This is from exercice 5, chap 2 from Atiyah and McDonald "Introduction to Commutative Algebra".

Let $A[x]$ be the ring of polynomials in one indeterminate over a ring $A$. Prove that $A[x]$ is a flat $A$-algebra.

Clearly, we notice that $\displaystyle A[x]=\bigoplus_{m=0}^\infty A\cdot (x^m)$.

We showed in the previous exercice that for any family $M_i$ ($i\in I$) of $A$-modules and $M$ their direct sum, then $M$ is flat iff each $M_i$ is flat.

Our problem is then reduced to showing that each $(x^m)$ is flat and that it is an $A$-algebra.

Some solutions on the internet require "Lang's Lemma" such that

[I]t only suffices to prove that the natural map $\phi : a \otimes (x^m ) \longrightarrow a(x^m )$ is an isomorphism for any ideal $a$ of $A$.

But we haven't seen that lemma previously in the book so is there another method ?


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    $\begingroup$ You probably mean $A[x] = \bigoplus_{m=0}^{\infty} A \cdot x^m$? $\endgroup$ – Martin Brandenburg Jan 8 '14 at 10:51
  • $\begingroup$ Indeed, coefficients are in A $\endgroup$ – ALM Jan 8 '14 at 10:56
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    $\begingroup$ And the degree is not bounded... $\endgroup$ – Martin Brandenburg Jan 8 '14 at 10:58
  • $\begingroup$ Edited, that's true :) $\endgroup$ – ALM Jan 8 '14 at 11:07

$A[x]$ is free as an $A$-module. Free modules are flat.

  • $\begingroup$ We need A to be isomorphic to each $A\cdot x^m$ then right ? Is it true ? I didn't know that property of free modules, thanks ! $\endgroup$ – ALM Jan 8 '14 at 10:56
  • $\begingroup$ Ok I get it, everything works then The property I was talking about is that free modules are flat. Thanks for your help $\endgroup$ – ALM Jan 8 '14 at 10:59

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