Showing isomorphism of field extensions I need to show that if $\alpha$ has minimal polynomial $t^2-2$ over $\mathbb{Q}$ and $\beta$ has minimal polynomial $t^2-4t+2$ over $\mathbb{Q}$, then the extensions $\mathbb{Q}(\alpha):\mathbb{Q}$ and $\mathbb{Q}(\beta):\mathbb{Q}$ are isomorphic. 
I want to say that somehow $t^2-2 \equiv t^2-4t+2$ modulo something and this will be of help in the proof? I see that $t^2-4t+2-(t^2-2)=-4t+4$, so maybe I should work modulo $1-t$? Does this make any sense and am I on the right track? 
Thanks.
 A: Prove the homomorphism $\mathbb Q(\alpha)\to \mathbb Q(\beta)$ determined by $\alpha\mapsto \beta-2$ is an isomorphism. How did I pick this map? I didn't pull it out of thin air. Compute the roots of the two polynomials. $\beta$ is a root of the second if and only if $\beta-2$ is a root of the first.
A: Write $\Bbb Q(\alpha)\cong\Bbb Q[t]/(t^2 - 2)$ and $\Bbb Q(\beta)\cong\Bbb Q[x]/(x^2 -4x + 2)$. The roots of $t^2 - 2$ are $\pm\sqrt{2}$ and the roots of $t^2 - 4t + 2$ are $2\pm\sqrt{2}$. Since $t$ in the first quotient represents $\sqrt{2}$ or $-\sqrt{2}$, and $x$ in the second represents either $2 + \sqrt{2}$ or $2 - \sqrt{2}$, does this tell you where to send $t$ to define a map $\Bbb Q[t]/(t^2 - 2)\to\Bbb Q[x]/(x^2 - 4x + 2)$?
A: Working in $\mathbb{C}$ (which includes the algebraic closure of $\mathbb{Q}$), you can see that
$$\alpha=\pm\sqrt{2}$$
$$\beta=\frac{4\pm\sqrt{16-8}}{2}=2\pm\sqrt{2}$$
hence $\mathbb{Q}(\alpha)=\mathbb{Q}(\beta)$.
If you want to reason purely algebraically, then you need to show that $\beta\in\mathbb{Q}(\alpha)$ (and viceversa for $\alpha$). You know that every element of $\mathbb{Q}(\alpha)$ is of the form $x\alpha+y$  for $x,y\in\mathbb{Q}$. Can you find suitable $x,y$ such that $\beta=x\alpha+y$?
Use the fact that $$0=\beta^2-4\beta+2=(x^2\alpha^2+y^2+2xy\alpha)-4(x\alpha+y)+2=(2x^2+y^2-4y+2)+\alpha(2xy-4x)$$
hence $x=1$ and $y=2$. Something similar shows that $\alpha\in\mathbb{Q}(\beta)$.
Edit: note that writing elements of $\mathbb{Q}(\beta)$ as elements of $\mathbb{Q}(\alpha)$ makes sense since we are viewing both fields as subfields of the same algebraic closure (which exists, assuming the axiom of choice).
