# Intersection of any family of subfields is itself a subfield

Prove that the intersection of any family of subfields is itself a subfield.

In the countable case:

Suppose that $\mathscr K$ is a field and consider $(\mathcal K_n)_{n\in\mathbb N}\subset \mathscr K$ a family of subfield of $\mathscr K$ . Let show that $\mathcal K:=\bigcap_{n\in\mathcal N}\mathcal K_n$ is still a field. Let $a\in \mathcal K$. Then for all $n\in\mathbb N$, there is a $b_n\in\mathcal K_n$ such that $a b_n=1$

But I can't conclude.

And if the family is uncountable, how can I do ?

## 2 Answers

The main point is that operations in subfields coincide with operations in the large field. That's the definition of subfield.

In the particular case of inverses, you use uniqueness:

Let $a\in \mathcal K = \bigcap_{i\in\mathcal I}\mathcal K_i$. Since $\mathcal K_i$ is field, there is $b_i \in \mathcal K_i$ such that $ab_i=1$. Since $\mathscr K$ is a field, there is $b \in \mathscr K$ such that $ab=1$. But then $b=b1=bab_i = 1b_i=b_i$. Thus, $b\in \mathcal K_i$ for all $i$ and so $b\in \mathcal K$.

The proof above works for arbitrary families; it does not depend on the family being countable.

hint: use that the inverse is unique and also that the inclusions map $1 \mapsto 1$