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Let $\alpha = 2^{1/3}$.

  1. Prove that the set $\mathbb{Q}[\alpha] = \{a+b\alpha+c\alpha^2 \mid a, b, c \in\mathbb{Q}\}$ is closed under addition and multiplication.

  2. Prove that if $z \in \mathbb{Q}[\alpha]$ is nonzero then there exists $z^{-1} $such that $z*z^{-1} = 1$

I have proved that $\mathbb{Q}[\alpha]$ is closed under addition, but I am having difficulty proving that it is closed under multiplication.

With all coefficients in $\mathbb{Q}$, let $a_1 + b_1\alpha + c_1\alpha^2 $ and $a_2 + b_2\alpha + c_2\alpha^2 \in \mathbb{Q}[\alpha]$. When multiplied, these two elements yield a fourth degree polynomial, and I do not know how to prove that it is an element of $\mathbb{Q}[\alpha]$

user26857
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kt046172
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    Then $\alpha^3 = 2$. So if you have $\alpha^4, \alpha^5, \cdots$ you can reduce to a polynomial of degree ${}\le 2$. – GEdgar Jan 23 '20 at 21:14

2 Answers2

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You should use the fact that $\alpha^3 = 2$. So then $$\begin{align*} (a_1+b_1\alpha + c_1\alpha^2)(a_2+b_2\alpha+c_2\alpha^2) &= (a_1a_2) + (a_1b_2+b_1a_2)\alpha +(a_1c_2+b_1b_2+c_1a_2)\alpha^2\\ &\qquad\mathop{+} (b_1c_2+c_1b_2)\alpha^3 + (c_1c_2)\alpha^4\\ &= (a_1a_2) + (a_1b_2+b_1a_2)\alpha + (a_1c_2+b_1b_2+c_1a_2)\alpha^2\\ &\qquad\mathop{+} (b_1c_2+c_1b_2)2 + (c_1c_2)\alpha^3\alpha\\ &= \Bigl( (a_1a_2)+2(b_1c_2+c_1b_2)\Bigr) + (a_1b_2+b_1a_2)\alpha\\ &\qquad\mathop{+} (a_1c_2+b_1b_2+c_1a_2)\alpha^2+ (c_1c_2)2\alpha\\ &\vdots \end{align*}$$ etc.

Arturo Magidin
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Hint $\,\alpha^3 \in V\! = \Bbb Q\langle1,\alpha,\alpha^2\rangle\Rightarrow\,\color{#c00}\alpha V\subseteq V\,\Rightarrow\, \color{#0a0}{\alpha^2}V\subseteq \color{#c00}\alpha V\subseteq V\,$ so $\,(a_1\! +\! b_1\color{#c00}\alpha\! +\! b_2 \color{#0a0}{\alpha^2})V\subseteq V$

To show $\,f(\alpha)\in V\,$ is invertible, note $\,x^3-2\,$ is irreducible over $\,\Bbb Q\,$ so $\,\gcd(f(x),x^3-2)=1\,$ since $\,\deg f \le 2,\,$ so we can invert by using the gcd Bezout identity - just like we invert in $\,\Bbb Z_n,\,$ i.e

$$ \begin{align} a(x) (x^3-2) +\ &b(x) f(x) = 1\\[.2em] \Rightarrow\ \bmod x^3\!-2\!:\ \ \ \ \ &b(x) f(x)\equiv 1\end{align}\qquad\ \ $$

which - when translated to our quotient ring - says $\, b(\alpha) f(\alpha) = 1\,$ in $\,\Bbb Q[\alpha]\cong \Bbb Q[x]/(x^3\!-2) $

Bill Dubuque
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