# Transitivity of integral extensions and prime ideals

The situation is as follows:

We have

• $$K$$ a field

• $$K[a_1, ..., a_n] \subseteq R$$ finite ring extension

• $$R \subset R'$$ integral ring extension of integral domains

Since $$R$$ is finite over $$K[a_1, ... , a_n]$$, it also is integral over it. By transitivity of integral extensions $$R'$$ is integral over $$K[a_1, ..., a_n]$$.

Sadly, in my lecture the proof the transitivity was omitted and I wonder whether it works via composition of the field extension homomorphism or somehow else.

The context is that I have prime ideals $$p_1 \subsetneq p_2 \subset R$$ and $$q_2 \subset R'$$ with $$q_2 \cap R=p_2$$ and I am trying to find a prime $$q_1 \subset R'$$ with $$q_1 \subsetneq q_2 \subset R'$$.

I have shown the existence by considering prime ideals $$p_1 \cap K[a_1, ... , a_n] \subsetneq p_2 \cap K[a_1, ... , a_n]$$ and using the Going Down theorem, as $$K[a_1, ... , a_n]$$ fulfills the requirements for it.

Now I was wondering whether it is guaranteed that $$q_1 \cap R=p_1$$. I have a strong feeling that it is not, but I do not know how the extension from $$R \subset R'$$ influences the extension from $$K[a_1, ... , a_n] \subset R'$$.

Any insight on this is welcome!