# Splitting field is countable

I'm trying to prove that if $K$ is a countable field then there exists a countable field $L$ containing $K$ such that every polynomial in $K[X]$ splits in $L$. I know that if $L$ is the splitting field for one such polynomial then $[L:K]$ is finite and so $L$ is countable. I know that $K[X]$ is countable if $K$ is countable. But I can't see why the field containing the roots of all polynomials over $K$ should be countable. Thanks.

• The algebraic closure of $K$ is countable if $K$ is countable, hint: copy the proof for the rational $Q$ – Tsemo Aristide Dec 28 '15 at 20:43
• I'm actually using this result on the way to proving that the algebraic closure is countable. – gj255 Dec 28 '15 at 20:46

HINT: Since $K$ is countable, order the polynomials in $K$ as $p_i$ (for $i\in\mathbb{N}$) and consider a tower of extensions $$K\subset L_0\subset L_1\subset ...$$ such that $p_i$ splits in $L_i$. Then:

• How big is each individual $L_i$?

• How big is the union $L=\bigcup L_i$?

• What is this $L$?

• OK thanks. My problem is now reduced to showing that the union is in fact a field. This isn't obvious to me, since for instance $\mathbb{Q}(\sqrt{2}) \cup \mathbb{Q}(\sqrt{3})$ is not a field. – gj255 Dec 28 '15 at 20:44
• @gj255 They're all contained in one another, so at any finite point the union is simply the same as the last one in the chain. – Arthur Dec 28 '15 at 20:48
• OK thanks, so perhaps this is trivial but can you explain to me how this implies that the infinite union is a field? If $J_n$ is the union of $L_i$ for $i \leq n$ then it's clear to me that $J_n$ is a field for all finite $n$. But is $J_\infty$ obviously then a field? – gj255 Dec 28 '15 at 21:15
• @gj255 Well, you have to show that each of the field axioms is satisfied in $J_\infty$. For instance, one of the axioms (well, not really an axiom, more a property of binary operations) is that if $a, b$ are elements, then so is $a+b$. (Note that this fails for $\mathbb{Q}(\sqrt{2})\cup\mathbb{Q}(\sqrt{3})$.) Well, if $a, b\in J_\infty$, then $a\in J_i$ and $b\in J_j$ for some $i, j$. Now either $i\le j$ or $j\le i$; suppose WLOG $i\le j$. Then $a\in J_i$, so $a\in J_j$. But then $a, b\in J_j$, so since $J_j$ is a field, $a+b\in J_j$. The other axioms are similar. – Noah Schweber Dec 28 '15 at 21:27
• @gj255 Any two elements of $J_\infty$ belong to some $J_n$ for $n$ large enough, so their sum/product/inverse are all defined there. So the sum/product/inverse belong to $J_n \subset J_\infty$. – Alex Provost Dec 28 '15 at 21:29