# Minimal polynomial of $\sqrt[3]{7-\sqrt{2}}$

I want to find the minimal polynomial (over $\mathbb{Q}$) of: $k:=\sqrt[3]{7-\sqrt{2}}$.

With simple 'tricks' I got that: $P=(x^3-7)^2+2$ is a polynomial such that $P(k)=0$.

But I don't know if, or how to prove that $P$ is the minimal polynomial. How can I prove this/find the minimal polynomial ?

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$P$ is the minimal polynomial iff it's irreducible. –  Qiaochu Yuan Mar 18 '12 at 16:51
Since $P(k) = 0$, the minimal polynomial necessarily divides $P$. If $P$ is irreducible, it can only have one irreducible factor, which must be the minimal polynomial. Otherwise, the minimal polynomial is irreducible so it can't be $P$. –  Qiaochu Yuan Mar 18 '12 at 16:53
You want $(x^3-7)^2-2=x^6-14x^3+47$; you have the polynomial for $(7-\sqrt{-2})^\frac13$. –  bgins Mar 18 '12 at 17:05
... and, according to Maple, $(x^3-7)^2-2$ is indeed irreducible over $\mathbb Q$ –  Robert Israel Mar 18 '12 at 17:10
@bgins : Even intuitively, the polynomial might factor into two cubics or a quadratic and a quartic, so this is not sufficient as a proof. –  Patrick Da Silva Mar 18 '12 at 17:44

Any irreducible polynomial having $\rm\:k\:$ as a root necessarily has minimal degree. This proof is easy: the set $\rm\:S\:$ of polynomials $\rm\:f \in \mathbb Q[x]\:$ such that $\rm\:f(k) = 0\:$ is closed under addition, and under multiplication by any $\rm\:g\in \mathbb Q[x],\:$ hence it is closed under gcd, since, by Bezout, the gcd of $\rm\:f,g\:$ is a linear combination $\rm\:a\:f + b\:g,\:$ for some $\rm\: a,b\in \mathbb Q[x].\:$ Suppose irreducible $\rm\:f \in S.\:$ Now $\rm\:g\in S\:$ $\Rightarrow$ $\rm\:gcd(f,g)\in S.\:$ Since $\rm\:f\:$ is irreducible, $\rm\:gcd(f,g) = 1\:$ or $\rm\:f.\:$ But $\rm\:1\not\in S\:$ since $\rm\:k\:$ is not a root of $\:1,\:$ so $\rm\:gcd(f,g) = f,\:$ i.e. $\rm\:f\ |\ g.\:$ Since $\rm\:f\:$ divides every $\rm\:g\in S,\:$ we infer $\rm\:f\:$ has minimal degree in $\rm\:S.$ The structure at the heart of this proof will become clearer once you learn some ideal theory.

In your case $\rm\:f(x) = (x^3-7)^2-2 = x^6 - 14\:x^3 + 47\:$ is irreducible over $\rm\mathbb Q$ since it is irreducible mod $5$. Hence it is the minimal polynomial of $\rm\:k\:$ over $\mathbb Q$.

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Is it easy to see that it is irreducible $\pmod 5$? I don't see any non-calculatory way to do this, but of course it's possible to compute all possibilities.. I wondered if you had seen a trick to work $\pmod 5$ and find irreducibility quick. –  Patrick Da Silva Mar 18 '12 at 17:57
Indeed, ideal theory makes your first paragraph a lot easier, and sums up to "Since $\mathbb Q$ is a field, $\mathbb Q[x]$ is an Euclidean domain, hence a P.I.D., so that every ideal is principal. Since the set of polynomials that vanish at $\rm k$ form an ideal, it is principal and generated by some polynomial." –  Patrick Da Silva Mar 18 '12 at 18:08
Why is it irreducible mod 5 ? –  Belgi Mar 18 '12 at 18:44

The only point of this answer is to supplement Bill Dubuque's answer, and to give a short argument proving that $f(x)$ is irreducible in the ring $F_5[x]$.

Here $f(x)\equiv x^6+x^3+2$. Because $g(x)=x^2+x+2$ is irreducible, it is not a factor of $$x^8-1=(x^4-1)(x^4+1)=(x-1)(x-2)(x-3)(x-4)(x^2-2)(x^2+2)$$ in the ring $F_5[x]$. Let $\alpha\in F_{25}$ be a zero of $g(x)$. Then $\alpha^{24}=1$, but $\alpha^8\neq1$. Thus the order of $\alpha$ is a multiple of three ($\alpha$ is actually a generator of $F_{25}^*$ but we won't need that fact).

Next consider $f(x)=g(x^3)$. Let $\beta$ be a zero of $f(x)$ in some extension field of $F_5$. Then $\beta^3$ is a zero of $g(x)$, so by our earlier observation $\beta$ has order that is a multiple of nine. But $F_{5^6}$ is the smallest extension of $F_5$ that has elements of order nine, because $k=6$ is the smallest positive integer with the property $5^k\equiv 1\pmod 9$. Therefore $f(x)$ is the minimal polynomial of $\beta$, and the claim follows.

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Let $\alpha = \sqrt[3]{7 - \sqrt{2}}$ and $P(x) = x^6 - 14x^3 + 47$. Since $P(\alpha) = 0$, in order to show $P$ is irreducible it suffices to show $[Q(\alpha):Q] = 6$. Since $7 - \alpha^3 = \sqrt{2}$, you have that $\sqrt{2} \in Q(\alpha)$ and thus $$[Q(\alpha):Q] =[Q(\alpha):Q(\sqrt{2})][Q(\sqrt{2}):Q] = 2[Q(\alpha):Q(\sqrt{2})]$$ Thus our goal is to show that $[Q(\alpha):Q(\sqrt{2})] = 3$. Since $\alpha$ satisfies the third-degree polynomial equation $Q(x) = x^3 - 7 + \sqrt{2} = 0$ over $Q(\sqrt{2})$, our task is equivalent to showing $Q(x)$ is irreducible over $Q(\sqrt{2})$.

But if $Q(x)$ were not irreducible over $Q(\sqrt{2})$, it would have a linear factor. Thus it would have a root $a + b\sqrt{2}$, with $a$ and $b$ rational. But the roots of $Q(x) = 0$ are the numbers $\alpha$, $e^{2\pi i \over 3}\alpha$, and $e^{4\pi i \over 3}\alpha$. Of these, only $\alpha$ is real, so we have $$\alpha = a + b\sqrt{2}$$ Cubing, we get $$7 - \sqrt{2} = (a + b\sqrt{2})^3$$ Thus we must have $$7 + \sqrt{2} = (a - b\sqrt{2})^3$$ Multiplying the above two equations together we get $$47 = (a^2 - 2b^2)^3$$ Since $47$ is not the cube of a rational number, we have arrived at a contradiction and we are done.

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Really? And what prime divides both 14 and 47? –  Najib Idrissi Mar 18 '12 at 17:18
this doesn't make sense to me. The constant term becomes $47^6 - 14\cdot 47^3+47$. How does one apply Eisenstein? –  Harry Mar 18 '12 at 17:30
Ok what I originally wrote was wrong. My new answer (hopefully) should work. –  Zarrax Mar 18 '12 at 19:16
@Zarrax - can you explain please how did you get $7+\sqrt2$ ? –  Belgi Mar 28 '12 at 20:35