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Let k Be a Field

Date: 04/20/99 at 00:18:28
From: Erin Gallagher
Subject: Abstract Algebra

Could you please help out if possible?

Let k be a field for 1 and 2

  1. Prove or disprove that a prime ideal I of the ring k[x] is a 
     maximal ideal.

  2. Prove that a nonzero polynomial in k[x] can not have more roots 
     in k than its degree.

  3. Let R be a ring and M be an ideal of R. Suppose that every  
     element of R that is not in M is a unit of R. Prove that M is a 
     maximal ideal and that it is the only maximal ideal of R.

Date: 04/20/99 at 15:34:40
From: Doctor Rob
Subject: Re: Abstract Algebra

Thanks for writing to Ask Dr. Math.

1. First of all, because there is a Euclidean Algorithm for k[x], you
can always find a single generator of every ideal (that is, every 
ideal is principal). Next, consider a generator f(x) of the ideal I.
f(x) cannot be factorable, because if f(x) = g(x)*h(x) with degrees
of g and h less than the degree of f, then, since I is a prime ideal,
either g or h would be in I.  Then since the degree of f is larger
than the degree of g, g cannot be a multiple of f, so f would not be
a generator of the ideal I, a contradiction. Thus f(x) must be

Now let M be an ideal properly containing I, and take any polynomial 
F(x) in M - I, that is, not a multiple of f(x). Then GCD(F(x),f(x)) = 
1, because f is irreducible and F is not a multiple of it. That means 
that there exist polynomials a(x) and b(x) in k[x] with a(x)*F(x) + 
b(x)*f(x) = 1. That means that 1 lies in the ideal M. That means that 
M is the whole ring k[x]. (Why?) You finish.

2. Proceed by induction on the degree of f(x). It is true for nonzero 
constants, which have degree 0 and no roots. Suppose it is true for 
all polynomials of degree k-1, and take f(x) of degree k. Let r be any 
root of f(x). Use the division algorithm to write

   f(x) = q(x)*(x-r) + s, degree(s) < degree(x-r) = 1,

so s is a constant. Use the fact that r is a root of f(x) to show
s = 0 (or use the Remainder Theorem, if you know that). Then

   f(x) = q(x)*(x-r),

and the degree of q(x) is k-1. Apply the induction hypothesis and
finish the proof.

3. Use the usual trick to prove M is maximal. Suppose it is in a 
larger ideal I, and pick r in I - M. Then r is a unit, so it has an
inverse s, so s*r = 1. Since r is in I, and s is in R, s*r = 1 is in 
...  you finish.

- Doctor Rob, The Math Forum   
Associated Topics:
College Modern Algebra

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