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Consider $\mathbb{F}_{5}(\sqrt{2},\sqrt{3})$.

If $\sqrt{2},\sqrt{3}$ were both algebraic then $\sqrt{2},\sqrt{3}\in A$, where $A$ is the field of all algebraic elements over $\mathbb{F}_{5}$ (with extension $\mathbb{F}_{5}(\sqrt{2},\sqrt{3})/\mathbb{F}_{5}$). However $(\sqrt{6}-1)(\sqrt{6}+1) = 5$, which is also algebraic, of course, but since $A$ is field then it is integral domain, so it must be that $5$ in $A$ is somehow different from $5=0$ in $\mathbb{F}_{5}\subset \mathbb{F}_{5}(\sqrt{2},\sqrt{3})$ and so $\sqrt{2},\sqrt{3}$ are not algebraic. Am I right or I am missing something?

matheg
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    $\sqrt{6} = \sqrt1 = \pm 1$ and $(\sqrt{6}-1)(\sqrt{6}+1) =5=0$. – reuns Dec 03 '21 at 20:58
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    Square roots are algebraic by definition ($\sqrt a$ is defined as a root of $X^2-a$). If your conclusion is that they aren't, then something in your argument must be wrong (which @reuns has pointed out already). – Vercassivelaunos Dec 03 '21 at 21:07
  • A different way of seeing the same thing. $A$ is a field, so there are no zero divisors. You correctly observed that $(\sqrt6-1)(\sqrt6+1)=0$ in $A$. Therefore one of $\sqrt6-1$ or $\sqrt6+1$ must be zero. We cannot tell which it is unless you specify exactly what you mean $\sqrt2,\sqrt3,\sqrt6$ as elements of $A$. – Jyrki Lahtonen Dec 04 '21 at 05:21
  • All these numbers are actually already in the smaller field $K=\Bbb{F}_{25}=\Bbb{F}_5(\sqrt2)$. In $\Bbb{F}_5$ we have $3=-2=4\cdot2=2^2\cdot 2$. Implying that $$\sqrt3=\pm2\sqrt2$$ in the field $K$. Here there is no natural way to select the sign, so you have to tell us which do you mean. If $\sqrt3=2\sqrt2$, then $$\sqrt3\cdot\sqrt2+1=2(\sqrt2)^2+1=4+1=0.$$ On the other hand, if you selected $\sqrt3=-2\sqrt2$, then $$\sqrt3\cdot\sqrt2-1=-2(\sqrt2)^2-1=0.$$ – Jyrki Lahtonen Dec 04 '21 at 05:26
  • In other words, unlike over $\Bbb{Q}$, the square roots $\sqrt2$ and $\sqrt3$ are not "independent" from each other, and only generate a degree two extension field. This happens with every prime $p$ and any pair of quadratic non-residues $a,b$. The field $\Bbb{F}_p(\sqrt a,\sqrt b)$ is the same field as $\Bbb{F}_p(\sqrt a)$. This is because both $\Bbb{F}_p(\sqrt a)$ and $\Bbb{F}_p(\sqrt b)$ have $p^2$ elements, and hence are isomorphic (and hence equal as subsets of $A$). There is (up to isomorphism or as a subset of $A$) only a single extension field of a given degree. – Jyrki Lahtonen Dec 04 '21 at 05:32
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    Let $F = \Bbb Z_5[\sqrt 2] := \Bbb Z_5[w]/(w^2-2).,$ In $F[x]$ we have $,x^2-3,$ is reducible, viz. $,x^2-3 = x^2-8 = x^2-4w^2 = (x-2w)(x+2w)$ so $,F[\sqrt 3]:= F[x]/(x^2-3)$ is not an integral domain (so not a field). $\ \ $ – Bill Dubuque Dec 04 '21 at 15:08
  • See also this answer – Bill Dubuque Dec 04 '21 at 15:18

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