Field extensions and radical ideals (2 problems)

[Seda]

PROBLEM 1:

SOLVED THANKS TO POSTER DICK

How can I prove these two field extensions are equal?

Q(√3, -√3, i, -i) = Q(√3+i) where Q is the field of the rational numbers.


I got Q(√3+i) ⊆ Q(√3, -√3, i, -i), that direction is easy:

Let m ∈ Q(√3+i). Therefore, m = q + p(√3+i) where q,p ∈ Q.

Let p = a-b, with a,b ∈ Q.

So m = q + a(√3) + a(i) + b(-√3) + b(-i)

Therefore m ∈ Q(√3, -√3, i, -i)



But how can I prove Q(√3, -√3, i, -i) ⊆ Q(√3+i)?

Let m ∈ Q(√3, -√3, i, -i)

Therefore m = q + a√3 + bi + c(-√3) + d(-i) where q,a,b,c,d ∈ Q

Let p = a-c and r = b-d. Therefore, p,r ∈ Q

Therefor m = q + p√3 + ri

But I have to prove m = q + p√3 + ri = q + (SOMETHING IN Q)* (√3+i)

Any hints?

PROBLEM 2:

Prove that if I is an Ideal for communitive ring R, prove Rad(I) is and Ideal of R and that I is an ideal of Rad(I)

For the first, I looked at for r ∈ R, a ∈ Rad(I). Therefore aⁿ∈ I for some positive integer n.

(ra)ⁿ = rⁿaⁿ ∈ I since rⁿ ∈ R and since I is an Ideal.

Therefore since (ra)ⁿ ∈ I, ra ∈ Rad(I). So Rad(I) is an Ideal of R.

But for the second, I let a ∈ I, b ∈ Rad(I), therefore bⁿ ∈ I, but I don't know how to prove ba ∈ I.

Thanks in advance!

 

[Dick]

For the first one you just have to prove sqrt(3) and i are in the extension field Q[sqrt(3)+i]. Try computing (sqrt(3)+i)^2 to start with. What does that imply? Then compute some more products.

 

[Seda]

For the first one you just have to prove sqrt(3) and i are in the extension field Q[sqrt(3)+i]. Try computing (sqrt(3)+i)^2 to start with. What does that imply? Then compute some more products.

2+2i√3 ...? It shows that the degree is the same for both?

 

[Dick]

2-2i√3 ...? It shows that the degree is the same for both?

I get 2+2i*sqrt(3). But it shows i*sqrt(3) is in Q[sqrt(3)+i]. That's a start. Now keep going. Try (sqrt(3)+i)^3.

 

[Seda]

I get 2+2i*sqrt(3). But it shows i*sqrt(3) is in Q[sqrt(3)+i]. That's a start. Now keep going. Try (sqrt(3)+i)^3.

Aw cool it's 8i! I'll try to figure that in...

Yeah sorry, I caught the sign error. Is the way i figured Q(√3+i) ⊆ Q(√3, -√3, i, -i) ok?

 

[Dick]

Yeah sorry, I caught the sign error. Is the way i figured Q(√3+i) ⊆ Q(√3, -√3, i, -i) ok?

Sure it's ok. Since sqrt(3) and i are in Q(√3, -√3, i, -i) then so is sqrt(3)+i. That's the easy direction as you said. Well, it was so easy I didn't really read the proof. All you need to say is that q + p(√3+i)=q+p*sqrt(3)+p*i. That's clearly in Q(√3, -√3, i, -i), isn't it?

 

[Seda]

Sure it's ok. Since sqrt(3) and i are in Q(√3, -√3, i, -i) then so is sqrt(3)+i. That's the easy direction as you said. Well, it was so easy I didn't really read the proof. All you need to say is that q + p(√3+i)=q+p*sqrt(3)+p*i. That's clearly in Q(√3, -√3, i, -i), isn't it?

yeah, I just wrote it in a way to overly-clearly include the -√3 and -i also, even though it could be considered implicit.

As for the "hard direction" am I wording this right?:

Obviously (√3 + i) ∈ Q(√3+ i)
Therefore (√3 + i)^3 = 8i ∈ Q(√3+ i). Similarily, (-8)(-i) ∈ Q(√3+ i). So i, -i ∈ Q(√3+ i).

How specifically do I get √3, -√3? Does that fall directly from knowing (√3+ i)^2 = 2+2√3i ∈ Q(√3+ i) therefore √3i ∈ Q(√3+ i) and knowing i ∈ Q(√3+ i), therefore √3 ∈ Q(√3+ i)?

 

[Dick]

yeah, I just wrote it in a way to overly-clearly include the -√3 and -i also, even though it could be considered implicit.

As for the "hard direction" am I wording this right?:

Obviously (√3 + i) ∈ Q(√3+ i)
Therefore (√3 + i)^3 = 8i ∈ Q(√3+ i). Similarily, (-8)(-i) ∈ Q(√3+ i). So i, -i ∈ Q(√3+ i).

How specifically do I get √3, -√3? Does that fall directly from knowing (√3+ i)^2 = 2+2√3i ∈ Q(√3+ i) therefore √3i ∈ Q(√3+ i) and knowing i ∈ Q(√3+ i), therefore √3 ∈ Q(√3+ i)?

You might be thinking too hard. If i is in Q[sqrt(3)+i] and sqrt(3)+i is in Q[sqrt(3)+i] then so is (sqrt(3)+i)-i. Or since you know i*sqrt(3) is in Q[sqrt(3)+i] as is i, then (i*sqrt(3))/i is also in Q[sqrt(3)+i]. There's more than one way to skin a cat.

 

[Seda]

You might be thinking too hard. If i is in Q[sqrt(3)+i] and sqrt(3)+i is in Q[sqrt(3)+i] then so is (sqrt(3)+i)-i. Or since you know i*sqrt(3) is in Q[sqrt(3)+i] as is i, then (i*sqrt(3))/i is also in Q[sqrt(3)+i]. There's more than one way to skin a cat.

Doh...well I'd rather be thinking too hard than thinking too little I guess. Thanks for your help! I'd hate to ask for more of your time, but do you have a hint for the second problem?

 

[Dick]

Isn't that the easy direction? If Rad(I) is contained in R, and I is an ideal of R, isn't I automatically an ideal of Rad(I)? a in I times any element of R is automatically in I, isn't it?

 

[Seda]

Isn't that the easy direction? If Rad(I) is contained in R, and I is an ideal of R, isn't I automatically an ideal of Rad(I)? a in I times any element of R is automatically in I, isn't it?

Um...yeah. I wasn't thinking that way. But that's true. Thanks