The field of quotients of any fieldDis isomorphic toD.Why?
Solution
Let K be the field of quotients of D.
Define the following function f: D --> K such that f(x)=x
First this is clearly a ring homomorphism:
f(x+y)=x+y=f(x)+f(y)
f(xy)=xy=f(x)f(y)
f(1)=1
It is also easily seen to be one to one:
if f(x)=f(y) then x=y
So far all this is saying is that D is contained in its field of quotients. But this is true for any
integral domain. Next we show f is onto which will say that F and K are actually the same field.
In other words the field of quotients of a field is just the field itself.
Let x be in K. Now by definition x=a/b for some elements a and b in D such that b is nonzero.
But if b is nonzero then since D is a field, b has a multiplicative inverse 1/b in D. Then
a*(1/b)=a/b is in D and f(a/b)=a/b=x. So f is onto.
In short, the field of quotients of an integral domain is the smallest field in which the integral
domain can be embedded. So if D is a field then it\'s field of quotients must be at least as big as
D (since D has to be embedded into it). But D is a field so it is thus the smallest field in which D
can be embedded. In other words, the field of quotients of a field F is just F.
Now of course you could say that they are not exactly the same field since the field of quotients
is actually comprised of equivalence classes of elements a/b where b is nonzero. But if D is a
field and a,b are elements of F with b nonzero then what is the equivalence class of a/b? Well,
suppose c/d is in the same equivalence class, then ad=bc. But b is nonzero so since we are in a
field we can multiply by its inverse to obtain.
(ad)/b = c
But d is also nonzero so multiplying by its inverse gives
a/b=c/d (now we are talking about elements of D and not equivalence classes)
So the equivalence class of a/b is really just a/b.
So D and its field of quotients really are the same field..
The field of quotients of any fieldDis isomorphic toD.WhySoluti.pdf
1. The field of quotients of any fieldDis isomorphic toD.Why?
Solution
Let K be the field of quotients of D.
Define the following function f: D --> K such that f(x)=x
First this is clearly a ring homomorphism:
f(x+y)=x+y=f(x)+f(y)
f(xy)=xy=f(x)f(y)
f(1)=1
It is also easily seen to be one to one:
if f(x)=f(y) then x=y
So far all this is saying is that D is contained in its field of quotients. But this is true for any
integral domain. Next we show f is onto which will say that F and K are actually the same field.
In other words the field of quotients of a field is just the field itself.
Let x be in K. Now by definition x=a/b for some elements a and b in D such that b is nonzero.
But if b is nonzero then since D is a field, b has a multiplicative inverse 1/b in D. Then
a*(1/b)=a/b is in D and f(a/b)=a/b=x. So f is onto.
In short, the field of quotients of an integral domain is the smallest field in which the integral
domain can be embedded. So if D is a field then it's field of quotients must be at least as big as
D (since D has to be embedded into it). But D is a field so it is thus the smallest field in which D
can be embedded. In other words, the field of quotients of a field F is just F.
Now of course you could say that they are not exactly the same field since the field of quotients
is actually comprised of equivalence classes of elements a/b where b is nonzero. But if D is a
field and a,b are elements of F with b nonzero then what is the equivalence class of a/b? Well,
suppose c/d is in the same equivalence class, then ad=bc. But b is nonzero so since we are in a
field we can multiply by its inverse to obtain.
2. (ad)/b = c
But d is also nonzero so multiplying by its inverse gives
a/b=c/d (now we are talking about elements of D and not equivalence classes)
So the equivalence class of a/b is really just a/b.
So D and its field of quotients really are the same field.
