Could someone point me to a proof which shows that an algebra over a ring can be presented as a quotient of a polynomial ring (in possibly infinitely many variables).
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$\begingroup$ Never mind, I figured it out. $\endgroup$– user3714Jul 24, 2011 at 17:26
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Let $A$ be an $R$ algebra. Let $X$ be a set of variables $x_a$ which arein bijection with the set $A$. Consider the unique map of $R$-algebras $f:R[X]\to A$ which maps $x_a$ to $a$ for all $a\in A$. This is clearly surjective, so $A\cong R[X]/\ker f$.
answered Jul 24, 2011 at 17:26
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1$\begingroup$ Never mind, the OP figured it out simultaneously with your answer :) $\endgroup$– t.b.Jul 24, 2011 at 17:38
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$\begingroup$ @Theo So it seems that we have just witnessed the hundredth monkey effect =P $\endgroup$ Jul 24, 2011 at 17:43
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$\begingroup$ @Theo: Well, that's usually the very best outcome :) $\endgroup$ Jul 24, 2011 at 17:44
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$\begingroup$ @Mariano: Thanks, that was my proof as well. The confirmation only helps :) $\endgroup$– user3714Jul 24, 2011 at 17:56
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1$\begingroup$ @user3714: notice that my construction uses many, many more variables than what is ever needed. Is is enough to use one variable per element in a generting set of $A$ as an $R$-algebra. $\endgroup$ Jul 24, 2011 at 18:06