Fractional field

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

Answers and Replies

  • #2
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3,297
Uuh, wouldn't that just be [itex]\mathbb{Q}[/itex]?

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
Yes, that's right -- I guess I don't fully understand how all of these things relate (integral domains, fraction fields, etc)
 
  • #4
How did you get that so quickly?
 
  • #5
22,129
3,297
How did you get that so quickly?

Uuuh, intuition I guess? After a lot of practising, these things come fast...
 
  • #6
Uuh, wouldn't that just be [itex]\mathbb{Q}[/itex]?

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
3,297
It could be, yes...
 
  • #9
22,129
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Doesn't the same theorem apply here?
 
  • #11
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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Uuh, I don't know any books that contain the proof. But the proof is a very good exercise. Why not try it for yourself? What do you know about fraction fields? Do you have characterizations for them?
 
  • #13
Just know the definition :

Fraction field for integral domain = {a/b | a, b are elements of D, b not equal to zero}
 
  • #14
22,129
3,297
Can you prove that the fraction field Q of an integral domain A is the smallest field that contains A?

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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