Prep for Algebra Comprehensive Exam #2

[BSMSMSTMSPHD]

2. Let R be an integral domain containing F as a subring. Suppose that R is finite-dimensional when viewed as a vector space over F. Prove that R is a field.

SOLUTION

I will show that \forall \ r \in R, (r \neq 0), r is a unit. Equivalently, \forall \ r \in R, (r \neq 0), \exists \ s \in R, (s \neq 0), such that <br /> rs = sr = 1.

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

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

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

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

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

Therefore, every element in R is a unit, and so R must be a field.

 

[Kummer]

Another approach is to show the maximal ideal of R is \{ 0\}. But this is not shorter than what you did.

By using the theorem.

Theorem: Let R be a commutative unitary ring. R is a field if and only if its maximal ideal is \{ 0 \}.

 

[morphism]

There's a problem with your proof: we don't know that a_0 \neq 0. Linear dependence doesn't imply that all the a_i's are nonzero - it implies that they're not all zero, leaving the possibility of some being zero as long as we have at least one nonzero a_j.

 

[BSMSMSTMSPHD]

Ah yes, you are right. I shouldn't have written that all of the coefficients were not zero. I suspect that we can say a_0 \neq 0 based on the second line:

a_1(r) + a_2(r^2) + ... + a_n(r^n) + a_{n+1}(r^{n+1}) = -a_0

Help?

 

[morphism]

One way you can overcome this is by letting j be the first index for which a_j \neq 0. Your expression will become:

a_j r^j + a_{j+1} r^{j+1} + \cdots + a_{n+1} r^{n+1} = 0

Then you can pull out r^j:

r^j \left(a_j + a_{j+1} r + \cdots + a_{n+1} r^{n+1-j}\right) = 0

Now your previous argument, together with the fact that R is an integral domain, can be used to finish things off.

 

[mathwonk]

injective linear maps of finite dimensional spaces are surjective. done.