Proposition 5.1 from Commutative Algebra by Atiyah and Macdonald:

$x∈B$ is integral over $A$,then $A[x]$ is a finitely generated $A$-module.

The elements in $A[x]$ are the set of all the sumenter image description here.

If $A[x]$ is a $A$-module,then it's the ideal of $A$.

But the sum may not in $A$, since $x$ may be the element of $B-A$.

Then I go to read the proof to looking for the answer.

I still don't figure out but more confused.

enter image description here

The new problem is why all positive powers of $x$ lie in the $A$-module generated by $1,x,...$..

If enter image description here,

then there may exist some $r < n$ such that enter image description here.

Hence there exist a positive powers of $x$ lie in the $A$-module doesn't generated by $1,x,...$..

I must misunderstand something, I have checked the concept, but still don't figure out where is the mistake.

That's all the question and what I have try.Could you tell me where is the mistake?

Thanks in advance

  • 1
    $\begingroup$ Be careful. $A$-modules aren't necessarily ideals of $A$. When you're considering $A$ as being a module over itself, then submodules of $A$ are ideals of $A$ - this is probably what you're thinking of. $\endgroup$
    – ah11950
    Apr 1, 2014 at 7:08

1 Answer 1


You say:"If $A[x]$ is a $A$-module,then it's the ideal of $A$." this is not true. ideals of $A$ are $A$-modules, but the converse is not always true.
for the 2nd question:
positive powers of $x$ are $\ge$ $n$ or are $1,2,...,n-1$. in each case they lie in the $A$-module generated by $1, x, .., x^{n-1}$ because $x^{n+r}=...$

  • $\begingroup$ Thank you for the clarification.I figure it out now. $\endgroup$
    – amateur
    Apr 2, 2014 at 5:23

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