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I have tried to solve an exercise regarding integral ring extensions which asked me to prove this:

If each of $\{R_i\mid i=1\ldots n\}$ is a family of integral ring extensions of a ring $R$, then $\prod_{i=1}^n R_i$ is an integral extension of $R$ as well.

It seems simple but I am not sure how I should prove that the polynomial is also monic

rschwieb
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Greg
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  • Are you asking this? : "The direct product of a finite family of rings which are integral extensions of a given ring $R$ is also integral over $R$." I am guessing the close voter had to do too many takes to figure out the question. – rschwieb Jan 09 '17 at 17:34
  • I am sorry, I tried so hard to write it down without the use of symbols and eventually messed it up. You are correct – Greg Jan 09 '17 at 17:44
  • "It seems simple but I am not sure how I should prove that the polynomial is also monic " What argument do you have in mind? Maybe write it down. Seprately there is not need to avoid symbols see http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference for details on typesetting. – quid Jan 09 '17 at 18:02
  • Its actually the Cartesian Product lets do it for n=2 p_1(x) \in R[X] p_1(s_1)=0 for every s_1\in S_1 The same goes for p_2 So I am think of choosing P=p_1*p_2 and s=(s_1,s_2) which is monic But here I am stuck because I do not know how I should substitute s in P – Greg Jan 09 '17 at 18:50

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