GRE9367 Problem 32

Forum for the GRE subject test in mathematics.
EmLasker
Posts: 24
Joined: Sun Oct 05, 2008 1:41 am

GRE9367 Problem 32

Postby EmLasker » Tue Oct 14, 2008 1:25 am

I'm a bit stuck on this one...

Which of the following are subfields of the reals:

1. {a/b : a and be are in Z and b is odd}
2. {a + b(2)^(1/2) : a and b are in Z}
3. {a + b(2)^(1/2) : a and b are in Q}
4. {a + b(2)^(1/4) : a and b are in Q}

(the answer is only 3)

I know that 1 isn't a subfield as it doesn't contain the multiplicative inverses of elements where a is even... I know from experience that 3 is a field but I can't remember what the multiplicative inverses are. I think if I knew what they were, it would be obvious that 2 wasn't a field, but I'm not sure why 4 isn't a field.

Kastro
Posts: 10
Joined: Mon Oct 13, 2008 10:36 pm

Postby Kastro » Tue Oct 14, 2008 2:07 am

Just multiply top and bottom by the conjugate to get the inverse in workable form.

EmLasker
Posts: 24
Joined: Sun Oct 05, 2008 1:41 am

Postby EmLasker » Tue Oct 14, 2008 2:57 am

Thank you for the help - I see it now, even though I feel kind of silly for not seeing it before... :oops:

EmLasker
Posts: 24
Joined: Sun Oct 05, 2008 1:41 am

Postby EmLasker » Tue Oct 14, 2008 3:02 am

4 is not a field because it isn't closed...

[a + b(2^(.25))] * [a - b(2^(.25))] = a^2 - (b^2)(2^(.5))

blp
Posts: 17
Joined: Sun Oct 19, 2008 5:24 am

Postby blp » Sun Oct 19, 2008 6:55 pm

I was going through some posts and noticed that many of you solve these problems in a much too complicated way. There's a very trivial theorem for verifying if something which is a subring of the complex numbers is a field. The general theorem is as follows.

Theorem: Let R be an integral domain that contains a field F. If R is finite dimensional as an F-vector space, then R is a field.

Proof: Choose an element r in R which is non-zero. Then the map R->R defined by x->rx is an F-linear map and because R is finite dimensional, it's a bijection. Hence there's an x in R that maps to 1, so we have an inverse for r. QED

The way that this can be used for anything that's contained under the complex numbers (which is the only thing you'll pretty much find on the test) is that a ring that is a subset of C is necessarily an integral domain (a subring of an integral domain is of course an integral domain). Hence something like

a+b*sqrt(3), a,b in Q

has to be an field as it has the basis (1,sqrt(3)) as a Q-vector space. Similarly something horrible like

a+b*3^(1/3)+c*3^(2/3)+d*sqrt(2)+e*sqrt(7)

is immediately seen to be a field (basis (1, 3^(1/3), 3^(2/3), sqrt(2), sqrt(7))), though it would be a horrible task to give a formula for the inverse for a general element of the ring.

EDIT: This theorem can immediately be used to confirm the affirmative of a claim. You still need to find counterexamples for the others. Usually it's just sufficient to realize that for something that's spanned only as a lattice (i.e. linear combinations of Z), you're automatically having Z itself as a subring and these elements can usually be immediately seen to have no inverse.

amateur
Posts: 42
Joined: Wed Sep 10, 2008 9:41 am

Postby amateur » Sun Oct 19, 2008 11:13 pm

Dear blp,

With due respect... I am of the opinion... that all your posts are simply unrealistic... That's what I meant when I said that you sound out of this planet.

Regarding your assertion that

a+b*3^(1/3)+c*3^(2/3)+d*sqrt(2)+e*sqrt(7)

is a field... it is simply not a field as it is not closed wrt multiplication. (Take a = 0, b = 0, c = 0 and see why if you know enough math).

blp
Posts: 17
Joined: Sun Oct 19, 2008 5:24 am

Postby blp » Mon Oct 20, 2008 5:21 am

Sorry, I was writing that in a hurry and should have added all the multiples too. ;) So

a+b*3^(1/3)+c*3^(2/3)+d*sqrt(2)+e*sqrt(7)+f*sqrt(14)+g*3^(1/3)*sqrt(2)+h*3^(2/3)*sqrt(2)+i*3^(1/3)*sqrt(7)+j*3^(2/3)*sqrt(7)+k*3^(1/3)*sqrt(14)+l*3^(2/3)*sqrt(14)

with all coefficient belonging to Q. That's a field by the theorem, though proving it directly would be a nightmare.




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