- #1

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How would I go about finding the fraction field of Z[1/2]?

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- Thread starter Metric_Space
- Start date

- #1

- 98

- 0

How would I go about finding the fraction field of Z[1/2]?

- #2

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In general, if R is an integral domain, and if Q is it's fraction field, then, if

[tex]R\subseteq S\subseteq Q[/tex]

then the fraction field of S is Q.

- #3

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- #4

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How did you get that so quickly?

- #5

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How did you get that so quickly?

Uuuh, intuition I guess? After a lot of practising, these things come fast...

- #6

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In general, if R is an integral domain, and if Q is it's fraction field, then, if

[tex]R\subseteq S\subseteq Q[/tex]

then the fraction field of S is Q.

Is this a theorem then?

- #7

- 22,129

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It could be, yes...

- #8

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What about Z[1/3]?

- #9

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Doesn't the same theorem apply here?

- #10

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yes, you're right

- #11

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It could be, yes...

do you know the name of this theorem so I could look it up and see the proof?

- #12

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- #13

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Fraction field for integral domain = {a/b | a, b are elements of D, b not equal to zero}

- #14

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I.e. Assume that K is a field such that

[tex]A\subseteq K[/tex]

then

[tex]Q\subseteq K[/tex]

Start by showing this...

Edit: I might have take [itex]\subseteq[/itex] a bit too liberal in the last equation. Formally, there only exists an injective ring morfism [itex]Q\rightarrow K[/itex]. But I see that as the same thing as a subset...

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