Determine if this Ring is a field

**mulaosmanovicben** · January 19th 2011, 09:12 AM

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

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

So I tried:

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

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

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

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

but then I get stuck after that

**DrSteve** · January 19th 2011, 09:54 AM

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.

**FernandoRevilla** · January 19th 2011, 10:07 AM

If $d\geq 0$ is a perfect square then $\mathbb{Q}[\sqrt{d}]=\mathbb{Q}$ and so $\mathbb{Q}[\sqrt{d}]$ is a field.

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

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

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

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

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

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

Fernando Revilla

**mulaosmanovicben** · January 19th 2011, 06:15 PM

Thanks guys I have it figured out

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