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 September 11th, 2012, 10:17 AM #1 Senior Member   Joined: Nov 2011 Posts: 100 Thanks: 0 Show R (comm. domain) over a field k is a field if dimR
 September 14th, 2012, 10:07 PM #2 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 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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