Re: Another algebra question

In article <1191170849.443132.285730@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Snis Pilbor <snispilbor@xxxxxxxxx> wrote:
On Sep 30, 9:10 am, quasi <qu...@xxxxxxxx> wrote:
On Sun, 30 Sep 2007 01:55:37 -0700, Snis Pilbor <snispil...@xxxxxxxxx>
wrote:

another thing I am stumped on. Suppose R is a ring in a field F, and
R is integrally closed in its field of fractions. I want to show that
an element x in F is integral over R (that is, it's annihilated by
some monic polynomial with coefficients in R), iff its minimal
polynomial has coefficients in R.

Hints:

Suppose x is integral over R. Let K be the quotient field of R.

(1) Let p be the minimal polynomial of x over K. By definition, p is
monic and the coefficients of p are in K.

(2) Argue that the coefficients of p are integral over R [think about
symmetric polynomials].

(3) But R is integrally closed, so ...

quasi

Interesting, but this only shows that the minimal polynomial of x over
K are elements of R. Couldn't x have a different minimal polynomial
over F?

Since x is assumed to be an element of F, its minimal polynomial over
F is X - x. Probably not the same one as over K (unless x happens to
lie in K as well)

But... So what? "Minimal polynomial" above refers to the minimal over
R (i.e., over K), not over F.


--
======================================================================
"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

.

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