# 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.

• What do you mean by "they are isomorphic"? As fields? No, they aren't. As rational linear spaces? Yes, they both are of dimension two. – DonAntonio Sep 11 '16 at 20:48
• Sorry, isomorphic field extensions. – Ldog327 Sep 11 '16 at 20:49
• @DonAntonio Actually they are isomorphic. – Matt Samuel Sep 11 '16 at 20:51
• @DonAntonio Why would you doubt it? The roots of the second polynomial are $2\pm \sqrt2$, the roots of the first are $\pm\sqrt2$. – Matt Samuel Sep 11 '16 at 20:54
• @MattSamuel Why wouldn't I doubt? I'm a mathematician: I must doubt. But when you pointed out what the roots are I realized I thought all the time the second polynomial was $\;t^2-4t-2\;$ , so I took a good second look...and I doubt no more anymore. ;) – DonAntonio Sep 11 '16 at 20:59

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).

• No you don't need the axiom of choice. Just consider them as subfields of $\mathbb{Q}(α,β)$. – user21820 Sep 12 '16 at 5:36
• @user21820 you're absolutely right, although my comment wanted to be more general than this case (for instance, an infinite sequence of algebraic extensions...) – Angelo Rendina Sep 12 '16 at 9:30
• Yes if you want to consider $\mathbb{Q}(S)$ and $\mathbb{Q}(T)$ for some infinite sets $S,T$ of algebraic numbers. But it's a very tiny bit misleading to use the algebraic closure when it's not necessary for the task at hand, in my opinion. =) – user21820 Sep 12 '16 at 9:52

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.

• Or observe that $t^2-4t+2=t^2-4t+4-2=(t-2)^2-2$. – egreg Sep 11 '16 at 21:55

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 $\Bbb Q[t]/(t^2 - 2)\to\Bbb Q[x]/(x^2 - 4x + 2)$?