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The following is a proposition from abstract algebra. The book gives this as a proposition and asks the reader to do it as an exercise. It seems it should be easy. (Assuming every ring has a $1$.)

Let $R$ be a subring of $S$. If $S \setminus R$ is closed under multiplication, then $R$ is integrally closed in $S$.

I went by contradiction. Let $s$ be in $R \setminus S$. Then, we get that $$ s^n+a_1s^{n-1} + \cdots + sa_{n-1}+a_{n}=0 $$ for some $n$ and the $a$'s are in $R$. Now I am thinking this $$ s^n=(-1)(a_1s^{n-1} + \cdots + sa_{n-1}+a_{n}) $$ Now multiply both sides by $s$ again to get $$ ss^n=(-s)(a_1s^{n-1} + \cdots + sa_{n-1}+a_{n}) $$ Now by closure, $ss^n$ is in $S \setminus R$ which means the rest $a_1s^{n-1} + \cdots + sa_{n-1}+a_{n}$ is in $S \setminus R$ which means the $a_i$'s are in it, which is a contradiction. Does this look correct?

Edit (not by OP): The above is also Exercise 7 from Chapter 5 of Atiyah and Macdonald's Introduction to Commutative Algebra.

user26857
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user224392
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  • Why to denote by $;s;$ an element in $;R\setminus S;$ ? To confuse? Anyway, if $;R\subset S;$ , then it must be $;s\in S\setminus R;$ . Finally, at the end you say "...which is a contradiction"? To what ? Where did you clearly state what is the assumption you make? I think I know what you meant, yet you must explicitly state this. – Timbuc Mar 17 '15 at 19:17
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    Welcome to [math.se]. Please don't edit out your question after you have received an answer/a hint, the purpose of the site is to have the questions and answers as resources for others having the same or a sufficiently similar problem in the future. Editing the question out to say thanks defeats that purpose. – Daniel Fischer Mar 17 '15 at 22:08

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Hint $\ $ Choose $\,n\,$ minimal. Then $\ s f(s) = s (s^{n-1}+ r_{n-1} s^{n-2} + \cdots + r_1)\, =\, -r_0.\,$ But $\,f(s)\not\in R\,$ by minimality of $\,n\,$ so $\,s,f(s)\in S\backslash R \,\Rightarrow\, sf(s) = -r_0\in S\backslash R ,\,$ contradiction.

Bill Dubuque
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