# Minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$

I had an example in the book given as follows:

Find the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ .

Solution: $~~~~~$$(\sqrt{2}+\sqrt{3})^2=5+2 \sqrt6 (\sqrt{2}+\sqrt{3})^4=49+20 \sqrt6 Then (\sqrt{2}+\sqrt{3})^4-10(\sqrt{2}+\sqrt{3})^2+1=0. Thus a=\sqrt{2}+\sqrt{3} satisfies f(x)=x^4-10x^2+1 over \mathbb Q. Let p(x) be the minimal polynomial of \sqrt{2}+\sqrt{3} over \mathbb Q.Then (\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3}) are also roots of p(x). So degree of p(x) is atleast 4. But f(a)=0 and f(x) \in \mathbb Q[x] \implies p(x) divides f(x) . So f(x) is minimal polynomial of \sqrt{2}+\sqrt{3}. But I can't get the step how are (\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3}) also roots of p(x). Kindly help with this. ## 3 Answers That fact comes from Galois Theory. Under the assumption that K/F is Galois, the (monic) minimal polynomial f of an element \alpha \in K (over F) is of the form$$f(x) = \prod_{\sigma \in \text{Gal}(K/F)} (x - \sigma \alpha).$$This implies that \sigma \alpha are all roots of the minimal polynomial for \alpha. In your case, let K = \mathbb{Q}(\sqrt{2}, \sqrt{3}) and F = \mathbb{Q}. Then K/F is Galois and there are 4 automorphisms in \text{Gal}(K/F): namely • identity on K • the one that sends \sqrt{2} \mapsto -\sqrt{2} keeping \sqrt{3} fixed: this automorphism sends \sqrt{2} + \sqrt{3} to -\sqrt{2} + \sqrt{3}; hence -\sqrt{2} + \sqrt{3} must be a root of p(x). • the one that keeps \sqrt{2} fixed and sends \sqrt{3} \mapsto -\sqrt{3}: like above, we deduce \sqrt{2} - \sqrt{3} is root of minimal polynomial p(x) • the one that negate both roots: finally - \sqrt{2} - \sqrt{3} must also be a root. • How do we know K/F Is Galois? How do we know these maps are automorphisms? Somewhere we need to use some facts about \sqrt{2} and \sqrt{3}. – 6005 Sep 3 '17 at 0:53 Good question: that is a big leap in the argument in my opinion. If we let K = \mathbb{Q}(\sqrt{2},\sqrt{3}), then \sqrt{2} + \sqrt{3}, \sqrt{2} - \sqrt{3}, -\sqrt{2} + \sqrt{3}, and -\sqrt{2} - \sqrt{3} all lie in K. In order to prove that they are all roots of p(x), the minimal polynomial of \sqrt{2} + \sqrt{3}, we follow An Hoa's idea: find automorphisms \sigma, \tau of K / F such that \sigma(\sqrt{2}) = -\sqrt{2}, \sigma(\sqrt{3}) = \sqrt{3}, \tau(\sqrt{2}) = \sqrt{2}, and \tau(\sqrt{3}) = -\sqrt{3}. The argument is then completed by noting that \sigma, \tau both preserve p(x) (since its coefficients are in K), so they preserve the roots of p(x). (If \alpha is a root of p(x), then \sigma(\alpha) is a root of \sigma(p(x)) = p(x) and \tau(\alpha) is a root of \tau(p(x)) = p(x).) Therefore what we need to show is that these automorphisms \sigma and \tau exist. K = \mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q} is the splitting field of (x^2 - 2)(x^2 - 3), so it is a Galois extension. To argue the extension has degree 4, further notice that it contains both \mathbb{Q}(\sqrt{2}) and \mathbb{Q}(\sqrt{3}). These two fields are distinct because one contains a square root of two and one doesn't (all elements of \mathbb{Q}(\sqrt{3}) can be written a + b\sqrt{3} and you can show that such a thing squared cannot be 2). So K has degree divisible by 2 and strictly greater than 2 over \mathbb{Q}, and it also has degree at most 4 since it is