# Thread: Prove the polynomial is irreducible..

1. ## Prove the polynomial is irreducible..

(a) Prove that the polynomial g(x) = x2 - 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.

2. ## 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 $\displaystyle a + b\sqrt(2) | a,b \in \mathbb{Q}$

3. ## 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 x2-√2 in Q(√2)[x]).

this means that a2+2b2 + (2ab)√2 = √2.

thus a2+2b2 = 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 a2 + 2/a2 = 0, so a4 = -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 a2+2b2 = 0, 2ab = 0, so Q(√2) contains no roots of x2-√2

which is thus irreducible over Q(√2).

now if we let u be a root of x2-√2, we see that dimQ(√2)(Q(√2)(u)) = 2, so dimQ(Q(√2)(u)) = 4.

so we have the basis {1,√2,u√2,u} of Q(√2)(u) over Q. note that u2 = √2, so we can also write this basis as {1,u,u2,u3}

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 dimQ(Q(u)) = 4, the minimal polynomial of u must have degree 4.

since x4 - 2 is a monic polynomial of degree 4 that u satisfies, it must be the minimal polynomial of u.

alternatively: suppose x4 - 2 has a root in Q. then since

x4 - 2 = (x2 - √2)(x2 + √2), one of these factors must have a root in Q.

but x2 + √2 > 0, and x2 - √2 has no roots in Q (as we saw above).

so if x4 - 2 factors over Q it must factor as:

x4 - 2 = (x2 + px + q)(x2 + 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 - p2 - 2/p = p2 - 4/p --> p3 = 4

now if p is rational, p = m/n with gcd(m,n) = 1, whence m3 = 4n3 <--these are integers.

if m is odd, m3 is odd, but 4n3 is even, a contradiction.
if m is even, then 8 divides m3, whence 8 divides 4n3, thus 2 divides n3, so 2 divides n, contradicting gcd(m,n) = 1.

so x4 - 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 x4 - 2: use eisenstein's criterion with p = 2.

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### x^4-2 is irreducible polynomial

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