Q32. Let R denote the field of real numbers, Q the field of rational numbers, and Z the ring of integers. Which of the following subsets F_i of R, 1<=i<=4, are subfields of R?

$$F_1=\{a/b: a,b \in Z \text{and b is odd}\}$$

$$F_2=\{a+b\sqrt{2}: a,b \in Z\}$$

$$F_3=\{a+b\sqrt{2}:a,b\in Q\}$$

$$F_4=\{a+b\sqrt[4]{2}:a,b\in Q\}$$

Answer is (B), F_3 only.

Can anyone explain to me briefly why?

(I have only studied groups before.)

Thanks a lot.

## 9367 Q32

### Re: 9367 Q32

A field consists of a set F and two operations, addition and multiplication with three properties:

1) F together with addition is a commutative group

2) Nonzero elements of F together with multiplication is a commutative group

3) Multiplication is distributive over addition

There are other definitions but this one uses only groups.

Example to keep in mind: the rationals with addition and multiplication are a field; the integers are not since 2) is not satisfied (try seeing exactly why).

Now a subfield of F is nothing more than a field that is contained in the larger field F. So you need to test 1) and 2) on your subset. If they are satisfied you have a subfield. Usually 1) works for most contenders but it is at 2) that they fail since most of these are not closed under multiplication or inverses. For example try multiplying two nonzero elements of F_4. Is the result still in F_4?

1) F together with addition is a commutative group

2) Nonzero elements of F together with multiplication is a commutative group

3) Multiplication is distributive over addition

There are other definitions but this one uses only groups.

Example to keep in mind: the rationals with addition and multiplication are a field; the integers are not since 2) is not satisfied (try seeing exactly why).

Now a subfield of F is nothing more than a field that is contained in the larger field F. So you need to test 1) and 2) on your subset. If they are satisfied you have a subfield. Usually 1) works for most contenders but it is at 2) that they fail since most of these are not closed under multiplication or inverses. For example try multiplying two nonzero elements of F_4. Is the result still in F_4?

Last edited by apap on Sun Jan 01, 2012 6:01 pm, edited 1 time in total.

### Re: 9367 Q32

$$(a+b\sqrt[4]{2})(c+d\sqrt[4]{2})=ac+ad\sqrt[4]{2}+bc\sqrt[4]{2}+bd\sqrt[4]{4}$$

Aha, so F_4 is not closed.

Does this count as a counter-example for F_1?

$$(1/3)+(-1/3)=(0/2)$$

$$(a+b\sqrt{2})(c+d\sqrt{2})=ac+ad\sqrt{2}+bc\sqrt{2}+2bd=(ac+2bd)+(ad+bc)\sqrt{2}$$

How do I construct a counter-example for F_2? Is division involved?

Thanks a lot.

Aha, so F_4 is not closed.

Does this count as a counter-example for F_1?

$$(1/3)+(-1/3)=(0/2)$$

$$(a+b\sqrt{2})(c+d\sqrt{2})=ac+ad\sqrt{2}+bc\sqrt{2}+2bd=(ac+2bd)+(ad+bc)\sqrt{2}$$

How do I construct a counter-example for F_2? Is division involved?

Thanks a lot.

### Re: 9367 Q32

Yes. The ordered pair would be (0,2), but 2 isn't odd.yoyobarn wrote:$$(a+b\sqrt[4]{2})(c+d\sqrt[4]{2})=ac+ad\sqrt[4]{2}+bc\sqrt[4]{2}+bd\sqrt[4]{4}$$

Aha, so F_4 is not closed.

Does this count as a counter-example for F_1?

$$(1/3)+(-1/3)=(0/2)$$

[/quote][/quote]$$(a+b\sqrt{2})(c+d\sqrt{2})=ac+ad\sqrt{2}+bc\sqrt{2}+2bd=(ac+2bd)+(ad+bc)\sqrt{2}$$

How do I construct a counter-example for F_2? Is division involved?

Sort of. You would just need to recognize that the integers don't have inverses, so this isn't going to work out. Of, you could take the long and tedious route (which obviously is not suggested) and suppose a+b(2) had an inverse, and then come to a contradiction. Obviously knowing that the ring your pulling from isn't a field helps in this situation.

### Re: 9367 Q32

Yoyobarn, your counterexample for F_1 does not work since 0/2=0/1=0 so it is contained in F_1. You need to look at what the multiplicative inverse looks like in this case. Is it of the same form? You cannot think of these fractions as ordered pairs since some of them are equivalent.

To figure out why F_2 doesn't have a multiplicative inverse and why F_4 has a multiplicative inverse just continue your calculation: $$(a+b\sqrt{2})(c+d\sqrt{2})=ac+ad\sqrt{2}+bc\sqrt{2}+2bd=(ac+2bd)+(ad+bc)\sqrt{2}$$ What if this is equal to 1? Express c and d in terms of a and b, this way you get a general formula for the inverse. Obviously you wouldn't have time to do this on the exam but it's worth doing since it gives more insight than just picking counterexamples.

Most of the times expressions involving integers do not have integer inverses. But there are exceptions, for example the group of integer matrices with determinant 1.

To figure out why F_2 doesn't have a multiplicative inverse and why F_4 has a multiplicative inverse just continue your calculation: $$(a+b\sqrt{2})(c+d\sqrt{2})=ac+ad\sqrt{2}+bc\sqrt{2}+2bd=(ac+2bd)+(ad+bc)\sqrt{2}$$ What if this is equal to 1? Express c and d in terms of a and b, this way you get a general formula for the inverse. Obviously you wouldn't have time to do this on the exam but it's worth doing since it gives more insight than just picking counterexamples.

Most of the times expressions involving integers do not have integer inverses. But there are exceptions, for example the group of integer matrices with determinant 1.

### Re: 9367 Q32

The inverse of a/b is b/a, and a might be even, so F_1 is not closed.

ac+2bd=1

ad+bc=0

c=-ad/b

d=(1-ac)/2b

c=-a(1-ac)/2b^2

2b^2(c)=-a+a^2(c)

c[2b^2-a^2]=-a

c=-a/[2b^2-a^2], might not be an integer

Ok thanks, I think I got it.

ac+2bd=1

ad+bc=0

c=-ad/b

d=(1-ac)/2b

c=-a(1-ac)/2b^2

2b^2(c)=-a+a^2(c)

c[2b^2-a^2]=-a

c=-a/[2b^2-a^2], might not be an integer

Ok thanks, I think I got it.

### Re: 9367 Q32

c=-a/[2b^2-a^2] always exists when a and b are in Q. This is why. F_3 is in fact a subdield. I had this very problem in my Abstract Algebra exam