Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have $a,b,c\in\mathbb{Q}$ not all zero. ($a^2+b^2+c^2\ne 0$), I want to show that the following determinant is then non-zero. I failed to arrive at an appropriate form of the polynomial. Help please. $$\left|\begin{bmatrix} a & 2c & 2b\\b & a & 2c\\ c & b & a\end{bmatrix}\right| = a^3+2 b^3-6 a b c+4 c^3$$


Second question, what is the easiest way to argue that $\{1,\sqrt[3]{2},(\sqrt[3]{2})^2\}$ is linearly independent in $\mathbb{Q}$?


Motivation:
Prove that $\mathbb{Q}[\sqrt[3]{2}] = \{a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2\;|\;a,b,c\in\mathbb{Q}\}$ forms a field.

Proof: Since $\mathbb{Q}[\sqrt[3]{2}] \subset \mathbb{R}$, we prove $\mathbb{Q}[\sqrt[3]{2}]$ is a subfield of $(\mathbb{R},+,\cdot)$

$\forall (a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2)\in \mathbb{Q}[\sqrt[3]{2}]\backslash\{0\}.$ We want to find $(a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2)$ such that

$(a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2)(d+e\sqrt[3]{2}+f(\sqrt[3]{2})^2) =$
$ (ad+2ec+2bf)+(ae+bd+2cf)\sqrt[3]{2}+(af+cd+be)(\sqrt[3]{2})^2 = 1$

Since $\{1,\sqrt[3]{2},(\sqrt[3]{2})^2\}$ is linearly independent (?) over $\mathbb{Q}$, we show there is unique solution to:
$$\begin{bmatrix} a & 2c & 2b\\b & a & 2c\\ c & b & a\end{bmatrix} \cdot \left[\begin{array}{l l} d\\e\\f \end{array}\right] = \left[\begin{array}{l l} 1\\0\\0 \end{array}\right] $$ Which is equivalent in showing the determinant is non-zero $$\left|\begin{bmatrix} a & 2c & 2b\\b & a & 2c\\ c & b & a\end{bmatrix}\right| = a^3+2 b^3-6 a b c+4 c^3=(?)$$

By subfield test, 1)2)3)4) is enough to say that $(\mathbb{Q}[\sqrt[3]{2}],+,\cdot)$ is a subfield of $(\mathbb{R},+,\cdot)$ therefore a field.

EDIT: If you have shorter way that prove the proposition without touching my 2 questions, that is even better.

share|improve this question
2  
Your two questions are quite independent. Why not ask two questions? –  Mariano Suárez-Alvarez Mar 11 '13 at 22:54
    
@MarianoSuárez-Alvarez because they are inherently from the same problem where I need to show $\{a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2\;|\;a,b,c\in\mathbb{Q}\}$ forms a field. –  mez Mar 11 '13 at 22:56
    
If you just need to show it forms a field you could just verify its axioms. –  user32847 Mar 11 '13 at 23:05
    
If they are connected in some way, tell us how. Why make people guess? –  Mariano Suárez-Alvarez Mar 11 '13 at 23:07
1  
@Alex Sure, but nevertheless these two things will still need to be shown along the way of verifying axioms, namely in the existence of multiplicative inverse. –  mez Mar 11 '13 at 23:21

5 Answers 5

up vote 4 down vote accepted

I'm not sure that's the best way to prove that $\mathbb{Q}[\sqrt[3]{2}] = \{a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2\;|\;a,b,c\in\mathbb{Q}\}$ is a field.

I would argue instead that if $a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2 \ne 0$, then the polynomial $f(x) = a+bx+c x^2$ is coprime to $x^{3} - 2$, so there are polynomials $u(x), v(x)$ such that $$ 1 = f(x) u(x) + (x^{3} - 2) v(x), $$ so that evaluating for $x = \sqrt[3]{2}$, $$ 1 = f(\sqrt[3]{2}) u(\sqrt[3]{2}) = (a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2) \cdot u(\sqrt[3]{2}), $$ and $u(\sqrt[3]{2}) \in \mathbb{Q}[\sqrt[3]{2}]$ is the required inverse.

share|improve this answer
    
That is short and nice. –  mez Mar 12 '13 at 18:42
    
What is the name of the theorem about $1 = px+qy$ in the context of UFD? –  mez Mar 12 '13 at 21:50
1  
@mezhang, in the context of an Euclidean domain like $\Bbb{Q}[x]$, it's usually called Bezout's Lemma. For UFD it does not hold in general, think of $2$ and $x$ in $\Bbb{Z}[x]$, their $\gcd$ is $1$, but you cannot write it as a linear combination of them. –  Andreas Caranti Mar 12 '13 at 21:55

Since $a,\ b$ and $c$ are rational, we may clear denominators in $$a^3 + 2b^3 -6abc +4c^3 = 0$$ The above equation is a homogenous equation of degree $3$ so we may cancel common factors. If there exists non-trivial solutions to the equation, we may therefore assume without loss of generality that $a,\ b$ and $c$ are integers with $\gcd(a,\ b,\ c)=1$.

