Re: Proof of Existence of Unique monic polynomial of minimal degree
- From: magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin)
- Date: Wed, 19 Mar 2008 14:05:02 +0000 (UTC)
In article <y8z63vj6v5v.fsf@xxxxxxxxxxxxxxxxxxxx>,
Bill Dubuque <wgd@xxxxxxxxxxxxxxxxxxxx> wrote:
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.
Duh. Quite right. Sorry about that.
Let g(x) be monic of minimal degree among polynomials in R[x] that
have s as a root. We claim g(x) is irreducible in F[x], where F is the
field of fractions in R[x]. Indeed, by Gauss's Lemma (which holds
since R is a UFD) any factorization g(x) = p(x)q(x) in F[x]
corresponds to a factorization g(x)=P(x)Q(x) with p(x),q(x) in R[x],
and with P(x)=ap(x), q(x) = bq(x) for some a,b in F. The leading
coefficients of P and Q must be units, so multiplying by an adequate
unit we may assume P(x) is monic; minimality of deg(g) now gives that
P(x)=1 (in which case Q(x)=g(x)) or Q(x) is constant in which case
P(x) = g(x). So g(x) is irreducible in R[x]. In particular, g(x) is
the irreducible of s in F[x], so if f(x) is any other monic polynomial
in R[x] with f(s) = 0 and deg(f)=deg(g), then f(x) must be an
F-multiple of g(x). Since they are both monic, f(x)=g(x).
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
.
- References:
- Proof of Existence of Unique monic polynomial of minimal degree
- From: TimmyTimTim
- Re: Proof of Existence of Unique monic polynomial of minimal degree
- From: matmackaizer
- Re: Proof of Existence of Unique monic polynomial of minimal degree
- From: Arturo Magidin
- Re: Proof of Existence of Unique monic polynomial of minimal degree
- From: Bill Dubuque
- Proof of Existence of Unique monic polynomial of minimal degree
- Prev by Date: Re: Try this one.. Really Interesting!!!
- Next by Date: Re: -- Hopelessly stuck on normed algebra proof (repost)
- Previous by thread: Re: Proof of Existence of Unique monic polynomial of minimal degree
- Next by thread: Re: Proof of Existence of Unique monic polynomial of minimal degree
- Index(es):
Relevant Pages
|