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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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    Never mind, I figured it out.2011-07-24

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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 $A\cong R[X]/\ker f$.

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    Never mind, the OP figured it out simultaneously with your answer :)2011-07-24
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    @Theo So it seems that we have just witnessed the [hundredth monkey effect](http://en.wikipedia.org/wiki/Hundredth_monkey_effect) =P2011-07-24
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    @Theo: Well, that's usually the very best outcome :)2011-07-24
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    @Mariano: Thanks, that was my proof as well. The confirmation only helps :)2011-07-24
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    @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.2011-07-24