# Prep for Algebra Comprehensive Exam #2

1. Jul 23, 2007

### 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 $$rs = sr = 1$$.

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:

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

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

Call 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.

Last edited: Jul 24, 2007
2. Jul 23, 2007

### 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 $$\{ 0 \}$$.

3. Jul 23, 2007

### 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.

4. Jul 24, 2007

### 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?

5. Jul 24, 2007

### 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.

6. Jul 30, 2007

### mathwonk

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