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

In article <c2f62a3e-7890-4e66-94b1-567f0d1b7861@xxxxxxxxxxxxxxxxxxxxxxxxxx>,
<matmackaizer@xxxxxxxx> wrote:

Please don't top-post.

http://www.xs4all.nl/~hanb/documents/quotingguide.html

http://www.caliburn.nl/topposting.html

http://www.html-faq.com/etiquette/?toppost

Edited to remove the top-posting...

On Mar 18, 11:44 pm, Gerry Myerson <ge...@xxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
In article
<10880484.1205894357093.JavaMail.jaka...@xxxxxxxxxxxxxxxxxxxxxx>,

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 r in R there is a UNIQUE monic polynomial P in
R[x] of MINIMAL degree, such that P(r) = 0.

Simple: x - r.

But maybe you meant for every s in S.

Yes, I meant for every s in S. How to prove 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 when R is UFD, S is integral domain that is
integral over R.

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.

Since we are in a domain, deg(f) = deg(g)+deg(q), so deg(q)=0, hence q
is constant. Since both f and g are monic, q=1, so f=g.

--
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Arturo Magidin
magidin-at-member-ams-org

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