(adsbygoogle = window.adsbygoogle || []).push({}); 2. Let [tex]R[/tex] be an integral domain containing [tex]F[/tex] as a subring. Suppose that [tex]R[/tex] is finite-dimensional when viewed as a vector space over [tex]F[/tex]. Prove that [tex]R[/tex] is a field.

SOLUTION

I will show that [tex] \forall \ r \in R, (r \neq 0), r[/tex] is a unit. Equivalently, [tex] \forall \ r \in R, (r \neq 0), \exists \ s \in R, (s \neq 0), [/tex] such that [tex]

rs = sr = 1[/tex].

Suppose [tex] dim_{F}(R)=n [/tex] and consider the subset [tex] \{ 1, r, r^2, r^3, ..., r^n, r^{n+1} \}[/tex]. This set is linearly dependent over [tex]F[/tex]. That is, [tex] \exists \ a_0, a_1, ..., a_{n+1} \in F - \{ 0 \}[/tex] such that:

[tex]a_0(1) + a_1(r) + a_2(r^2) + ... + a_n(r^n) + a_{n+1}(r^{n+1}) = 0[/tex]

Then, [tex]a_1(r) + a_2(r^2) + ... + a_n(r^n) + a_{n+1}(r^{n+1}) = -a_0[/tex]

Since [tex]F[/tex] is a field, and [tex]a_0 \neq 0[/tex], it follows that [tex]a_0[/tex] must be a unit. So, we can divide the last equation by [tex]-a_0[/tex] producing:

[tex]- \frac{a_1}{a_0}(r) - \frac{a_2}{a_0}(r^2) - ... - \frac{a_n}{a_0}(r^n) - \frac{a_{n+1}}{a_0}(r^{n+1}) = 1[/tex]

[tex]r(- \frac{a_1}{a_0} - \frac{a_2}{a_0}(r) - ... - \frac{a_n}{a_0}(r^{n-1}) - \frac{a_{n+1}}{a_0}(r^n)) = 1[/tex]

Call the expression in the parentheses [tex]s[/tex]. We have shown that [tex]rs=1[/tex]. Additionally, [tex]s \neq 0[/tex] since [tex]R[/tex] in n-dimensional over [tex]F[/tex]. Finally, since [tex]R[/tex] is an integral domain, [tex]rs = sr = 1[/tex].

Therefore, every element in [tex]R[/tex] is a unit, and so [tex]R[/tex] must be a field.

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