# Is $\mathbb Z/p\mathbb Z$ a subfield of every finite field?

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 $K$.

• Strictly speaking, the author should say a subfield isomorphic to $\mathbb{Z}/p\mathbb{Z}$. Oct 29, 2013 at 13:29
• 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. Oct 29, 2013 at 13:30
• @GA316 If by $Q$ you mean the rationals, then that is not correct. Oct 29, 2013 at 13:31
• @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 $\;\Bbb Q\;$ . Oct 29, 2013 at 13:33
• @Casteels, that's speaking too strictly, imo. Oct 29, 2013 at 13:33

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 sub*field* isomorphic to $\mathbb{Q}$.

Otherwise the kenerl 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).

• thank you, so the statement would be true if the author had written "isomorphic". Oct 29, 2013 at 14:10
• Yes, but it's probably a deliberate omission, and this is common practice in abstract algebra, so be aware. Oct 29, 2013 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, 2017 at 20:35
• @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.) Dec 31, 2017 at 0:47

$\mathbb Z / 2\mathbb Z$ contains $K=\{0,1\}$. It means $\mathbb Z / 2\mathbb Z$ and $K=\{0,1\}$ are isomorphism.