Proposition 4

— Let E, F be two extensions of K.

a) Let $(L, u, v)$ be a composite extension of E and F; then the kernel p of the homomorphism $u * v$ of $E \otimes_K F$ into L is a prime ideal.

b) Let p be a prime ideal of $E \otimes_K F$; then there exists a composite extension $(L, u, v)$ of E and F such that p is the kernel of $u * v$, and any two such composite extensions are isomorphic.

Assertion a) follows from the fact that the kernel of a homomorphism of a ring into a field is a prime ideal (I, p. 116-117).

Let p be a prime ideal of $E \otimes_K F$, A the quotient ring $(E \otimes_K F)/p$ and L the field of fractions of A. For $x \in E$ (resp. $y \in F$) we denote by $u(x)$ (resp. $v(y)$) the residue class mod p of $x \otimes 1$ (resp. $1 \otimes y$). Then $u$ (resp. $v$) is a K-homomorphism of E (resp. F) into L and $u(E) \cup v(F)$ generates A as a ring, hence L as a field. Therefore $(L, u, v)$ is a composite extension of E and F; we see at once that $u * v$ is the canonical homomorphism of $E \otimes_K F$ into L, and its kernel is thus equal to p.

Let $(L', u', v')$ be a composite extension of E and F such that the kernel of $u' * v'$ is equal to p. Since $u * v$ and $u' * v'$ have the same kernel, there exists an isomorphism $\psi$ of A onto the image $A'$ of $u' * v'$, characterized by $u' * v' = \psi \circ (u * v)$. But $A'$ is the subring of $L'$ generated by $u'(E) \cup v'(F)$, hence L' is the field of fractions of A'. Therefore $\psi$ extends to an homomorphism $\varphi$ of L onto L' and it is clear that $\varphi$ is an isomorphism of $(L, u, v)$ onto $(L', u', v')$.