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I'm trying to prove the following result. Let $R_1\subseteq R_2\subseteq R_3$ be integral domains such that $R_1\subseteq R_2$ and $R_2\subseteq R_3$ are algebraic extensions (note: I do not want to assume that they are integral). In this case I want to prove that the extension $R_1\subseteq R_3$ is also algebraic.

My idea is to use something like the Tower Law for field extensions, but using rank instead of dimension. That is, given an extension of integral domains $R\subseteq S$ I will write $[S:R]$ for the rank of $S$ as an $R$-module. Then I want to use the hypotheses above to show that $[R_3:R_1]<\infty$ and hence that $R_3$ is algebraic over $R_1$.

Any advice would be appreciated.

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  • $\begingroup$ What is an algebraic extension of rings? $\endgroup$
    – Qiaochu Yuan
    Commented Apr 6, 2016 at 2:08
  • $\begingroup$ Say that a ring extension $R\subseteq S$ is algebraic if every element $s\in S$ satisfies a polynomial equation $f(s)=0$ for $0\neq f(x)\in R[x]$. $\endgroup$
    – Drew Armstrong
    Commented Apr 6, 2016 at 2:34
  • $\begingroup$ Can't you reduce it to the tower law for their quotient fields? $\endgroup$
    – darij grinberg
    Commented Apr 6, 2016 at 5:59
  • $\begingroup$ And by "tower law", I mean "an algebraic extension of an algebraic extension is algebraic", not the formula for the dimension. An algebraic extension can have infinite dimension. $\endgroup$
    – darij grinberg
    Commented Apr 6, 2016 at 6:00
  • $\begingroup$ Why not? If $s \in S $ is a root of a polynomial with coefficients in $\operatorname{Frac}R $, then can't we just multiply the polynomial by the common denominator of its coefficients? $\endgroup$
    – darij grinberg
    Commented Apr 6, 2016 at 19:23

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