Prove the polynomial is irreducible..

(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.

Re: Prove the polynomial is irreducible..

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 $\displaystyle x^2 = \sqrt(2) $ which would mean i need the fourth root of 2 which you can easily see is not in $\displaystyle \mathbb{Q}[\sqrt{2}] $

Elements look like $\displaystyle a + b\sqrt(2) | a,b \in \mathbb{Q} $

Re: Prove the polynomial is irreducible..

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:

$\displaystyle u = \sqrt[4]{2}, u^2 = \sqrt{2}, u^3 = \sqrt[4]{8}$.

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.

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even shorter proof of the irreducibility of x^{4} - 2: use eisenstein's criterion with p = 2.