How to find a compositum of two fields

Question

Let $k_0$ be a field, $k$ its algebraic closure, and $K$ a field extension of $k_0$. Let $P_1$ and $P_2$ be prime ideals of $K \otimes_{k_0} k$. I have two questions about this.

1) I want to prove that there exists a field $\Omega$ extending $K$ containing both integral domains $(K \otimes_{k_0} k)/P_1$ and $(K \otimes_{k_0} k)/P_2$. How do I know such a big field exists?

2) Next let $\alpha_i: k \to (K \otimes_{k_0} k)/P_i$ defined by $x \mapsto (1 \otimes x) + P_i$. I want to show that $\alpha_1(k) = \alpha_2(k)$ in $\Omega$. How can I prove this?

Any explanation would be appreciated. Thank you!

Ps I have fixed the question and also added more to the question.

Answer 1

There are many different ways to construct a field $\Omega$; even if we require the field to be minimal with respect to the demanded properties, it will in general not be uniquely determined. As GreginGre says, one could for example set $L_1$ and $L_2$ to be the respective fraction fields of $(K \otimes_{k_0} k)/P_1$ and $(K \otimes_{k} k_0)/P_2$. Then $L_1$ and $L_2$ are both $K$-algebas, whereby also $L_1 \otimes_K L_2$ is in a natural way a $K$-algebra. Fixing a maximal ideal $\mathfrak{m}$ of $L_1 \otimes_K L_2$, $(L_1 \otimes_K L_2)/\mathfrak{m}$ is (in a natural way) a field extension of $K$ containing $(K \otimes_{k_0} k)/P_1$ and $(K \otimes_{k_0} k)/P_2$.

The second part will not be true exactly as you state it, but it will be true once we identify $k_0$ with a subfield of $\Omega$, i.e. if we demand that $\alpha_1$ and $\alpha_2$ coincide on $k_0$. I will sketch why this is a necessary extra condition in the end, but let us first see why it suffices.

So let $\Omega$ be any field extending both $(K \otimes_{k_0} k)/P_1$ and $(K \otimes_{k_0} k)/P_2$ such that the following diagram commutes: $$ \require{AMScd} \begin{CD} k_0 @>{\alpha_1}>> (K \otimes_{k_0} k)/P_1\\ @VV{\alpha_2}V @VVV \\ (K \otimes_{k_0} k)/P_2 @>>> \Omega \end{CD} $$ Then in particular, $\Omega$ will contain the algebraic closure $k$ of $k_0$. Certainly $\alpha_1(k)$ and $\alpha_2(k)$ consist only of elements which are algebraic over $k_0$. And conversely, if an element $x$ of $\Omega$ is algebraic over $k_0$, then it must already lie in $\alpha_1(k)$ and $\alpha_2(k)$. Hence, $\alpha_1(k) = \alpha_2(k)$, because both are equal to the relative algebraic closure of $k_0$ in $\Omega$.

If we do not fix an identification of $k_0$ as a subfield of $\Omega$, then certainly the statement could fail, even for $K = k_0$ and $P_1 = P_2$. Indeed, there are two at least two different ways to embed $\mathbb{C}(X)$ into $\mathbb{C}(X, Y)$ such that the images of the embeddings will be different, and the same for their algebraic closures.

Answer 2

Too long for a comment.

Let $L_i$ the field of fractions of $K\otimes_{k_0}/P_i$. Let $M$ be a maximal ideal of $L_1\otimes_{k_0} L_2$, and set $\Omega$ to be the corresponding quotient ring.

Then we have canonical maps $K\otimes_{k_0}/P_i\to L_i\to\Omega$, which are both injective, so the composite map is also injective. Then $\Omega$ "contains" $K\otimes_{k_0}/P_i$ ,up to canonical isomorphism. I doubt you can impose an true inclusion here, so in my opinion question 2) does not make sense.

I've asked a very similar question and did get any satisfactory answer (at least to my taste). I tried to do apply some classical set theory tricks to get two inclusions and didn't succeed (namely transporting structures). Even if we could, the construction of $\Omega$ would'nt be very natural...