Re: math -- algebraic degree vs integral degree

Let R, S be integral domains, where R is contained in
S.

Suppose s in S is such that s is integral over R.

Let J = {f in R[x] | f(s) = 0}

Let m be the "algebraic degree of s over R", that is,


m = min {deg(f) | f in J, f =/= 0}

and let n be the "integral degree of s over R", that
is,

n = min {deg(f) | f in J, f monic}

Clearly, we always have m <= n.

In the thread

"Proof of existence of unique monic polynomial of
of minimal degree",

I proved that if R is a UFD, then m = n.

Prove or disprove:

If R is integrally closed, then m = n.

quasi


The algebraic degree equals the degree of the minimal
polynomial f \in K[x] of s over K:=Frac(R). Hence if
we show f \in R[x] we are through.

1. By assumption R can be expressed as an intersection of
valuation rings O_v of K. Let V be the family of valuations
of K appearing in this representation.

2. A valuation v of K can be extended to the polynomial
ring K[x] resp. its field of fractions K(x) by setting

v_x(p):=min (v(c) : c runs through the coefficients of p)

where p \in K[x].
(To prove that v_x is a valuation one needs a generalization
of the so-called Lemma of Gauss for polynomials over factorial
rings.)

3. A polynomial p \in K[x] lies in R[x] if and only if v_x(p)>=0
for all valuations v appearing in V.

4. Let g \in R[x] be a monic polynomial such that g(s)=0 and let
f \in K[x] be the (monic) minimal polynomial of s over K.
Then g=f*h for some h \in K[x]. Since g and f are monic, h is
monic too.
For every valuation v \in V one has

v_x(g)=v_x(f)+v_x(h)

Since g is monic and in R[x]: v_x(g)=0.
Since f and h are monic: v_x(f)<=0 and v_x(h)<=0.
Thus v_x(f)=0.

H
.

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