I'll answer a), basically a quadratic polynomial is reducible over a field if and only if there are its roots in that field. so i need which would mean i need the fourth root of 2 which you can easily see is not in
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(a) Prove that the polynomial g(x) = x^{2} - sqrt (2) is irreducible over Q(sqrt(2)). It follows that g is the minimal polynomial for ^{4} (sqrt(2)) <--- (the fourth root of the square root of two)
Hint: What are the elements of Q(sqrt(2))?
(b) Use part (a) to show that the elements of Q(^{4} sqrt(2)) are of the form a + b ^{4} (sqrt(2)) + c sqrt(2) + d ^{4} (8) for a,b,c,d in Q, uniquely.
(c) Find the minimal polynomial for ^{4} (sqrt (2)) over Q, and show that it is minimal.
I'll answer a), basically a quadratic polynomial is reducible over a field if and only if there are its roots in that field. so i need which would mean i need the fourth root of 2 which you can easily see is not in
Elements look like
let's pretend we never heard of the fourth root of two, just the square root.
we know that Q(√2) is a vector space of dimension 2 over Q, since {1,√2} is a basis. this means the elements of Q(√2) are of the form a +b√2, for a,b in Q.
now, suppose there WAS some a+b√2 with:
(a+b√2)^{2} = √2 (that is: a+b√2 is a root of x^{2}-√2 in Q(√2)[x]).
this means that a^{2}+2b^{2} + (2ab)√2 = √2.
thus a^{2}+2b^{2} = 0 and 2ab = 1.
but if we know that Q(√2) is an ordered field (as a subfield of R), then the first equation implies a = b = 0, contradicting the second equation.
even if we do not know Q(√2) is an ordered field, we have that:
b = 1/(2a), so that a^{2} + 2/a^{2} = 0, so a^{4} = -1/4, which is impossible for a rational number a, since Q IS an ordered field:
(p/q < p'/q' iff pq' < p'q).
so there do not exists any rational a,b with a^{2}+2b^{2} = 0, 2ab = 0, so Q(√2) contains no roots of x^{2}-√2
which is thus irreducible over Q(√2).
now if we let u be a root of x^{2}-√2, we see that dim_{Q(√2)}(Q(√2)(u)) = 2, so dim_{Q}(Q(√2)(u)) = 4.
so we have the basis {1,√2,u√2,u} of Q(√2)(u) over Q. note that u^{2} = √2, so we can also write this basis as {1,u,u^{2},u^{3}}
and also since √2 is in Q(u), Q(√2)(u) = Q(u).
in normal notation:
.
since dim_{Q}(Q(u)) = 4, the minimal polynomial of u must have degree 4.
since x^{4} - 2 is a monic polynomial of degree 4 that u satisfies, it must be the minimal polynomial of u.
alternatively: suppose x^{4} - 2 has a root in Q. then since
x^{4} - 2 = (x^{2} - √2)(x^{2} + √2), one of these factors must have a root in Q.
but x^{2} + √2 > 0, and x^{2} - √2 has no roots in Q (as we saw above).
so if x^{4} - 2 factors over Q it must factor as:
x^{4} - 2 = (x^{2} + px + q)(x^{2} + rx + s), so:
rs = 2
qr + ps = 0
q + pr + s = 0
p + r = 0
so r = -p, when s = 2/r = -2/p, so
0 = qr + ps = -pq - 2 --> q = -2/p
0 = q - p^{2} - 2/p = p^{2} - 4/p --> p^{3} = 4
now if p is rational, p = m/n with gcd(m,n) = 1, whence m^{3} = 4n^{3} <--these are integers.
if m is odd, m^{3} is odd, but 4n^{3} is even, a contradiction.
if m is even, then 8 divides m^{3}, whence 8 divides 4n^{3}, thus 2 divides n^{3}, so 2 divides n, contradicting gcd(m,n) = 1.
so x^{4} - 2 has no rational linear or quadratic factors in Q[x], so is irreducible over Q.
*********
even shorter proof of the irreducibility of x^{4} - 2: use eisenstein's criterion with p = 2.