the splitting field of a degree-4 polynomial, so it has degree exactly 4. Now that we know K has degree 4 over \mathbb{Q} and is Galois, it must have four automorphisms (this is the definition of Galois, that the number of automorphisms equals the degree). These automorphisms permute the roots of (x^2 - 2)(x^2 - 3) and are defined by where they send these roots. They must send roots of (x^2 - 2) to roots of (x^2 - 2) and roots of (x^2 - 3) to roots of (x^2 - 3). But there are only 4 different ways to do this. Therefore, every possible way of sending \sqrt{2} \mapsto \pm \sqrt{2} and \sqrt{3} \mapsto \pm \sqrt{3} is an automorphism. In particular, \sigma and \tau are automorphisms. \square • But couldn’t OP have done it by noticing that the original computation, but replacing \sqrt2 with -\sqrt2, would give the same conclusion? similarly for replacing \sqrt3 with -\sqrt3; and for making both replacements at once? – Lubin Sep 4 '17 at 22:11 • @Lubin No. It is true that all four of those values are roots of (x^2 - 5)^2 - 24 = 0. But what you have to show is that this is actually the minimal polynomial of \sqrt{2} + \sqrt{3}. The minimal polynomial of \alpha is the smallest degree monic polynomial p such that p(\alpha) = 0. So just knowing that \alpha satisfies that particular polynomial doesn't show it's minimal. – 6005 Sep 4 '17 at 22:46 • You are quite right: I read too carelessly, and thought OP was referring to his f, not p. My apologies. – Lubin Sep 5 '17 at 0:30 Assume that m and n are square-free coprime integers. Then, if we prove that Q(\sqrt m +\sqrt n) = Q(\sqrt m, \sqrt n), we can conclude that that f(x) = x^4 -2(m+n)x +(m-n)^2 must be the minimal polynomial of \sqrt m +\sqrt n since Q(\sqrt m +\sqrt n) is a four-dimensional Q-vector space. But if \xi = (\sqrt m + \sqrt n), then {\frac {{\xi}^2-(m+n)} {2}}={\sqrt m}{\sqrt n}. Thus, ({\frac {{\xi}^2-(m+n)} {2}})\xi = m{\sqrt n}+n{\sqrt m}. Hence,$$({\frac {{\xi}^2-(m+n)} {2}})\xi -n{\xi} = (m-n){\sqrt n},$$and$$({\frac {{\xi}^2-(m+n)} {2}})\xi -m{\xi} = (n-m){\sqrt n}.$$Thus,$Q(\sqrt m +\sqrt n)$contains$\{\sqrt m, \sqrt n \}$, and so$Q(\sqrt m +\sqrt n)\supset Q(\sqrt m ,\sqrt n)$. The reverse set inclusion is immediate. The case$m =2$, and$n = 3$is of course what inspired this train of thought which is hopefully correct. • This is a proof that$x^4 - 10x^2 + 1$has those roots, not that the minimal polynomial has those roots. You have to prove that the minimal polynomial has those other roots. – 6005 Sep 3 '17 at 0:52 •$x^2-3$is the minimal polynomial of$\sqrt{3}$over$\mathbb{Q}(\sqrt{2})$thus$(x-\sqrt{2})^2-3$is the minimal polynomial of$\sqrt{3}+\sqrt{2}$over$\mathbb{Q}(\sqrt{2})$. Can you finish ? – reuns Sep 3 '17 at 1:53 • This is definitely taking the "low road" but$f(x) = x^4 -10x^2 +1$must be a multiple of the minimal polynomial of$\sqrt 2 +\sqrt 3$, and as$f(x)$factors into four distinct linear factors$(x-(\sqrt 2 +\sqrt 3))$,$(x+(\sqrt 2 +\sqrt 3))$,$(x-(\sqrt 2 -\sqrt 3))$, and$(x+(\sqrt 2 -\sqrt 3))$, and since all 6 combinations into products of 2 factors do not yield a factorization of$f(x)$as a product of two quadratics in$Q[x]$, we conclude that$f(x)$must be irreducible in$Q[x]$, and must therefore be the minimal polynomial of$\sqrt 2 +\sqrt 3\$. – student Sep 4 '17 at 2:26
• "reuns" Does the above edit make sense? – student Sep 4 '17 at 18:25