Let $F$ be a field, $E_1$ and $E_2$ are two distinct extension fields of $F$. Is it the case that we can always somehow find a field $G$ that contains both $E_1$ and $E_2$? In other words, could extensions of fields have different 'direction's such that they are incompatible?
Edit: I began to think about this problem while reading a proof. $F$ is a field. $a$ and $b$ are algebraic over $F$. $p(x)$ and $q(x)$ are two polynomials in $F[x]$ of minimum degree that respectively make $a$ and $b$ a zero. The proof claims that there is an extension $K$ of $F$ such that all distinct zeros of $p(x)$ and $q(x)$ lie in $K$. For a single polynomial, I know this kind of field exists because of the existence of splitting field, why it is true for two polynomials?
Best Answer
Consider field extensions $E_1/F$ and $E_2/F$. Then the tensor product $A=E_1\otimes_F E_2$ is a commutative ring, not necessarily a field though. Non-trivial commutative rings have maximal ideals, by a Zorn's lemma argument. Let $I$ be a maximal ideal of $A$. Then $K=A/I$ is a field. The map $x\mapsto \overline{x\otimes 1}\in A/I$ is a ring homomorphism $E_1\to K$. As $E_1$ is a field, this is an injective homomorphism, so we can think of $E_1$ being "contained" in $K$. Likewise $E_2$ is "contained" in $K$.
Beware though, the ideal $I$ may not be unique.