Thread: 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.

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 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 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 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:

.

since dim_Q(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.