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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.

spectraa
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3 Answers3

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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.
An Hoa
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  • 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 $\sqrt{3}$. – Caleb Stanford Sep 03 '17 at 00:53
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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. $\square$

Caleb Stanford
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  • 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 04 '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. – Caleb Stanford Sep 04 '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 05 '17 at 00:30
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    Wouldn't the following argument work: it is clear that the computed polynomial $f$ kills $x = \sqrt{2} + \sqrt{3}$. Thus the minimal polynomial $p$ of $x$ must divide $f$. But factoring $f$ over $\mathbb{R}$ we see that that $f$ is irreducible over $\mathbb{Q}$. Thus $f = p$. – gen Mar 13 '22 at 16:04
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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.

student
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    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. – Caleb Stanford Sep 03 '17 at 00:52
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    $x^2-3$ 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 03 '17 at 01:53
  • 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 04 '17 at 02:26
  • "reuns" Does the above edit make sense? – student Sep 04 '17 at 18:25