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Math Help - Fermat's theorem on sums of two squares

  1. #1
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    Fermat's theorem on sums of two squares

    Hello,
    I have to write an essay about Fermat's theorem on sums of two squares. Now I found a proof in a book of the following theorem:
    Let p \neq 2 a prime number. Then the ideal p\mathbb Z[i] is a prime ideal in \mathbb Z[i] if and only if p \equiv 3\ (\text{mod}\ 4).
    Ok, now I have a question about the proof. The ideal p\mathbb Z[i] in \mathbb Z[i] is a prime ideal if and only if the quotient ring \mathbb Z[i]/p \mathbb Z[i] is a integral domain (this is clear).

    "Now there is a isomorphism \mathbb Z[i]/p \mathbb Z[i]\cong \mathbb F_p[X]/(X^2-1)\mathbb F_p[X] and p\mathbb Z[i] is a prime ideal in \mathbb Z[i] if and only if (X^2-1) is irreducible in \mathbb F_p, which is equivalent that -1 is no square mod p."

    Is there a typing error? Is (X^2-1) correct or should it be correctly (X^2+1)?

    And what about " -1 is no square mod p"? I don't unterstand it.

    Thanks in advance,

    Bye,
    Lisa
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  2. #2
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    It's a typo. Clearly x^2-1 is always reducible as (x+1)(x-1).

    The second part is talking about -1 being a quadratic residue modulo p, ie, does there exist an x with x^2 = -1 (mod p)? For example, -1 is a quadratic residue mod 5 and 13 (since 2^2 = -1 (mod 5) and 5^2 = -1 (mod 13)).

    In general, if p is an odd prime, then -1 is a quadratic residue mod p if and only if p = 1 (mod 4).
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  3. #3
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    Hello.

    Ok, then the correct equivalences are:
    (X^2+1) is irreducible in \mathbb F_p \Leftrightarrow -1 is no quadratic residue mod p

    This means, that there is no x \in \mathbb F_p with x^2\equiv -1\ (\mathrm{mod}\ p)

    But I don't understand the equivalence above. How to proof it?!

    Bye,
    Lisa
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  4. #4
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    Suppose -1 is a quadratic residue mod p. Then there exists some a \in \mathbb{F}_p with a^2 \equiv -1. We then have

    (x+a)(x-a) \equiv x^2 - a^2 \equiv x^2+1,

    and x^2+1 is reducible.

    Now the converse. If x^2+1 is reducible, then x^2+1 \equiv f(x)g(x) for some non-constant elements f and g of \mathbb{F}_p[x]. Since p is prime \mathbb{F}_p has no zero-divisors, hence f and g must be linear. Thus f must have a root; call it a. a is also a root of x^2+1, so a^2 = -1 (mod p) and -1 is a quadratic residue of p.
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