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Math Help - prime field

  1. #1
    ux0
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    prime field

    I need to show that the prime field \mathbb{R} and the prime field \mathbb{C} are in \mathbb{Q}

    could i get help showing one of these (the harder one) then ill try doing the other one... ??
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  2. #2
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    I assume you mean the prime subfield of \mathbb{R}, notice that by definition this is the intersection of all subfields of \mathbb{R} ie. P(\mathbb{R})= \bigcap_{B\subset \mathbb{R} : \ B \ field } B. Since 1 \in P(\mathbb{R}) and it is a field, it contains a copy of \mathbb{Z}. Now \mathbb{Q} is the field of fractions of \mathbb{Z} so \mathbb{Q} is the smallest field containing \mathbb{Z}, so \mathbb{Q} \subset P(\mathbb{R} ) and since P(\mathbb{R}) \cap \mathbb{Q} = P(\mathbb{R})...
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  3. #3
    ux0
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    wow i was completely off on the question.. sorry.

    i) Show that every subfield of \mathbb{C} contains \mathbb{Q}

    I thought \mathbb{Z}[i] was a subfield of \mathbb{C} but there is no \mathbb{Q} subfield of \mathbb{Z}[i]... but wouldn't that disprove this statement... (\mathbb{Z}[i] units are 1,-1,i,-i )

    ii) Show that the prime field of \mathbb{R} is \mathbb{Q}

    iii) Show that the prime field of \mathbb{C} is \mathbb{Q}
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  4. #4
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    Read carefully my first post and you'll notice that it answers all three questions.

    As for \mathbb{Z} [i] this is, by definition, the smallest ring that contains both \mathbb{Z} and i, and \mathbb{Z} (i) is the aboves fraction field, and since it contains \mathbb{Z} it must contain \mathbb{Q}
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  5. #5
    Senior Member Shanks's Avatar
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    Yeah, Jose27 is quite right.
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  6. #6
    ux0
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    Quote Originally Posted by Jose27 View Post
    I assume you mean the prime subfield of \mathbb{R}, notice that by definition this is the intersection of all subfields of \mathbb{R} ie. P(\mathbb{R})= \bigcap_{B\subset \mathbb{R} : \ B \ field } B. Since 1 \in P(\mathbb{R}) and it is a field, it contains a copy of \mathbb{Z}. Now \mathbb{Q} is the field of fractions of \mathbb{Z} so \mathbb{Q} is the smallest field containing \mathbb{Z}, so \mathbb{Q} \subset P(\mathbb{R} ) and since P(\mathbb{R}) \cap \mathbb{Q} = P(\mathbb{R})...

    Ohhhh so what your saying in words is that


    Because the intersection of the prime subfiled of \mathbb{R} and \mathbb{Q} is just the prime subfield of \mathbb{R} , we can conclude that \mathbb{Q}   = P(R) ,

    and the proof is the same thing for \mathbb{C} to show the prime subfiled of \mathbb{C} is \mathbb{Q} ??? (basically just change the R's to C's)
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