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Prep for Algebra Comprehensive Exam #2

  1. Jul 23, 2007 #1
    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.
     
    Last edited: Jul 24, 2007
  2. jcsd
  3. Jul 23, 2007 #2
    Another approach is to show the maximal ideal of [tex]R[/tex] is [tex]\{ 0\}[/tex]. But this is not shorter than what you did.

    By using the theorem.

    Theorem: Let [tex]R[/tex] be a commutative unitary ring. [tex]R[/tex] is a field if and only if its maximal ideal is [tex]\{ 0 \}[/tex].
     
  4. Jul 23, 2007 #3

    morphism

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    There's a problem with your proof: we don't know that a_0 [itex]\neq[/itex] 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.
     
  5. Jul 24, 2007 #4
    Ah yes, you are right. I shouldn't have written that all of the coefficients were not zero. I suspect that we can say [tex]a_0 \neq 0[/tex] based on the second line:

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

    Help?
     
  6. Jul 24, 2007 #5

    morphism

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

    [tex]a_j r^j + a_{j+1} r^{j+1} + \cdots + a_{n+1} r^{n+1} = 0[/tex]

    Then you can pull out r^j:

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

    Now your previous argument, together with the fact that R is an integral domain, can be used to finish things off.
     
  7. Jul 30, 2007 #6

    mathwonk

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