Is this polynomial: $irr(\sqrt{3 -\sqrt{6}}, \mathbb{Q})$ irreducible?
Here is what I did $ a = \sqrt{3 -\sqrt{6}}$
$a^2 = 3 - \sqrt{6}$
$a^2 - 3 = -\sqrt{6}$
$(a^2 - 3)^2 = 6$
Our polynomial ends up being $a^4 - 6a^2 + 9$ Which I can obviously factor into 2 nonconstant polynomials. Am I misunderstanding the definiton? My problem set says it's irreducible
The polynomial $f(x)=irr(\sqrt{3-\sqrt{6}},\Bbb Q)$ is irreducible almost by definition, i.e., this is the smallest polynomial (in the degree sense) such that $f\left(\sqrt{3-\sqrt{6}}\right)=0$, if you could factor $f(x)$ as
$$f(x)=g(x)h(x)$$
with $g,h$ non-constant polynomials, then
$$f\left(\sqrt{3-\sqrt{6}}\right)=g\left(\sqrt{3-\sqrt{6}}\right)\cdot h\left(\sqrt{3-\sqrt{6}}\right)=0$$
Can you see what contradiction arises here? I suggest you finish the proof.
Now, let's focus on your proceeding. The relation you've found is correct:
$$(a^2-3)^2=6$$
If you expand that square, you get
$$a^4-6a^2+9=6$$
Which is the polynomial that $a$ annihilates? It's not
$$f(x)=x^4-6x^2+9$$
Since $f(a)=6$, the polynomial is actually
$$f(x)=x^4-6x^2+3$$
This polynomial is irreducible over $\Bbb Q$.