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September 11th, 2012, 10:17 AM   #1
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Show R (comm. domain) over a field k is a field if dimR<inft

I'm working on the following problem:

If , where is a field, and is a commutative domain with , then is a field and is algebraic.

Since is already a commutative domain, I need only show every nonzero element of has a multiplicative inverse in . So, take any nonzero element . If , I'm done, so say . As , I think this means I can think of as , for some (?). If this is so, , for . But then, using the fact that is a domain, I seem to obtain that , implying that , which I don't feel is correct.

Any suggestions or hints of what I'm missing? Thanks!
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September 14th, 2012, 10:07 PM   #2
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Re: Show R (comm. domain) over a field k is a field if dimR<

first of all, note R is, in fact, a vector space over k:

(R,+) is clearly an abelian group (since R is a ring).

for a in k, and r in R, we can define a.r = ar (the LHS is the scalar multiplication in the vector space, the RHS is the product in the ring R).

(a+b).r = (a+b)r = ar + br = a.r + b.r (since a ring satisfies the distributive laws).

a.(r+s) = a(r+s) = ar + as = a.r + a.s

a.(b.r) = a.(br) = a(br) = (ab)r = (ab).r (since the product in a ring is associative).

1.r = 1r = r

now suppose r ? 0 in R, with dim_k(R) = n. then the set {1,r,r^2,...,r^n} is linearly dependent, so:

c0 + c1r + c2r^2 +...+cnr^n = 0 for some {c0,c1,...,cn} where not all the cj are 0. let m = max{j: cj ? 0}.

then r satisfies the polynomial:

p(x) = c0 + c1x + c2x^2 +...+ cmx^m

so r is algebraic over k. therefore there is a monic polynomial of least degree, m(x), for which m(r) = 0.

i claim that m(x) is irreducible in k[x]. for suppose not:

m(x) = f(x)g(x) where deg(f),deg(g) < deg(m).

then 0 = m(r) = f(r)g(r), and since R is a domain, either f(r) = 0, or g(r) = 0, contradicting the minimal degree of m(x).

i further claim that the constant term of m(x) is non-zero, for if not then x is a factor of m(x), contradicting the irreducibilty of m(x).

so we have:

c0 + c1r +...+c(t-1)r^(t-1) + r^t = 0 (where t = deg(m)), and c0 ? 0.

therefore:

1 = -(c1/c0)r - (c2/c0)r^2 -....- (c(t-1)/c0)r^(t-1) - r^t = r[-(c1/c0) - (c2/c0)r -...- (c(t-1)/c0)r^(t-2) - r^(t-1)],

which show r is a unit, thus R is a field.
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