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Is $(\mathbb{Q} (\sqrt[3]{2}), +, \cdot)$ a field ? I came across this question while doing homework, but it was unlike any other problem I faced. For instance, how would one show that : $(a+b\sqrt[3]{2})\cdot(c+d \sqrt[3]{2})$ belongs to $\mathbb{Q} (\sqrt[3]{2})$.

Thank you

Martin Sleziak
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    What happened when you attempted to multiply out the expression $(a+b\sqrt[3]{2})\cdot(c+d \sqrt[3]{2})$? Did you get stuck or lost? – Lee Mosher Nov 18 '16 at 14:16
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    You should clarify whether by $\mathbb Q(\sqrt[3]2)$ you mean ${a+b\sqrt[3]2; a,b\in\mathbb Q}$ or ${a+b\sqrt[3]2+c\sqrt[3]{2^2}; a,b,c\in\mathbb Q}$. The former seems closer to your attempt, the latter is the usual definition. – Martin Sleziak Nov 18 '16 at 14:43
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    Some related posts: http://math.stackexchange.com/questions/533655/is-mathbbq%CE%B1-ab%CE%B1c%CE%B12-a-b-c-%E2%88%88-mathbbq-with-%CE%B1-sqrt32-a-fiel http://math.stackexchange.com/questions/294993/show-mathbbq-sqrt32-is-a-field-by-rationalizing http://math.stackexchange.com/questions/1298030/multiplicative-inverse-element-in-mathbbq-sqrt32 http://math.stackexchange.com/questions/1717974/prove-that-mathbb-q-sqrt32-is-a-field – Martin Sleziak Nov 18 '16 at 14:49

1 Answers1

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$\mathbb{Q}(\sqrt[3]{2})$ is by definition, the smallest field containing $\mathbb{Q}$ and $\sqrt[3]{2}$.

Thus, there is nothing to prove.

The fact that $\mathbb{Q}[\sqrt[3]{2}]$ is a field may need further proof. Note the difference between round bracket and square bracket!

yoyostein
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