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


"Arturo Magidin" <magidin@xxxxxxxxxxxxxxxxx> a écrit dans le message de
news: frr6ie$14uv$1@xxxxxxxxxxxxxxxxxxxxx
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;


Also Q(x) is monic ?

minimality of deg(g) now gives that P(x)=1 (in which case Q(x)=g(x))>
Ok.

or Q(x) is constant in which case
P(x) = g(x). So g(x) is irreducible in R[x].

I do not understand this cas.

Thanks for your enligthments.


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