Subfield of a finite field

Question

I have started studying field theory and i have a question.somewhere i saw that a finite field with $p^m $ elements has a subfield of order $p^m $ where $m$ is a divisor of $n $.My question that if it is a field then how can it have a proper subfield.because since it is field it doesnt have any proper ideal.how can it have a subfield

Answer 1

You're assuming that a subfield has to be an ideal. This is not true. The conditions for being an ideal are in some respects harder to satisfy than being a subring/subfield -- more precisely, an ideal must contain a product $ab$ if even one of $a$ and $b$ is in the ideal, whereas a subring/subfield only needs to contain $ab$ if both $a$ and $b$ are in the subring.

(On the other hand, in other respects being a subfield is also harder than being a ideal, because ideals don't need to be closed under reciprocals).

If your reasoning were true, subfields could not exist at all -- but it should be well known that, for example, $\mathbb Q$ is a proper subfield of $\mathbb R$.

Answer 2

In a ring an ideal is something you can quotient by. For instance, $2\Bbb{Z}$ is an ideal in $\Bbb{Z}$ and we can construct the quotient $\Bbb{Z}/2\Bbb{Z}$ and that thing is still a ring.

Subfields are subsets that are fields. As one commenter pointed out, $\Bbb{Q} \subset \Bbb{R} \subset \Bbb{C}$ is a nested sequence of subfields. No one claims any of $\Bbb{C}/\Bbb{R}$, $\Bbb{C}/\Bbb{Q}$, or $\Bbb{R}/\Bbb{Q}$ are again fields or even rings. (They're not. Respectively: not closed under multiplication, not closed under addition, not closed under addition.)

If you try to write down a principal ideal for a field, it is either $(0)$ or $(r)$ for $r \neq 0$. In the first case, you get the ideal $\{0\}$ and in the second case, since $1/r$ is an element of the field, you generate $1$ and hence the whole field. General ideals are ideal sums of these, so any ideal in a field is not proper. The only thing you can mod out of a field, $F$, is either $\{0\}$ (which produces the same structure, just with "$+0F$" ($=0$) attached to everything) or $\{F\}$, which leaves the much simpler object $\{0\}$.

So, subfields exist, but the only subfields that you can mod out by are not proper.