Computing the Minimal polynomial - Ring Theory

[chwala]

TL;DR

See attached

Am going through this notes...kindly let me know if there is a mistake on highlighted part. I think it ought to be;

$α^2=5+2\sqrt{6}$

 

[fresh_42]

Summary: See attached

Am going through this notes...kindly let me know if there is a mistake on highlighted part. I think it ought to be;

$α^2=5+2\sqrt{6}$

View attachment 315473

You are right. It is a mistake in the book and should be $\alpha^2=5+2\sqrt{6}.$ It is the correct value on the list at the end again. The procedure itself is correct.

 

[chwala]

You are right. It is a mistake in the book and should be $\alpha^2=5+2\sqrt{6}.$ It is the correct value on the list at the end again. The procedure it self is correct.

Thanks...let me peruse through...

 

[WWGD]

The text from Chwala provides a nice, clean proof of the Irrationality of $\sqrt 2$, though Eisenstein theorem. Per that theorem, $x^2 -2$ has no Rational solution. A nice, handwavy proof.

 

[fresh_42]

The text from Chwala provides a nice, clean proof of the Irrationality of $\sqrt 2$, though Eisenstein theorem. Per that theorem, $x^2 -2$ has no Rational solution. A nice, handwavy proof.

Yes, but Eisenstein uses the fact that $\mathbb{Z}$ is a UFD and $2$ is prime. With that, you don't need Eisenstein anymore:
\begin{align*}
\sqrt{2}=\dfrac{m}{n} \Longrightarrow 2n^2=m^2 \Longrightarrow 2\,|\,m \Longrightarrow 4\,|\,m^2\Longrightarrow 2\,|\,n^2
\end{align*}
contradicting the assumption that $\dfrac{m}{n}$ was cancelled.

Hence, you do not use Eisenstein, you use its conditions.