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I am struggling in finding the minimal polynomial of $\sqrt{3}+\sqrt{5}\in \mathbb C$ over $\mathbb Q$

Any ideas? I tried to consider its square but it did not helped..

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  • $\begingroup$ Try computing it's cube and its fourth power. $\endgroup$ – Alex Wertheim Nov 29 '14 at 17:02
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    $\begingroup$ Let $\alpha =\sqrt 3+\sqrt 5$. Square both sides then isolate surds and square again. $\endgroup$ – Git Gud Nov 29 '14 at 17:03
  • $\begingroup$ But how do I know that there is no other minimal polynomial with smaller degree? $\endgroup$ – Duke Nov 29 '14 at 17:06
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    $\begingroup$ You can apply irreducibility criterion. Alternatively, note that $\mathbb{Q}(\sqrt{3}+\sqrt{5})$ is a subfield of $\mathbb{Q}(\sqrt{3}, \sqrt{5})$. Since the latter is a degree $4$ extension ($x^{2}-5$ is irreducible over $\mathbb{Q}(\sqrt{3})$), $\mathbb{Q}(\sqrt{3}+\sqrt{5})$ must be a degree 2 or 4 extension of $\mathbb{Q}$. Hence, the minimal polynomial must have degree $2$ or degree $4$. But you've shown it can't have degree $2$. $\endgroup$ – Alex Wertheim Nov 29 '14 at 17:06
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As was mentioned in the other answers, $x^4-16x^2+4$ is a polynomial with root $\sqrt{3}+\sqrt{5}$, so it suffices to prove that this polynomial is irreducible. There are several methods:

1) Calculation of the quadratic factors, see Lubin's answer.

2) Field theory: $\mathbb{Q}(\sqrt{3},\sqrt{5}) = \mathbb{Q}(\sqrt{3})(\sqrt{5})$ has degree $2 \cdot 2=4$ over $\mathbb{Q}$ essentially because $\sqrt{5} \notin \mathbb{Q}(\sqrt{3})$ (easy computation!) and hence $X^2-5$ stays irreducible over $\mathbb{Q}(\sqrt{3})$. So it suffices to prove that $\mathbb{Q}(\sqrt{3},\sqrt{5})=\mathbb{Q}(\sqrt{3}+\sqrt{5})$ (because then $\sqrt{3}+\sqrt{5}$ has degree $4$ over $\mathbb{Q}$). Since $\supseteq$ is trivial, it suffices to check $\subseteq$. Notice that rationalisiation yields $$\frac{1}{\sqrt{3}+\sqrt{5}}=\frac{\sqrt{3}-\sqrt{5}}{3-5}.$$ Hence, $\sqrt{3}-\sqrt{5} \in \mathbb{Q}(\sqrt{3}+\sqrt{5})$. But then we also have $2 \sqrt{3} = (\sqrt{3}-\sqrt{5}) + (\sqrt{3}+\sqrt{5}) \in \mathbb{Q}(\sqrt{3}+\sqrt{5})$. This shows $\sqrt{3} \in \mathbb{Q}(\sqrt{3}+\sqrt{5})$ and then we also get $\sqrt{5} \in \mathbb{Q}(\sqrt{3}+\sqrt{5})$. $\checkmark$

The same proof shows: If $p,q$ are distinct prime numbers, then $\mathbb{Q}(\sqrt{p},\sqrt{q})=\mathbb{Q}(\sqrt{p}+\sqrt{q})$ has degree $4$ over $\mathbb{Q}$. The minimal polynomial of $\sqrt{p}+\sqrt{q}$ is $(x^2-p-q)^2 - 4 pq$.

3) Kummer theory: It is a special case of Kummer theory that for distinct prime numbers $p_1,\dotsc,p_n$ (actually pairwise coprime square-free integers also work) $\mathbb{Q}(\sqrt{p_1},\dotsc,\sqrt{p_n})$ has degree $2^n$ over $\mathbb{Q}$, has primitive element $\alpha=\sqrt{p_1} + \dotsc + \sqrt{p_n}$ and that its minimal polynomial therefore has degree $2^n$. It is given by the product of all $X-\beta$, where $\beta$ runs through the conjugates $\pm \sqrt{p_1} \pm \dotsc \pm \sqrt{p_n}$. (It is quite obvious that this polynomial is rational and has $\alpha$ as a root, but its irreducibility is not so easy to prove directly, and this is where Kummer theory helps.)

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  • $\begingroup$ Good and complete. $\endgroup$ – Lubin Nov 29 '14 at 21:51
  • $\begingroup$ This is why I love MSE. Thank you, Martin! $\endgroup$ – Prism Nov 30 '14 at 10:40
  • $\begingroup$ Sir, I get that $\sqrt{3}$,$\sqrt{2}$, $\sqrt{3}-\sqrt{5}$, $\sqrt{3}+\sqrt{5}$ are in $\mathbb{Q}(\sqrt{3}+\sqrt{5})$ " but how and why they are roots of minimal polynomial of $\sqrt{3}+\sqrt{5}$ over $\mathbb{Q}$ ? $\endgroup$ – Akash Patalwanshi Feb 8 '19 at 6:39
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Another method is to write down all four conjugates of $\sqrt3+\sqrt5$, namely the obvious $\pm\sqrt3\pm\sqrt5$. Then expand $$ (X-\sqrt3-\sqrt5)(X-\sqrt3+\sqrt5)(X+\sqrt3-\sqrt5)(X+\sqrt3+\sqrt5)\,. $$ It comes right out.

