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Thread: Determine if this Ring is a field

  1. #1
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    Determine if this Ring is a field

    Can you help me prove that this ring is a field?

    R = Q[$\displaystyle \sqrt(d)$] = { a+b$\displaystyle \sqrt(d)$ ; a,b are rational}

    So I tried:

    We must show that every element of R is a unit. So for some element (a+b*$\displaystyle \sqrt(d)$) there must be some (e+c*$\displaystyle \sqrt(d)$) such that

    (a+b*$\displaystyle \sqrt(d)$)*(e+c*$\displaystyle \sqrt(d)$) = 1

    so (ae+bed) + (be + ac)*$\displaystyle \sqrt(d)$ = 1

    so ae-bed = 1 and be+ac =0

    but then I get stuck after that
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  2. #2
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    I believe that the first equation should be ae+bcd=1. This is just a system of equations in the unknowns e and c which you can solve. Just be careful of the special cases a=0 and possibly b=0.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    If $\displaystyle d\geq 0$ is a perfect square then $\displaystyle \mathbb{Q}[\sqrt{d}]=\mathbb{Q}$ and so $\displaystyle \mathbb{Q}[\sqrt{d}]$ is a field.

    If $\displaystyle d\geq 0$ is not a perfect square, then, $\displaystyle \sqrt{d}$ is irrational.

    In this case choose $\displaystyle a+b\sqrt{d}\neq 0$. Then

    $\displaystyle (a+b\sqrt{d})(x+y\sqrt{d})=1\Leftrightarrow \begin{Bmatrix} ax+bdy=1\\bx+ay=0\end{matrix}$

    Now, using that $\displaystyle \sqrt{d}$ is irrational, prove that

    $\displaystyle \det \begin{bmatrix}{a}&{bd}\\{b}&{a}\end{bmatrix}=a^2-b^2d\neq 0$

    which implies that the system has a solution $\displaystyle (x,y)=\ldots$ ( besides, unique ).


    Fernando Revilla
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  4. #4
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    Thanks guys I have it figured out
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