# Minimal Polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}$

To find the above minimal polynomial, let $$x=\sqrt{2}+\sqrt{3}+\sqrt{5}$$ $$x^2=10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}$$ Subtracting 10 and squaring gives $$x^4-20x^2+100=4(31+2\sqrt{60}+2\sqrt{90}+2\sqrt{150})$$ $$x^4-20x^2+100=4(31+4\sqrt{15}+6\sqrt{10}+10\sqrt{6})$$ $$x^4-20x^2-24=40\sqrt{6}+24\sqrt{10}+16\sqrt{15}$$ $$x^4-20x^2-24=8(2\sqrt{6}+2\sqrt{10}+2\sqrt{15})+24\sqrt{6}+8\sqrt{10}$$ $$x^4-20x^2-24=8(x^2-10)+24\sqrt{6}+8\sqrt{10}$$ $$x^4-28x^2-104=24\sqrt{6}+8\sqrt{10}$$ Again, squaring both sides $$x^8-56x^6+576x^4+5428x^2+10816=4096+765\sqrt{6}$$ But if I square again, I will get a degree 16 polynomial. Mathematica says the minimal polynomial is degree 8, which would make sense since elements of $\mathbb{Q}[\sqrt{2},\sqrt{3},\sqrt{5}]$ look like $$a+b\sqrt{2}+c\sqrt{3}+d\sqrt{5}+e\sqrt{6}+f\sqrt{10}+g\sqrt{15}+h\sqrt{30}$$ Where am I making mistakes?

• Try a little galois theory: what do you get if you take all products of $x\pm\sqrt{2}\pm\sqrt{3}\pm\sqrt{5}$? Commented Jun 5, 2015 at 23:14
• you would get a degree 8 polynomial...I'm still working towards understanding Galois and self learning some Field Theory. I showed a colleague how do it for $\sqrt{2}+\sqrt{3}$ and then we tried this...but we got stuck. Commented Jun 5, 2015 at 23:20
• This example is a poster-child for not thinking that mathematics is formulas! :) Commented Jun 5, 2015 at 23:55
• There are many answers here which compute a polynomial which has $\sqrt{2}+\sqrt{3}+\sqrt{5}$ as a root, but it would be nice to see irreducibility proofs, too! Only then we know that it is the minimal polynomial. Commented Jun 7, 2015 at 12:37
• By the way, to find the minimal polynomial of, say, $\sqrt[3]2+\sqrt[3]3$, you need to deal with cube roots of unity… Specifically, it turns out to be $\prod(\zeta_1\sqrt[3]2+\zeta_2\sqrt[3]3)$, where the $\zeta$s range over the cube roots of unity. (This gives you a degree $9$ polynomial. Be careful, though — the similar-looking $\sqrt[3]2+\sqrt[3]4$ is only degree $3$! Why is it different?) Commented Jun 7, 2015 at 18:41

In general, if $p_1,\dotsc,p_n$ is any list of integers, then the polynomial $$f=\prod_{e_1,\dotsc,e_n \in \{\pm 1\}} (x+e_1 \sqrt{p_1}+\dotsc+e_n \sqrt{p_n})$$ has coefficients in $\mathbb{Z}$, which follows by induction from the observation

$p \in \mathbb{Z},\,g \in \mathbb{Z}[x] \Rightarrow g(\sqrt{p}) \cdot g(-\sqrt{p}) \in \mathbb{Z}[x]$.

You can prove this using the automorphism group of $\mathbb{Z}[\sqrt{p}]$. But you can also just verify it via some direct calculation, which has the advantage that you really see why the $\sqrt{p}$-terms vanish and that you may compute $f$ faster.

Clearly, $f$ is monic, has degree $2^n$, and has $\sqrt{p_1}+\dotsc+\sqrt{p_n}$ as a root. So this is quite elementary and for any fixed $n$, you can also compute $f$. What is not so easy to prove is that if $p_1,\dotsc,p_n$ are square-free and pairwise coprime integers, then $f$ is irreducible. Equivalently, the degree of $\sqrt{p_1}+\dotsc+\sqrt{p_n}$ equals $2^n$. You can find the proof here ("square-root-extension.pdf").

For the numbers $p_1,p_2,p_3 = 2,3,5$, we get $f = x^8 - 40 x^6 + 352 x^4 - 960 x^2 + 576$.