EDIT: Let me add an argument for irreducibility, prompted by Barry Cipra’s question below. The only factors to check would be quadratic, thus one wants to verify that none of $(X-\sqrt3-\sqrt5)(X-\sqrt3+\sqrt5)$, $(X-\sqrt3-\sqrt5)(X+\sqrt3+\sqrt5)$, or $(X-\sqrt3-\sqrt5)(X=\sqrt3-\sqrt5)$ has rational coefficients. For instance, the first of these multiplies out to $X^2-2\sqrt3X-2$. The story is similar for the other two. So irreducible.

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    $\begingroup$ Don't you also need to argue that no subset of these four factors multiply out to a polynomial with all coefficients rational? $\endgroup$ – Barry Cipra Nov 29 '14 at 17:34
  • $\begingroup$ @BarryCipra, that’s asking about the irreducibility of the polynomial, an issue I didn’t address. I was merely worried about an efficient way of finding the polynomial. You can’t show irreducibility by looking modulo $p$ for any single $p$, for reasons I don’t want to go into. Probably an argument prompted by galois theory would be best. $\endgroup$ – Lubin Nov 29 '14 at 18:14
  • $\begingroup$ Nice method! $\;\!$ $\endgroup$ – goblin Jan 7 '18 at 8:07
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$$\begin{align}\alpha &= \sqrt{3} + \sqrt{5} \Rightarrow \alpha^2 = 3 + 2\sqrt{3}\sqrt{5} + 5 \\&\Rightarrow (\alpha^2 - 8)^2 = 4\dot\ 15 \Rightarrow \alpha^4 - 16\alpha^2 +64 -60 = 0 \Rightarrow \alpha^4 - 16\alpha^2 +4 = 0 \end{align}$$

Take $f(X) = X^4 - 16X^2 + 4$. Notice that $f(X) \in \mathbb{Z}[x]$ and is irreducible over $\mathbb{Z}$ (by the arguments of Timbuc and Lubin). So using Gauss Theorem, it's irreducible over $\mathbb{Q}$.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – user642796 Dec 5 '14 at 9:44
  • $\begingroup$ Very helpful. +1 $\endgroup$ – mick Jan 20 '15 at 23:23
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You should be able to find a monic quartic $p(x)\in\Bbb Q[x]$ having $\sqrt3\pm\sqrt5$ and $-\sqrt3\pm\sqrt5$ as its roots. There are only $3$ monic quadratic and $3$ monic cubic factors of $p(x)$ of which $\sqrt3+\sqrt5$ is a root, so it shouldn't be tricky to check whether any of those factors is in $\Bbb Q[x]$. In fact, you shouldn't even have to check the cubic factors. (Do you see why?)

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  • $\begingroup$ Right. I did just that in detail in my response to Barry Cipra, above. $\endgroup$ – Lubin Nov 29 '14 at 18:37
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I propose the following way to prove the polynomial $\;x^4-16x^2+4\;$ is irreducible over $\;\Bbb Q\;$ . First, factor it over the reals:

$$x^4-16x^2+4=(x^2-2\sqrt5\,x+2)(x^2+2\sqrt5\,x+2)$$

(this is way easier than what can thought at first, at least in this an other similar cases).

From here, it's clear the polynomial cannot be factores any other essentially different way as $\;\Bbb R[x]\;$ is a UFD and, of course, $\;\Bbb Q\subset\Bbb R\;$ . End the argument now.

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    $\begingroup$ But neither of these factors is $\mathbb R$-irreducible, as you can see by calculating the discriminant $b^2-4ac$, positive in each. So I think your argument falls. $\endgroup$ – Lubin Nov 29 '14 at 18:20
  • $\begingroup$ @Lubin, you have a point there, yet that wasn't precisely my point but the following: as we already know the polynomial has no rational roots, the only non-trivial factorization left is in two quadratics. Since the above one is the factorization in two quadratics over the reals, if there were such a factorization over the rationals we'd get a contradiction to UFD. Now, even considering your objection, the OP can easily calculate the roots of each quadratic factor above and do with them whatever he please. Thank you. $\endgroup$ – Timbuc Nov 29 '14 at 18:23
  • $\begingroup$ It's kind of fascinating that the quadratic factors for a minimal polynomial of an irrational of the form $\sqrt p+\sqrt q$ has coefficients, in this case, that involve only one of the square roots. Until I checked it, I was sure those two $\sqrt5$'s were typos for $\sqrt{15}$! $\endgroup$ – Barry Cipra Nov 29 '14 at 19:21
  • $\begingroup$ @BarryCipra I didn't think of it before, but indeed interesting. $\endgroup$ – Timbuc Nov 29 '14 at 19:22
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    $\begingroup$ @Timbuc, oh, here's the explanation: You can also factor it as $$(x^2-2\sqrt3 x-2)(x^2+2\sqrt3 x-2)$$ $\endgroup$ – Barry Cipra Nov 29 '14 at 19:25
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$\mathbb Q(\sqrt 3 + \sqrt 5)/\mathbb Q$ is a finite separable normal extension whose Galois group consists of the four $\mathbb Q$-automorphisms $$ \sqrt 3 + \sqrt 5 \mapsto \pm\sqrt 3 \pm \sqrt 5. $$ Since there are precisely $4$ such $\mathbb Q$-automorphisms, by the Fundamental Theorem of Galois Theory, $[\mathbb Q(\sqrt 3 + \sqrt 5):\mathbb Q] = 4$, so the minimal polynomial has degree $4$.

The elements $\pm\sqrt 3 \pm \sqrt 5$ are all distinct and $\sqrt 3 + \sqrt 5$ is one root of the minimal polynomial, so the minimal polynomial is simply the product of the linear factors of Galois conjugates $$ [x-(\sqrt 3 + \sqrt 5)][x - (\sqrt 3 - \sqrt 5)][x - (-\sqrt 3 + \sqrt 5)][x - (-\sqrt 3 - \sqrt 5)]. $$ This is computed to be $x^4 - 16x^2 + 4$.

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