# Order Of The Intersection of Two Subfields.

Let $$E$$ and $$F$$ be subfields of $$GF(p^n)$$. If $$|E| = p^r$$ and $$|F| = p^s$$, what is the order of $$E\cap F$$?

I read a corollary that "A finite field of order $$p^n$$ contains a unique subfield of order p^m for each $$m$$ | $$n$$ and no other subfields.

If that's the case, wouldn't that mean if two subfields have different orders, their intersection is $$0$$? So in this case, the order of $$E \cap F$$ = $$0$$?

• The intersection can't be zero since it contains the prime subfield $GF(p)$! And anyway, $1 \in E, F$! – Robert Lewis Apr 6 '14 at 6:04
• Very good point, thank you for the insight. – James Apr 6 '14 at 6:33
• If you think about that corollary, you should be able to answer your question. – Gerry Myerson Apr 6 '14 at 6:35
• Any source of this corollary in any text or notes will be helpful!. – BAYMAX Dec 23 '17 at 3:25

So if $t\mid r$, and $t\mid s$, then, by the result you quoted, both $E$ and $F$ contain a unique copy of $GF(p^t)$. But also $GF(p^n)$ contains a unique copy of $GF(p^t)$, which means that the copies of $GF(p^t)$ inside $E$ (resp. $F$) must coincide with the one in $GF(p^n)$, and thus be contained in the intersection $E\cap F$.
In view of this the intersection $E\cap F$ is a copy of $GF(p^\ell)$, where $\ell=\gcd(r,s)$.
• So if $E$ and $F$ contain a unique copy of $GF(p^t)$ and $GF(p^n)$ also contains a unique copy of $GF(p^t)$ then $|E \cap F|$ $\geq$ $p^t$, and the intersection also contains the prime subfield $GF(p)$ as well. So... $|E \cap F| = p^t + p$ in this case? – James Apr 6 '14 at 6:43
• @James: The prime field is contained in all the copies of $GF(p^t)$. You should also realize that the intersection has to be a subfield. The question is: which subfield? Or what is the biggest common subfield? – Jyrki Lahtonen Apr 6 '14 at 6:51
• so with that. Using your example $GF(p^t)$, wouldn't the biggest subfield ($E \cap F$) just be the biggest $t$ that divides $r$, $s$ and $n$? – James Apr 6 '14 at 7:08