Algebra over a ring

Question

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).

Never mind, I figured it out. – user3714 Jul 24 '11 at 17:26 — user3714, Jul 24 '11 at 17:26

score 5 · Accepted Answer · answered Jul 24 '11 at 17:26

5

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 '11 at 17:26

Mariano Suárez-Álvarez

135,076

1

Never mind, the OP figured it out simultaneously with your answer :) – t.b. Jul 24 '11 at 17:38
@Theo So it seems that we have just witnessed the hundredth monkey effect =P – Adrián Barquero Jul 24 '11 at 17:43
@Theo: Well, that's usually the very best outcome :) – Mariano Suárez-Álvarez Jul 24 '11 at 17:44
@Mariano: Thanks, that was my proof as well. The confirmation only helps :) – user3714 Jul 24 '11 at 17:56
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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. – Mariano Suárez-Álvarez Jul 24 '11 at 18:06

Algebra over a ring

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