Q Field Questions

**Samson** · July 7th 2010, 05:19 AM

Hello everybody,

I reached this long section in my book about Q Fields, and after reading through this section, a couple of the examples have me feeling less than enlightened. Here they are:

1. If alpha = a + b*Sqrt(d) (which exists in) Q[Sqrt(d)], an expression for a/a_bar in terms of a,b, and d can be found, assuming d is not a perfect square. I read this and was like what????

My work thus far: I don't know what they mean by a/a_bar. I know I did Sqrt[(alpha-a)/b)=d and somehow we force d not to be an integer... perhaps they mean a_bar = alpha-a ? No clue

2. Does anyone know if every element of Q[Sqrt(2)] have a square root in Q[Sqrt(2)] ? Can this be proven/disproven with a counterexample?

I just have problems entirely on where to go with that one.

I really appreciate any and all help! Thank you!

**Amer** · July 7th 2010, 08:07 AM

Originally Posted by Samson

1. If alpha = a + b*Sqrt(d) (which exists in) Q[Sqrt(d)], an expression for a/a_bar in terms of a,b, and d can be found, assuming d is not a perfect square?

do you mean
an expression for $\alpha\cdot \overline{\alpha}$

**undefined** · July 7th 2010, 08:13 AM

Originally Posted by Samson

Hello everybody,

I reached this long section in my book about Q Fields, and after reading through this section, a couple of the examples have me feeling less than enlightened. Here they are:

1. If alpha = a + b*Sqrt(d) (which exists in) Q[Sqrt(d)], an expression for a/a_bar in terms of a,b, and d can be found, assuming d is not a perfect square. I read this and was like what????

My work thus far: I don't know what they mean by a/a_bar. I know I did Sqrt[(alpha-a)/b)=d and somehow we force d not to be an integer... perhaps they mean a_bar = alpha-a ? No clue

2. Does anyone know if every element of Q[Sqrt(2)] have a square root in Q[Sqrt(2)] ? Can this be proven/disproven with a counterexample?

I just have problems entirely on where to go with that one.

I really appreciate any and all help! Thank you!

1. I do not know what a_bar could refer to. My guess is that the alpha symbol looks enough like "a" that you couldn't distinguish them.

$\displaystyle \frac{\alpha}{\overline{\alpha}} = \frac{a+b\sqrt{D}}{a-b\sqrt{D}}=$

$\cdots$

$\displaystyle \frac{a^2+b^2D}{a^2-b^2D} + \frac{2ab}{a^2-b^2D}\sqrt{D}$

2. (Should be fixed now compared with what I initially wrote) This is pretty easily disproven. Consider that

$(a + b\sqrt{D})^2 = a^2+b^2D + 2ab\sqrt{D}$

Suppose that $\sqrt{3+0\sqrt{2}}\in\mathbb{Q}(\sqrt{2})$ . So

$a^2+2b^2=3$

$2ab = 0$

Clearly a and b can't both equal 0. Suppose a = 0, then

$2b^2=3$

$b^2=3/2$

b is not rational, contradiction.

Suppose b = 0, then a is the square root of 3, also not rational, thus contradiction. QED.

**Samson** · July 7th 2010, 08:14 AM

Originally Posted by Amer

do you mean
an expression for $\alpha\cdot \overline{\alpha}$

Almost, except $\alpha / \overline{\alpha}$

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