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- Jul 13th 2006, 07:21 PMharoldFinding a poly
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- Jul 13th 2006, 07:33 PMThePerfectHackerQuote:

Originally Posted by**harold**

$\displaystyle \mathbb{Q}(\sqrt{5})=\{ f(\sqrt{5})| f(x)\in \mathbb{Q}[x]\}$

Basically, all rational polynomials evaluated at $\displaystyle \sqrt{5}$. - Jul 13th 2006, 07:38 PMThePerfectHackerQuote:

Originally Posted by**harold**

$\displaystyle \mbox{irr}(\sqrt{2}+\sqrt{5},\mathbb{Q})$

You procedded correctly,

$\displaystyle x=\sqrt{2}+\sqrt{5}$

$\displaystyle x^2=2+5+2\sqrt{10}$

$\displaystyle x^2-7=2\sqrt{10}$

$\displaystyle x^4-14x^2+49=40$

$\displaystyle x^4-14x^2+9$

You have shown that there exists a polynomial,

$\displaystyle x^4-14x^2+9\in \mathbb{Q}[x]$ which has,

$\displaystyle \sqrt{2}+\sqrt{5}$ as a zero. Now you need to show that this polynomial is irreducible over $\displaystyle \mathbb{Q}$. - Jul 13th 2006, 08:06 PMThePerfectHacker
To show that,

$\displaystyle x^4-14x^2+9$ is irreducible is going to be ugly. Firstly, just because it has no rational roots does not mean it is irreducible (you can only use that fact for 2nd and 3rd degree). But what you do know is that it has no linear factors. Thus, if it is reducible it must be,

$\displaystyle x^4-14x^2+9=(x^2+ax+b)(x^2+cx+d)$

Open right hand,

$\displaystyle x^4+ax^3+bx^2+cx^3+acx^2+bcx+dx^2+adx+bd$

Now,

$\displaystyle x^4+(a+c)x^3+(b+ac+d)x^2+(ad+bc)x+bd$

Finally conclude by definition of polynomial equality,

$\displaystyle \left\{ \begin{array}{c}a+c=0\\ b+ac+d=-14\\ ad+bc=0\\ bd=9$

Since,

$\displaystyle a=-c$

Substitute that,

$\displaystyle \left\{ \begin{array}{c} b-c^2+d=-14\\ c(b-d)=0 \\ bd=9$

Since, $\displaystyle c(b-d)=0$ one of them is zero (property of integral domain) thus assume $\displaystyle c=0$

Then we have,

$\displaystyle b+d=-14$

$\displaystyle bd=9$

Which is impossible in integers.

And if I assume that, $\displaystyle b=d$ in $\displaystyle c(b-d)=0$ I have that $\displaystyle b=d=3$ in last equation. Which would mean in the first equation that,

$\displaystyle 3-c^2+3=-13$ thus, $\displaystyle c^2=19$ has no solution in integers.

Thus, what I have shown is that,

$\displaystyle x^4-14x^2+9$ is irreducible over $\displaystyle \mathbb{Z}$. But what about $\displaystyle \mathbb{Q}$? No problem. There is a theorem that says a polynomial reducible in $\displaystyle \mathbb{Z}[x]$ if and only if $\displaystyle \mathbb{Q}[x]$

And we have finally shown that this polynomial is irreducible, answering your question. - Jul 13th 2006, 08:11 PMThePerfectHackerQuote:

Originally Posted by**harold**

See the definition that I posted is the proper way. The one you wrote looked unusual to me. I believe that though look different represent the same field. (I did not write out the formal prove but I can see it involving the basis $\displaystyle \{ 1,\sqrt{5}\}$ as a vector space for $\displaystyle \mathbb{Q}(\sqrt{5})$ over $\displaystyle \mathbb{Q}$) - Jul 13th 2006, 08:21 PMThePerfectHackerQuote:

Originally Posted by**harold**

Thus, since you have a 2nd degree polynomial it is the**minimal**degree. And remember as I said before minimial and irreducible mean the same thing. Thus, the polynomial is irreducible.

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I realized my solutions are out of order, try to manage anyways. - Jul 13th 2006, 11:04 PMBubbleBrain_103
just one quick note, about the definition of Q(sqrt(5)), you are both right... it is equal to all rational polynomials evaluated at sqrt(5), and {a+b*sqrt(5):a,b in Q}. Since sqrt(5) is of degree two, Q(sqrt(5)) is a two dimensional vector space over Q, and {1, sqrt(5)} is a fine basis for it.

- Jul 14th 2006, 06:05 AMThePerfectHackerQuote:

Originally Posted by**BubbleBrain_103**

- Jul 15th 2006, 06:31 PMThePerfectHacker
Seems okay to me.

- Jul 15th 2006, 07:11 PMThePerfectHackerQuote:

Originally Posted by**harold**

**coefficients**are ELEMENTS of the field. In this case $\displaystyle \mathbb{Q}(\sqrt{5})$ is the field. So yes, you finaly polynomial's coefficients are elements of this field. Any rational number is good because $\displaystyle q\in \mathbb{Q}(\sqrt{5})$ and $\displaystyle \sqrt{5}\in \mathbb{Q}(\sqrt{5})$ so the answer seems okay to me.

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It seems to me that you do not know what a field is, do you? How about simple extension?