• In this case each of the 8 roots is clearly of degree 8 and this is enough to assure f is irreducible. Commented Jun 5, 2015 at 23:50
• So then the minimal polynomial is the same for all 8 sign permutations of the root above? This was insightful, thank you! Commented Jun 6, 2015 at 1:03
• @LuisGomezSanchez: This is not clear at all - it requires a proof! A priori, it is only clear that $\sqrt{p_1}+\dotsc+\sqrt{p_n}$ has degree $\leq 2^n$. For a proof of equality, see fed.matheplanet.com/mprender.php?stringid=18752154 (that's in german). I have already posted an english version of this somewhere on math.SE, but for some reason I cannot find it right now. Commented Jun 6, 2015 at 7:45
• I've added an english version. Commented Jun 7, 2015 at 12:30
• @Martin Branderburg: I told about $x=\pm\sqrt2\pm\sqrt3\pm\sqrt5$ in which are all the eight conjugates. Taking anyone of them,say $x=\sqrt2-\sqrt3-\sqrt5$ you get the minimal polynomial of degree eight.If f where reducible, say f(x)= g(x)h(x), the degree of some root would be of degree less than 8. Besides, because of the simplicity of Galois of quadratic extensions (just one non-trivial Q-automorphism) I think it goes also easily for $\sqrt{p_1}+\dotsc+\sqrt{p_n}$. Your answer deserves the distinction of the best one. Best regards. Commented Jun 7, 2015 at 15:53

The 'easy' way to get at the minimal polynomial in this case is to take the product of all the terms $x\pm\sqrt2\pm\sqrt3\pm\sqrt5$; this is, in essence, because the Galois group over $\mathbb{Q}$ in this case is just $(\mathbb{Z}/2\mathbb{Z})^3$, with each of the three copies of $\mathbb{Z}/2\mathbb{Z}$ corresponding to a sign change on one of the three square root terms. The simple approach (take the product of $(x-\alpha)$ over all of the possible $\alpha$ obtained by applying the automorphisms in the Galois group to $\sqrt2+\sqrt3+\sqrt5$) works here because $\sqrt2$, $\sqrt3$ and $\sqrt5$ are 'mutually irreducible' over $\mathbb{Q}$ (in the sense that $\sqrt2\not\in\mathbb{Q}(\sqrt{3}, \sqrt{5})$, etc.) , so we can compose extensions in a clean fashion.

• Linear independence does not suffice. And of course, that the Galois group is $(\mathbb{Z}/2)^3$, requires a proof. (It is rather trivial that it embeds into this group.) Commented Jun 5, 2015 at 23:39
• I didn't mean linearly independent; I've swapped that terminology out for 'mutually irreducible', which I think gets more cleanly to the heart of things. Commented Jun 5, 2015 at 23:49

I started out by observing that $2+3 = 5$; I thought that might make it easier. Then

$$x-\sqrt{5} = \sqrt{2}+\sqrt{3}$$ $$x^2-2\sqrt{5}x+5 = 2\sqrt{6}+5$$ $$x^2-2\sqrt{5}x = 2\sqrt{6}$$ $$x^4-4\sqrt{5}x^3+20x^2 = 24$$ $$x^4+20x^2-24 = 4\sqrt{5}x^3$$ $$x^8+40x^6+352x^4-960x^2+576 = 80x^6$$ $$x^8-40x^6+352x^4-960x^2+576 = 0$$

(Or, of course, you can do it more systematically via the Galois theory!)

• Oh, that's clever. Commented Jun 7, 2015 at 1:17

I thought I'd get rid of the surds one-by-one: first get rid of $\sqrt2$, then $\sqrt3$, and finally $\sqrt5$. To get rid of a surd, I would get everything with that surd in it to one side, and then square. (Note that $\sqrt{15}$ has a $\sqrt3$ in it.)

\begin{align} x&=\sqrt2+\sqrt3+\sqrt5\\ x-\sqrt3-\sqrt5&=\sqrt2\\ (x-\sqrt3-\sqrt5)^2&=2\\ x^2+(-2\sqrt5-2\sqrt3)x+(2\sqrt{15}+8)&=2\\ x^2+(-2\sqrt5-2\sqrt3)x+(2\sqrt{15}+6)&=0\\ x^2-2\sqrt5x+6&=2\sqrt3x-2\sqrt{15}\\ (x^2-2\sqrt5x+6)^2&=(2\sqrt3x-2\sqrt{15})^2\\ x^4-4\sqrt5x^3+32x^2-24\sqrt5x+36&=12x^2-24\sqrt5x+60\\ x^4-4\sqrt5x^3+20x^2-24&=0\\ x^4+20x^2-24&=4\sqrt5x^3\\ (x^4+20x^2-24)^2&=80x^6\\ x^8+40x^6+352x^4-960x^2+576&=80x^6\\ x^8-40x^6+352x^4-960x^2+576&=0 \end{align} This, of course, is equal to the Galois theory answer of $\displaystyle\prod(x\pm\sqrt2\pm\sqrt3\pm\sqrt5)$, where you multiply every possible choice of signs together.

• I recognize that the Galois theory way is much better. Commented Jun 5, 2015 at 23:56

An elementary way

$(x-\sqrt2)^2=(\sqrt3+\sqrt5)^2 \implies x^2+2-2x\sqrt2=8+2\sqrt{15}\implies(x^2-6-2x\sqrt2)^2=((x^2-6)^2-4x(x^2-6)\sqrt2+8x^2)^2=60\implies((x^2-6)^2+8x^2-60)^2=(4x(x^2-6)\sqrt2)^2$

This resulting polynomial of $\mathbb{Q}[x]$ is obviously of degree 8