Reducing modulo $2$, we find that $a\equiv 0\pmod 2$. Let $a=2\alpha$. Making the substitution and cancelling common factors, we arrive at $$4\alpha^3 +b^3 - 6\alpha bc + 2c^3 = 0$$ Reducing mod $2$ again, we get $b\equiv 0\pmod2$. So let $b=2\beta$ to obtain $$2\alpha^3 + 4\beta^3 - 6\alpha\beta c + c^3 = 0$$ Reducing modulo $2$ one last time gives $c\equiv 0\pmod 2$. This contradicts the fact that $\gcd(a,\ b,\ c)=1$. Therefore there are no non-trivial integer solutions to the above equation. It follows that the determinant is non-zero since $a,\ b$ and $c$ are not all zero.

To show the linear independence of $\left\{1,\ \sqrt[3]{2},\ \left(\sqrt[3]{2}\right)^2\right\}$ in $\mathbb{Q}$, suppose to the contrary that there exists some non-trivial rational linear combination such that $$r_0 + r_1\sqrt[3]{2} + r_2\left(\sqrt[3]{2}\right)^2 = 0$$ Then clearing denominators, there exists a non-trivial integral linear combination of the above set to $0$. Specifically, there exists an integral polynomial $p(x)$ of degree $2$ such that $\sqrt[3]{2}$ is a root. But the minimal polynomial of $\sqrt[3]{2}$ is $x^3 - 2$. This is a contradiction.

share|improve this answer

Edited in accordance with comment from Marc van Leeuwen:

Suppose $$a+b\root3\of2+c(\root3\of2)^2=0$$ with $a,b,c$ rational. Then $\root3\of2$ is a root of the polynomial $$f(x)=a+bx+cx^2$$ Now, $\root3\of2$ is also a root of $$g(x)=x^3-2$$ So $\root3\of2$ is a root of the gcd of $f$ and $g$. But $g$ is irreducible over the rationals, and $f$ has degree smaller than $g$ has, so $f$ is identically zero or the gcd is a nonzero constant. It isn't a nonzero constant, since it has to vanish at $\root3\of2$, so $f$ is the zero polynomial, so $$a=b=c=0$$ so $$\{{\,1,\root3\of2,(\root3\of2)^2\,\}}$$ is a linearly independent set over the rationals.

share|improve this answer
2  
Note however that since $f$ turned out to be $0$, one has $\gcd(f,g)=\gcd(0,x^3-2)=x^3-2$, it is not a constant after all! ($0$ is not a divisor of $x^3-2$ at all, so if you define $\gcd(0,x^3-2)$ at all, it has to be $x^3-2$.) The argument used that the degree of a $\gcd$ cannot exceed the degree of its operands is invalid if one of the operands in $0$. Of course the argument you give is basically valid, but you should not formulate it quite like you have done. –  Marc van Leeuwen Mar 12 '13 at 14:33

Let $r=\root3\of2$. The other answerers have shown that $1,r,r^2$ are linearly independent over $\mathbb{Q}$. Eu Yu's answer has also nicely shown that the determinant in question is nonzero. I don't have a better answer than his. However, since this question is, after all, one about determinant, I can't resist the temptation to solve it in (guise of) a matrix theoretical way.

Let $\omega$ be a primitive cube root of unity. Then your matrix is $\mathbb{C}$-similar to $$ A=\begin{pmatrix} a &r^2\omega ^2c &r\omega b\\ r\omega b &a &r^2\omega ^2c\\ r^2\omega ^2c &r\omega b &a \end{pmatrix}. $$ This is a circulant matrix. So, its eigenvalues are (see wikipedia): $$ \begin{cases} \lambda_1 = a+r^2\omega^2c+r\omega b &= a+r^2\omega^2c+r\omega b,\\ \lambda_2 = a+r^2\omega^2c\,\omega +r\omega b\,\omega^2 &= a+r^2c+rb,\\ \lambda_3 = a+r^2\omega^2 c\,\omega^2 +r\omega b\,\omega &= a+r^2\omega c+r\omega^2 b. \end{cases} $$ Note that $\lambda_2\neq0$ because $1,r,r^2$ are linearly independent over $\mathbb{Q}$. It follows that if both $\lambda_1$ and $\lambda_3$ have nonzero imaginary parts, $\det A\neq0$. Yet, if one of them is real, by inspecting its imaginary part, we get $b=rc$. So $b=c=0$ and $\det A=a^3$ is still nonzero.

share|improve this answer
    
This is very nice. Thank you. I didn't accept that answer yet thought it was complete in hope to see something like this. –  mez Mar 12 '13 at 14:07

Just for the linear independence part over $\Bbb Q$. I suppose you know that $\alpha=\sqrt[3]2$, which is by definition the (positive) real root of $X^3-2$, is irrational. But then $X^3-2$ has no rational roots, and (being of degree $3$) is irreducible over $\Bbb Q$, which means it is the minimal polynomial over $\Bbb Q$ of any of its (complex) roots, since such a minimal polynomial has to divide $X^3-2$. In particular the minimal polynomial of $\alpha$ has degree $3$, which implies that $1,\alpha,\alpha^2$ are $\Bbb Q$-linearly independent.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.