Find the minimal polynomial for $\alpha=\sqrt{3-2\sqrt{2}}$ over $\mathbb{Q}$

$\alpha=\sqrt{3-2\sqrt{2}} \implies -\alpha^2-3=2\sqrt{2} \implies (\frac{-\alpha^2-3}{2})^2-2=0$

$\implies (\frac{-\alpha^2-3}{2})^2-2=\frac{\alpha^4+6\alpha^2+9}{4}-2=\alpha^4+6\alpha^2+1=0$

So I believe $P=\alpha^4+6\alpha^2+1$ is a candidate for a minimal polynomial

Let $x=\alpha^2 \implies P=x^2+6x+1=(x+3)^2-8$

Could P be the minimal polynomial? It clearly has $\alpha$ as root, but I am not sure if there is another polynomial of lower degree also having $\alpha$ as a root

Would appreciate your guidance on this

  • 2
    $\begingroup$ Compare with this question. $\endgroup$ – Dietrich Burde Apr 13 '16 at 11:20
  • $\begingroup$ But $P$ has roots equal to $-3\pm2\sqrt{2}$ and not $\alpha$. $\endgroup$ – B. Pasternak Apr 13 '16 at 11:20
  • 1
    $\begingroup$ P is irreducible by Perron's criterion, which states that if the second highest coefficient is greater in absolute value than the sum of the modulus of all the other coefficients except the largest coefficient plus one, then the polynomial is irreducible (here, 6>1+1=2). However the square root can actually be simplified, in which case this isn't quite the minimal polynomial, it will be the minimal polynomial of $1-\sqrt{2}$. $\endgroup$ – астон вілла олоф мэллбэрг Apr 13 '16 at 11:20
  • $\begingroup$ $3-2\sqrt 2$ is a square in $\mathbb Q(\sqrt 2)$ $\endgroup$ – Piquito Jun 1 '16 at 23:30

Hint: Note that $(1-\sqrt 2)^2=3-2\sqrt{2}$.

Therefore, $\alpha = \pm(1-\sqrt 2)$. Since $\alpha>0$, we must have $\alpha = -1+\sqrt2$.

Then $\alpha^2 = 3-2\sqrt{2} = -2\alpha +1$ and so the minimal polynomial of $\alpha$ is $x^2+2x-1$.

Clearly, $\mathbb Q(\alpha)=\mathbb Q(\sqrt 2)$, because $\alpha = -1+\sqrt2 \in \mathbb Q(\sqrt 2)$ and $\sqrt 2 = \alpha+1 \in \mathbb Q(\alpha)$.

  • $\begingroup$ thanks, so is the answer $(x^2-(1-\sqrt{2}))(x^2+(1+\sqrt{2}))$? $\endgroup$ – thinker Jun 1 '16 at 14:24
  • $\begingroup$ And would the corresponding extension field be $\mathbb{Q(1-\sqrt{2})}$ $\endgroup$ – thinker Jun 1 '16 at 14:25
  • $\begingroup$ How do we know that $\mathbb{Q}(\alpha)=\mathbb{Q}(\sqrt{2})$? That's not clear to me... is it because $\sqrt{2}$ is the only surd in the expression? $\endgroup$ – thinker Jun 1 '16 at 19:54
  • $\begingroup$ @thinker, see my edited answer. $\endgroup$ – lhf Jun 1 '16 at 21:55
  • $\begingroup$ Thanks.. so in general if $\alpha$ and $\beta$ are linearly dependent, i.e. $\alpha=\beta + k $ where $k \in \mathbb{Q}$ then $\mathbb{Q}(\alpha)=\mathbb{Q}(\beta)$? Also, your answer makes sense, just the first line, how is it obvious that $(1-\sqrt{2})^2=3-2\sqrt{2}$? I am concerned I wouldn't be able to spot this, say, in an exam situation $\endgroup$ – thinker Jun 2 '16 at 1:51

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