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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,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 $\displaystyle \mathbb{Q}$.
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.
It might. Does it represent a simple extension?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 $\displaystyle \mathbb{Q}$)
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 $\displaystyle \mathbb{Q}(\sqrt{5})$ which would have $\displaystyle \sqrt{2}+\sqrt{5}$ as a root would imply that $\displaystyle \sqrt{2}\in \mathbb{Q}(\sqrt{5})$, but that is not true.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.
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.
It means that the 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.Originally Posted by harold
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It seems to me that you do not know what a field is, do you? How about simple extension?