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A question about fields and rational functions... Watch

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    If k is a field, prove that \sqrt{1-x^2} \not\in k(x), where k(x) is the field of rational functions.

    Hint: Mimic a proof that \sqrt{2} is irrational.

    Answer from answer sheet:

    The problem, in other words, so to show that there does not exist an element f \in k(x) such that f^2 = 1 - x^2 . This is certainly false if 2=0 in the field k, for then 1-x^2 = 1+x^2 = (1-x)^2 . So let's suppose 2 \not= 0 in k; this implies that 1-x \not= 1+x.

    I don't understand why it's "certainly false if 2=0"

    Suppose such an element f exists and write f=p/q with p,q in k[x] relatively prime. Then p^2 = (1-x^2)q^2 in k[x]. By unique factroization in k[x] we see that p^2 is divisible by (1-x)^2 and (1+x)^2, so p is divisible by 1-x and 1+x.

    I don't understand why p^2 = (1-x^2)q^2 in k[x] implies that p^2 is divisible by (1-x)^2 and (1+x)^2. Didn't the solution already assume taht 1-x is not equal to 1+x. Then how is p^2 divisible by both of them? Or are they just saying that p_1 = q(1-x)^2 ?


    Since 2 \not= 0 these irreducibles are distinct, and we write p=(1-x^2)p_1. Then p_1^2(1-x^2) = q^2, so q is also divisble by 1-x^2. This contradicts our assumption that p,q were relatively prime.

    Thanks in advance
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    (Original post by Artus)
    I don't understand why it's "certainly false if 2=0"
    Well, it tells you. Suppose 2 = 0 (think of \mathbb{Z}/2\mathbb{Z}), so that

    (1 - x)^2 = 1 - 2x + x^2 = 1 + x^2 = 1 - x^2

    because -1 = 1.

    (Original post by Artus)
    I don't understand why p^2 = (1-x^2)q^2 in k[x] implies that p^2 is divisible by (1-x)^2 and (1+x)^2. Didn't the solution already assume taht 1-x is not equal to 1+x. Then how is p^2 divisible by both of them? Or are they just saying that p_1 = q(1-x)^2 ?
    This should read: .. p^2 is divisible by 1 + x and 1 - x ..

    Now, 1 - x^2 = (1 - x)(1 + x) where 1 - x and 1 + x are clearly irreducible.

    Notice that k[x] is a PID, so these are in fact prime. Hence,

    1 \pm x \mid p^2\ \Rightarrow\ 1 \pm x \mid p

    by primality.
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    (Original post by jack.hadamard)
    Well, it tells you. Suppose 2 = 0 (think of \mathbb{Z}/2\mathbb{Z}), so that

    (1 - x)^2 = 1 - 2x + x^2 = 1 + x^2 = 1 - x^2

    because -1 = 1.
    Thanks, but what's wrong when (1-x)^2 = 1-x^2?
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    (Original post by Artus)
    Thanks, but what's wrong when (1-x)^2 = 1-x^2?
    You would like to show that there is no element r \in k(x) such that r^2 = 1 - x^2.
    The fact that (1 - x)^2 = 1 - x^2 shows that 1 - x = \sqrt{1 - x^2} \in k(x).

    In other words, if the field is has characteristic two, then 1 - x^2 has a square root.
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    (Original post by jack.hadamard)
    You would like to show that there is no element r \in k(x) such that r^2 = 1 - x^2.
    The fact that (1 - x)^2 = 1 - x^2 shows that 1 - x = \sqrt{1 - x^2} \in k(x).

    In other words, if the field is has characteristic two, then 1 - x^2 has a square root.
    Oh, so the answer is telling us that the question is wrong when 2=0, so we just assume that 2 is not equal to 0?
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    (Original post by Artus)
    Oh, so the answer is telling us that the question is wrong when 2=0, so we just assume that 2 is not equal to 0?
    Yes. The statement is false for fields of characteristic two.
 
 
 
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