Re: Proof of Existence of Unique monic polynomial of minimal degree

magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin) wrote:
<matmackaizer@xxxxxxxx> wrote:
Gerry Myerson <ge...@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
TimmyTimTim <matmackai...@xxxxxxxx> wrote:

Suppose R is a unique factorization domain, and suppose
S is an integral domain and which is integral over R.
Show that for every element s in S there is a UNIQUE
monic polynomial P in R[x] of MINIMAL degree, such that P(s) = 0.

Well, the fact that there is a least degree among all monic
polynomials in R[x] for which s is a root is of course trivial.
Suppose f and g are two monic polynomials with f(s)=g(s)=0, of minimal
degree among all monic polynomials h(x) in R[x] with h(s)=0. Since g
is monic, we can divide f by g to get

f(x) = q(x)g(x) + r(x) with r(x)=0 or deg(r)<deg(g).

Then 0 = f(s) = q(s)g(s) + r(s) = r(s). Thus, r(s)=0; by the
minimality of the degree of g, r(x) = 0. [...]

But r(x) needn't be monic so the "proof" is incorrect.

--Bill Dubuque
.

Relevant Pages

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  • Re: Proof of Existence of Unique monic polynomial of minimal degree
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