Question about subfield of order p

Question

I have seen this fact:-

The minimal subfield of a field F of characteristic p is the field of p-elements.

To me when I hear a finite field my mind directly go to think about the field of $\mathbb{Z}/p \mathbb{Z}$ and its clear that this fact above is not work with it ( at least for non-trivial subfield).

I wonder what other field do we have such that $Cha(\mathbb{F})=p$ and have non-trivial subfield of order $p$.

Answer 1

For every integer $n > 0$ and prime $p$ there is [up to isomorphy] exactly one field of size $p^n$. This field is written as $\mathbb{F}_{p^n}$ [see here for a construction] and it has $\mathbb{F}_p \cong \mathbb{Z}/p\mathbb{Z}$ as subfield. If a field has finite characteristic $p$, then the smallest subfield it contains is isomorphic to $\mathbb{Z}/p\mathbb{Z}$.

There are also infinite fields that have finite characteristic. Take for example the field $\mathbb{F}_{p^n}(X)$ of rational functions in one variable with coefficients in $\mathbb{F}_{p^n}$.

Answer 2

For any given $p$, find a polynomial of degree $2$ that is irreducible mod $p$, and call it $f(T)$. Define $\mathbb{F}_{p^2} :=(\mathbb{Z}/p\mathbb{Z})[T]/(f(T))$. Then $\mathbb{F}_{p^2}$ is a field of order $p^2$, with characteristic $p$ and has $\mathbb{F}_p$ as a subfield.

Answer 3

Do you know about extension fields?

Here is the simplest example with $p = 2$. Consider the polynomials, namely $0, 1, x, 1+x$, of degree 1 or less with coefficients in $\mathbb F_2 = \mathbb Z/2\mathbb Z$,. Let us define multiplication of these polynomials as polynomal multiplication modulo $x^2+x+1$ so that $$x\star x = x^2 \equiv x+1, x\star(1+x) = x+x^2 \equiv 1, (1+x)\star (1+x) = 1+x^2 \equiv x. $$ Then, these 4 polynomials with multiplication as defined here are $\mathbb F_4$, the field of 4 elements. The characteristic is $2$, and the elements $0$ and $1$ constitute the subfield $\mathbb F_2$.