyn Heydari
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