I have been asked to prove that $[\mathbb{Q}(\sqrt2,\sqrt{1+i}):\mathbb{Q}]=4$. I have some problems believing this to be true and have the following argument that assumes it to be true and seemingly derives a contradiction. If someone could verify whether in the following argument I have made any incorrect assertions or otherwise, I would be grateful.

  1. By the Tower Law for field extensions, we have: $[\mathbb{Q}(\sqrt2,\sqrt{1+i}):\mathbb{Q}]= [\mathbb{Q}(\sqrt2,\sqrt{1+i}):\mathbb{Q}(\sqrt2)]\cdot[\mathbb{Q}(\sqrt2):\mathbb{Q}]$. We have that $[\mathbb{Q}(\sqrt2):\mathbb{Q}]=2$. I have previously shown this result explicitly and have no problem working with this easy vector space over $\mathbb{Q}$.
  2. So, if the question is correct, then $[\mathbb{Q}(\sqrt2,\sqrt{1+i}):\mathbb{Q}(\sqrt2)]=2$. For a simple field extension $K(\alpha)$ over $K$, with $\alpha$ algebraic over $K$, we have $[K(\alpha):K]=$ deg(mipo$_K(\alpha))$. We need to show $[\mathbb{Q}(\sqrt2,\sqrt{1+i}):\mathbb{Q}(\sqrt2)]=[\mathbb{Q}(\sqrt2)(\sqrt{1+i}):\mathbb{Q}(\sqrt2)]=2$. Therefore, by the result just stated, with $\mathbb{Q}(\sqrt2)$ as our base field and $\sqrt{1+i}$ as the algebraic element we adjoin, we have that deg(mipo$_{\mathbb{Q}\sqrt2}(\sqrt{1+i}))=2$.
  3. If this is the case, then we have that for some $p,q\in \mathbb{Q}(\sqrt2)$ that $p+q\sqrt{1+i}+(1+i)=0$. But there are no non-trivial solutions to this equation, since writing $\sqrt{1+i}$ explicitly, i.e. $\sqrt{1+i}=\frac{\sqrt{1+\sqrt2}}{\sqrt2}+i\frac{\sqrt{-1+\sqrt2}}{\sqrt2}$ (I derived this form in radicals and it can be verified on Wolfram), we have that this equation is equivalent to $(a+b\sqrt2)+(c+d\sqrt2)(\frac{\sqrt{1+\sqrt2}}{\sqrt2}+i\frac{\sqrt{-1+\sqrt2}}{\sqrt2})+(1+i)=0$, where $a,b,c,d,e,f\in\mathbb{Q}$.
  4. Since the equation equals $0$, in particular the imaginary component vanishes, i.e. $(c+d\sqrt2)(\frac{\sqrt{-1+\sqrt2}}{\sqrt2})+1=0$. Ultimately, no consistent solution exists to this equation.
  5. Rearranging we obtain, $\sqrt{-1+\sqrt2}=\frac{-\sqrt2}{c+d\sqrt2}$, and after some rationalising it follows that $\sqrt{-1+\sqrt2}=\frac{2d}{c^{2}-2d^{2}}-\frac{c}{c^{2}-2d^{2}}\sqrt2$.
  6. Hence, $\sqrt{-1+\sqrt2}=p+q\sqrt2$, for $p,q \in \mathbb{Q}$ as in the previous step. But, by first squaring both sides, a contradiction is derived. After squaring, we have the quadratic equation: $-1+\sqrt2=p^{2}+2q^{2}+2pq\sqrt2$
  7. From this, we obtain two equations, which are: $p^{2}+2q^{2}=-1$ and $2pq=1$. Solving does not lead to any solutions in $\mathbb{Q}$ as, for instance $q^2=\frac{-4\pm\sqrt{-16}}{16}$, contradicting $p,q\in\mathbb{Q}$. Indeed, $p^{2}+2q^{2}\neq-1$ in $\mathbb{Q}$.
  8. Therefore, by our chain of reasoning, deg(mipo$_{\mathbb{Q}\sqrt2}(\sqrt{1+i}))\neq2$. Hence by the earlier result, $[\mathbb{Q}(\sqrt2,\sqrt{1+i}):\mathbb{Q}(\sqrt2)]\neq2$ and thus by the first point (using the Tower Law), $[\mathbb{Q}(\sqrt2,\sqrt{1+i}):\mathbb{Q}]\neq4$.

Please note that I have updated this solution after the feedback. Thank you all for your initial feedback (particularly T. Bongers). However, after writing the minimal polynomial in degree 2 (initially I had written it in degree 1), I believe that a contradiction also arises. I would appreciate any feedback as to whether the updated reasoning here is sound.

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    $\begingroup$ There is a problem in your argument at step 3. If there were p and q as you describe, that would show that $\sqrt{1+i}$ was in $\mathbb Q(\sqrt{2})$. $\endgroup$ – Hugh Thomas Nov 2 '13 at 18:55
  • $\begingroup$ And you should also be positive about what $\sqrt{1+i}$ means. It denotes one of two possible complex numbers and (maybe) it can happen that the statement is true for one and not for the other. $\endgroup$ – Git Gud Nov 2 '13 at 18:57
  • $\begingroup$ Dear @GitGud: no, it cannot happen because if a field contains an element it also contains the opposite of that element. $\endgroup$ – Georges Elencwajg Nov 2 '13 at 20:40
  • $\begingroup$ @GeorgesElencwajg Thank you. For some reason I assumed that if $w$ was a square root of $1+i$, then so would $\overline w$ be. $\endgroup$ – Git Gud Nov 2 '13 at 20:46
  • $\begingroup$ By the way, could someone show why $[\mathbb{Q}(\sqrt2,\sqrt{1+i}):\mathbb{Q}]=4$ ? I think it is true and easy but I can't prove it (which is not very flattering for me since I taught Galois theory!) $\endgroup$ – Georges Elencwajg Nov 2 '13 at 21:21

Your step three isn't correct. It should read

There exist $p, q \in \mathbb{Q}(\sqrt{2})$ with $$\sqrt{1 + i}^2 + p \sqrt{1 + i} + q = 0$$

Your step uses a minimum polynomial of degree less than $2$.

  • $\begingroup$ Thank you. I have made this mistake before. For some reason, I seem to read the degree as referring to a polynomial of degree 1 lower than it should be. I will have another look at the question. $\endgroup$ – JWeissman Nov 2 '13 at 19:03

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