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Math Help - Prove the polynomial is irreducible..

  1. #1
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    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.
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  2. #2
    Senior Member jakncoke's Avatar
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    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  x^2 = \sqrt(2) which would mean i need the fourth root of 2 which you can easily see is not in  \mathbb{Q}[\sqrt{2}]

    Elements look like  a + b\sqrt(2) | a,b \in \mathbb{Q}
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  3. #3
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    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:

    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.

    *********

    even shorter proof of the irreducibility of x4 - 2: use eisenstein's criterion with p = 2.
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