- #1
Seda
- 71
- 0
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!
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!
Last edited: