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A is an integral domain/ring, $x \in Frac(A)$. Show that if $A[x]$ is a finite A-module, then $\exists P \in A[X]$ unitary (i.e of the form $P(X)=X^n+a_{n-1} X^{n-1}+...$) such that $P(x) = 0$

Im really stuck there. I wrote that there are elements $e_1, ..., e_n$ that generate $A[x]$ and I would like to "invert the matrix" ($x^i$ function of the $e_j$) but modules are not like vector spaces so I don't know what to do from there. Thanks!

Thomas
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1 Answers1

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Since $A [x]$ is finite, there exist $f (X),...,f (X)\in A [X]$, the polynomial ring in one variable over $A $, such that $A [x]=<f_1(x), ...,f_n(x) >$, as an $A-$module, and set $n=Max\{deg(f_1(X)),..., (f_n(X))\}$. Now since $x^{ n+1 }\in A [ x ]$, there exist $a_1,...,a_n\in A $ such that $x^{n+1}=a_1f_1(x)+...+a_nf_n(x)$. Set $ P (X)=X^{ n+1 }-a_1f_1(X)-....-a_nf_n(X) $. Clearly, $P (x)=0$ as desired.

E.R
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