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On 29 Apr 2006 18:38:06 -0700, "Gmath" <kannnku@xxxxxxxxxxx> wrote:
>Is there anyone can help me prove this problem? " Let S be an integral
>extention over ring R. Then the polynomial ring S[x] is an integral
>extension of R[x]" . S is integral over ring R iff for each s in S
>there is a monic polynomail f(y) in R[y] such that f(s)=0. Thank you.
I'll assume you know the following basic property of integral
elements.
Let A be a subring of B and let H be the set of elements of B which
are integral over A. Then H is a subring of B and H contains A.
So for your problem, let H be the set of elements of S[x] which are
integral over R[x].
Since S is integral over R, S is integral over R[x], hence S is a
subset of H.
Also x is in R[x], so x is in H.
But the smallest subring of S[x] containing S and x is S[x].
It follows that H=S[x], so S[x] is integral over R[x].
quasi
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