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I Field of zero characteristics

  1. Jun 14, 2018 #1
    I am interested in the following theorem:
    Every field of zero characteristics has a prime subfield isomorphic to ℚ.
    I am following the usual proof, where we identify every p∈ℚ as a/b , a∈ℤ,beℕ, and define h:ℚ→P as h(a/b)=(a*1)(b*1)-1 (where a*1=1+1+1... a times) I have worked out the multiplicative rule for this homomorphism, but I am not sure how to prove the additive. I am also interested in what does the notation
    (a*1)(b*1)-1 exactly mean. Thanks.
     
  2. jcsd
  3. Jun 14, 2018 #2

    fresh_42

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    All we know about the prime field is its characteristic and ##0,1 \in \mathbb{P}##. ##a\in \mathbb{P}## is not sure. However ##\underbrace{1+\ldots +1}_{a \text{ times}}=a \,\cdot \, 1## for ##a \in \mathbb{N}## is, and likewise for negative ##a##. That's why ##a## times ##1## is used instead of ##a##. We only have isomorphic images of ##a## in ##\mathbb{P}##.

    For the homomorphism, I guess ##\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+bc}{bd}## will be needed.
     
  4. Jun 14, 2018 #3

    mathwonk

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    this probably won't help much until you have some commutative algebra but it gives the outline. First of all since the field F is an abelian group, there is exactly one additive homomorphism from the integers Z into F for each choice of the image of 1 from Z, so choose that image to be the unit element 1 in P. Now you have the unique additive homomorphism from Z to P that fresh described sending each positive integer n to 1+...+1 (n times). By definition of the multiplication in Z this is also a multiplicative map hence a ring map. Then since each non zero integer goes to an invertible element of F, (because F has characteristic zero), there is a unique extension of the previously defined ring homomorphism Z-->F to a ring homomorphism Q --> F. Then since Q is a field, this ring map is necessarily injective, hence defines an isomorphism onto the smallest subfield of F, i.e. the prime field.
     
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