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!


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