# Finding a poly

• Jul 13th 2006, 07:21 PM
harold
Finding a poly
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• Jul 13th 2006, 07:33 PM
ThePerfectHacker
Quote:

Originally Posted by harold
Given $\mathbb{Q}(\sqrt {5}) = {a + b\sqrt{5} | a,b\in \mathbb{Q}}$

I believe you made a mistake,
$\mathbb{Q}(\sqrt{5})=\{ f(\sqrt{5})| f(x)\in \mathbb{Q}[x]\}$
Basically, all rational polynomials evaluated at $\sqrt{5}$.
• Jul 13th 2006, 07:38 PM
ThePerfectHacker
Quote:

Originally Posted by harold
Find the monic polynomial with coefficients in $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{5})$ of smallest possible degree that has $\sqrt{2} + \sqrt {5}$ as a root.

There is a theorem in field theory that the smallest degree if and only if irreducible. Thus, what you are really trying to find is,
$\mbox{irr}(\sqrt{2}+\sqrt{5},\mathbb{Q})$
You procedded correctly,
$x=\sqrt{2}+\sqrt{5}$
$x^2=2+5+2\sqrt{10}$
$x^2-7=2\sqrt{10}$
$x^4-14x^2+49=40$
$x^4-14x^2+9$
You have shown that there exists a polynomial,
$x^4-14x^2+9\in \mathbb{Q}[x]$ which has,
$\sqrt{2}+\sqrt{5}$ as a zero. Now you need to show that this polynomial is irreducible over $\mathbb{Q}$.
• Jul 13th 2006, 08:06 PM
ThePerfectHacker
To show that,
$x^4-14x^2+9$ 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,
$x^4-14x^2+9=(x^2+ax+b)(x^2+cx+d)$
Open right hand,
$x^4+ax^3+bx^2+cx^3+acx^2+bcx+dx^2+adx+bd$
Now,
$x^4+(a+c)x^3+(b+ac+d)x^2+(ad+bc)x+bd$
Finally conclude by definition of polynomial equality,
$\left\{ \begin{array}{c}a+c=0\\ b+ac+d=-14\\ ad+bc=0\\ bd=9$
Since,
$a=-c$
Substitute that,
$\left\{ \begin{array}{c} b-c^2+d=-14\\ c(b-d)=0 \\ bd=9$

Since, $c(b-d)=0$ one of them is zero (property of integral domain) thus assume $c=0$
Then we have,
$b+d=-14$
$bd=9$
Which is impossible in integers.

And if I assume that, $b=d$ in $c(b-d)=0$ I have that $b=d=3$ in last equation. Which would mean in the first equation that,
$3-c^2+3=-13$ thus, $c^2=19$ has no solution in integers.

Thus, what I have shown is that,
$x^4-14x^2+9$ is irreducible over $\mathbb{Z}$. But what about $\mathbb{Q}$? No problem. There is a theorem that says a polynomial reducible in $\mathbb{Z}[x]$ if and only if $\mathbb{Q}[x]$
And we have finally shown that this polynomial is irreducible, answering your question.
• Jul 13th 2006, 08:11 PM
ThePerfectHacker
Quote:

Originally Posted by harold
Does this change anything?

It might. Does it represent a simple extension?

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 $\{ 1,\sqrt{5}\}$ as a vector space for $\mathbb{Q}(\sqrt{5})$ over $\mathbb{Q}$)
• Jul 13th 2006, 08:21 PM
ThePerfectHacker
Quote:

Originally Posted by harold
Also, do I need to show irreducibility over $\mathbb{Q}$ for $\mathbb{Q}(\sqrt{5})$ as well?

You got it. Cuz, how do you know that the polynomial you have it indeed irreducible? In this case it is easier. If you can find a linear polynomial in $\mathbb{Q}(\sqrt{5})$ which would have $\sqrt{2}+\sqrt{5}$ as a root would imply that $\sqrt{2}\in \mathbb{Q}(\sqrt{5})$, but that is not true.
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 PM
BubbleBrain_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 AM
ThePerfectHacker
Quote:

Originally Posted by BubbleBrain_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.

Thank you BubbleBrain but you should have read the post on top, it says the same thing ;)
• Jul 15th 2006, 06:31 PM
ThePerfectHacker
Seems okay to me.
• Jul 15th 2006, 07:11 PM
ThePerfectHacker
Quote:

Originally Posted by harold
Just to double check with you TPH, is that what it means to evaluate it at $\mathbb{Q}(\sqrt {5})$ ?

It means that the coefficients are ELEMENTS of the field. In this case $\mathbb{Q}(\sqrt{5})$ is the field. So yes, you finaly polynomial's coefficients are elements of this field. Any rational number is good because $q\in \mathbb{Q}(\sqrt{5})$ and $\sqrt{5}\in \mathbb{Q}(\sqrt{5})$ 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?