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Let $F$ be a field, and let $K_1$, $K_2$ be two splitting fields over $F$ (Suppose they are contained in a larger field $K$). Is $K_1\cap K_2$ necessarily a splitting field over $F$?

The statement is true if $K_1$ and $K_2$ are finite extensions of $F$, however I'm not sure how to prove (or disprove) the statement in the general case.


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up vote 6 down vote accepted

The following is standard; for example, it appears in Lang's Algebra.

Proposition. Let $K$ be an algebraic extension of $F$, contained in some algebraic closure $\overline{F}$ of $F$. The following are equivalent:

  1. Every embedding of $K$ into $\overline{F}$ over $F$ is an automorphism of $K$.

  2. $K$ is as splitting field over $F$.

  3. Every irreducible polynomial $f(x)\in F[x]$ that has at least one root in $K$ splits over $K$.

Proof. 1$\Rightarrow$2,3: Let $a\in K$, and let $f(x)$ be its irreducible polynomial over $F$. If $b$ is any root of $f(x)$ in $\overline{F}$, then the isomorphism $F(a)\to F(b)$ that maps $a$ to $b$ extends to an embedding of $K$ over $F$; by 1, it is an automorphism of $K$, so $b\in K$. Thus, an irreducible polynomial in $F[x]$ that has at least one root in $K$ has all its roots in $K$, proving 3. Letting $S$ be the collection of irreducible polynomials over $F$ of elements of $K$ shows that $K$ is a splitting field over $F$.

2$\Rightarrow$1: Let $S=\{f_i\}$ be a family of polynomials such that $K$ is the splitting field of $S$ over $F$. If $a$ is a root of some $f_i$ and $a\in K$, then for any embedding of $K$ into $\overline{F}$, $a$ must map to a root of $f_i$; but the roots of $f_i$ are all in $K$. Moreover, since $K$ is generated over $F$ by the roots of the $f_i$, then this means that every element of $K$ maps to an element of $K$ under the embedding. So $K$ satisfies condition 1.

3$\Rightarrow$1: Let $\sigma\colon K\to\overline{F}$ be an embedding over $F$. Let $a\in K$, and let $f(x)$ be the monic irreducible of $a$ over $F$. If $\sigma\colon K\to\overline{F}$ is any embedding over $F$, then $\sigma$ maps $a$ to a root of $f(x)$; but by assumption, all the roots of $f(x)$ are in $K$, so $\sigma(a)\in K$. Thus, $\sigma$ maps $K$ into itself, proving 1. QED

From this, it is easy to verify that the intersection of two splitting fields is indeed a splitting field.

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nice answer! – Rudy the Reindeer Mar 24 '11 at 20:36

It is proved in Serge Lang's book 'Algebra' page no 237 that an algebraic field extension $K/k$ is splitting field if and only if every irreducible polynomial of $k[X]$ which has a root in $K$ splits into linear factors in $K$. Use this result to answer your question.

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Saying "Serge Lang's book" is like saying "Euler's paper", or "the proposition in Euclid." – Arturo Magidin Mar 24 '11 at 20:50
That made me think of this joke posted on MO: "The primary reason Bourbaki stopped writing books was the realization that Lang was one single person." – Myself Mar 26 '11 at 11:58

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