I want to find a polynomial whose roots adjoin to a field $\mathbb Q(\sqrt6)$ to give an extension field $\mathbb Q(\sqrt2,\sqrt3)$such that $[\mathbb Q(\sqrt2,\sqrt3):\mathbb Q(\sqrt6)]=2$.
Is that the same polynomial $x^4-5x^2+6$ that extends subfield $\Bbb Q$ to an extension field $\Bbb Q(\sqrt2,\sqrt3)$ with the degree of extension $4$.
If so, how to write the degree of extension in this case & how to show it is $2$.
I'm lost, please help.
On
I've outlined a solution below. I left some of the details as exercises because (1) I don't want to bore you with details that you already understand, and (2) I wanted to focus on the main ideas. If you would like solutions to any of these exercises, or if you have any other questions, then let me know. I'll be happy to help.
Exercise: Show that $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2},\sqrt{6})$.
Let $K=\mathbb{Q}(\sqrt{6})$. Let $p(x)=x^2-2\in K[x]$. Note that the roots of $p(x)$ are $\pm\sqrt{2}$. Hence if we adjoin a root of $p(x)$ to $K$, then we obtain $\mathbb{Q}(\sqrt{2},\sqrt{6})$.
Since $\sqrt{2}$ is a root of $p(x)=x^2-2\in K[x]$, and $\deg p(x)=2$, it follows that $[\mathbb{Q}(\sqrt{2},\sqrt{6}):K]\le2$, and the following are equivalent:
$$[\mathbb{Q}(\sqrt{2},\sqrt{6}):K]=2$$ $$p(x)\text{ is irreducible over }K[x]$$ $$\sqrt{2}\notin K$$
We'll show that $[\mathbb{Q}(\sqrt{2},\sqrt{6}):K]=2$ by showing that $\sqrt{2}\notin K$.
Suppose that $\sqrt{2}\in K$. Then $\sqrt{2}=a+b\sqrt{6}$ for some $a,b\in\mathbb{Q}$.
Squaring both sides we have $2=a^2+6b^2+2ab\sqrt{6}$. It follows that $2=a^2+6b^2$ and $2ab=0$.
Exercise: Show that there is no $a,b\in\mathbb{Q}$ with $2=a^2+6b^2$ and $2ab=0$.
Hence we have a contradiction. So we've shown that $\sqrt{2}\notin K$. Hence $[\mathbb{Q}(\sqrt{2},\sqrt{6}):K]=2$.
So if we adjoin a root of $x^2-2$ to $K=\mathbb{Q}(\sqrt{6})$, then we obtain $\mathbb{Q}(\sqrt{2},\sqrt{6})=\mathbb{Q}(\sqrt{2},\sqrt{3})$. Furthermore, we have that $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{6})]=[\mathbb{Q}(\sqrt{2},\sqrt{6}):\mathbb{Q}(\sqrt{6})]=2$.
On
$x^4-4x^2+1$ does the trick. All the square roots 1 2 3 6 are available.
For this number system (I call it Z12: the span of chords of the dodecagon), it is possible to arrange a counting-board, with the two rules 2(a1) = (a3), and (c1) = (b2)+(a1). The numbers in brackets are cell numbers (col a, row 1 etc). The array stretches both ways.
It is possible to set, relative to (a1), the sqrt(2) = b1, sqrt3 = (a1)+(b2), and sqrt(6) = (b1)+(z1) (x,y,z,a,b).
To directly derive $Q(1,\sqrt 6)$ and then $Z(\sqrt 2, \sqrt 3)$, it suffices to use the polynomial $x^2 = 5 + 2\sqrt 6$, both of these are in $Q(1, \sqrt 6)$.
On
The polynomial $x^2-2$ is going to work since this polynomial is irreducible in $\mathbb{Q}(\sqrt{6}) $ and
$\mathbb{Q}(\sqrt{6})[x]/(x^2-2) \cong \mathbb{Q}(\sqrt{2},\sqrt{3}) $ since $\frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3} $
or you can use the $x^2-3$
On
Your question is rather imprecise, since you don't specify in what field you take the coefficients of the polynomial you want. To fix ideas, let us work inside the field $K=\mathbf Q(\sqrt 2, \sqrt 3)$ and list all its subfields. The extension $K/\mathbf Q$ is biquadratic i.e. the compositum of two linearly disjoint quadratic subextensions (here, $\mathbf Q(\sqrt 2)$ and $\mathbf Q(\sqrt 3)$). It is normal, with Galois group $G\cong C_2\times C_2$. This direct product admits exactly three subgroups of order $2$. Two of them are already given, which are Gal($K/\mathbf Q(\sqrt 2))$ and Gal($K/\mathbf Q(\sqrt 3))$. Another quadratic subfield is $\mathbf Q(\sqrt 6)$. The point is to show that it is distinct from the two given ones. It is classically known (and easily shown) that $\mathbf Q(\sqrt a)=\mathbf Q(\sqrt b)$ iff $ab\in {\mathbf Q^*}^2$ (NB: this remains true when replacing $\mathbf Q$ by any field of characteristic $\neq 2$). Here $6\notin {\mathbf Q^*}^2$ (because of the uniqueness of prime factorization in $\mathbf Z$), so we have found all the strict subfields of $K$, which are $\mathbf Q, \mathbf Q(\sqrt 2), \mathbf Q(\sqrt 3),\mathbf Q(\sqrt 6)$.
The three quadrartic subfields play ing symmetric roles, it suffices to look at what happens over $\mathbf Q(\sqrt 6)$ (for example). To get $K$, it suffices clearly to add $\sqrt 2$ (minimal polynomial $X^2-2$) or $\sqrt 3$ (min. pol. $X^2-3$). If you want a min. pol. of degree $4$, you must take it with coefficients in $\mathbf Q$. The knowledge of such an irreducible polynomial is equivalent to the knowledge of a primitive element, i.e. an element $\alpha$ of $K$ s.t. $K=\mathbf Q(\alpha)$ (NB: such a primitive element is not unique). Here a natural primitive element is $\sqrt 2 + \sqrt 3$. For details, see e.g. https://math.stackexchange.com/a/3325514/300700
You know that $[\mathbb Q(\sqrt2,\sqrt3):\mathbb Q(\sqrt6)]=2$ so you also know that $\sqrt{6} \in \mathbb Q(\sqrt2,\sqrt3)$. Therefore you know that $\mathbb Q(\sqrt2,\sqrt3) = \mathbb Q(\sqrt2,\sqrt3,\sqrt6)$ and hence
$$ [\mathbb Q(\sqrt2,\sqrt3,\sqrt6):\mathbb Q(\sqrt6)] = [\mathbb Q(\sqrt2,\sqrt3,\sqrt6):\mathbb Q(\sqrt2,\sqrt6)][\mathbb Q(\sqrt2,\sqrt6):\mathbb Q(\sqrt6)] = 2. $$
So one of these extensions must be degree $2$ and the other degree $1$.
Pause and think for a minute about which one of these extensions is the degree $1$ extension.
I claim that $\sqrt{2}$ is not in $\mathbb Q(\sqrt6)$ from which it follows that $\sqrt{3}$ is in $\mathbb Q(\sqrt2,\sqrt6)$. Can you see how to get $\sqrt3$ from $\sqrt2$ and $\sqrt6$?
Therefore if we add $\sqrt{2}$ to $\mathbb Q(\sqrt6)$ via its minimal polynomial $x^2 - 2$, then we can also get $\sqrt3$ and hence we have all of $\mathbb Q(\sqrt2,\sqrt3,\sqrt6) = \mathbb Q(\sqrt2,\sqrt3)$.