Determine the minimal polynomial of $\sqrt 3+\sqrt 5$

Question

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..

Answer 1

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.)

Answer 2

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.

Answer 3

$$\begin{align}\alpha &= \sqrt{3} + \sqrt{5} \Rightarrow \alpha^2 = 3 + 2\sqrt{3}\sqrt{5} + 5 \\&\Rightarrow (\alpha^2 - 8)^2 = 4\dot\ 15 \Rightarrow \alpha^4 - 16\alpha^2 +64 -60 = 0 \Rightarrow \alpha^4 - 16\alpha^2 +4 = 0 \end{align}$$

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 $\mathbb{Q}$.

Answer 4

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.

Answer 5

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?)

Answer 6

$\mathbb Q(\sqrt 3 + \sqrt 5)/\mathbb Q$ 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 $x^4 - 16x^2 + 4$.