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Is $\mathbb{Q}(\sqrt[3]3)$ = $\{a+b\sqrt[3]3+c\sqrt[3]9 | a,b,c \in \mathbb{Q}\}$ a field?

My first attempt was to prove that $\{a+b\sqrt[3]3\}$ was a field. But, I ran across a problem where I am unable to find an inverse for every element.

I know that the inverse would have to be of the form $\frac{1}{a+b\sqrt[3]3}$. I then multiplied top and bottom by the conjugate and got $\frac{a-b\sqrt[3]3}{a^2-b^2\sqrt[3]9}$. But, this isn’t of the form $\{a+b\sqrt[3]3\}$. Had I found an inverse of the proper form, I would’ve used induction to prove that my initial question was a field.

mathnoob
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  • Yes, it is a field. – markvs Apr 11 '22 at 20:08
  • If $\alpha\in\mathbb{C}$ is algebraic over $\mathbb{Q}$ and its minimal polynomial has degree $n$ then ${b_0+b_1\alpha+b_2\alpha^2+...+b_{n-1}\alpha^{n-1}: b_0,...,b_{n-1}\in\mathbb{Q}}$ is a subfield of $\mathbb{C}$, denoted by $\mathbb{Q}(\alpha)$. The minimal polynomial of $\sqrt[3]{3}$ has degree $3$ (it is $x^3-3$) and so the set you described is a field. – Mark Apr 11 '22 at 20:12
  • for reciprocal, all you need is that $x+y\sqrt[3]3+z\sqrt[3]9$ is an explicit factor of rational-valued $x^3 +3y^3 +9z^3 - 9xyz,$ the quotient being an expression using $x^2, y^2, z^2, yz, zx, xy$ with coefficients involving your cube roots. – Will Jagy Apr 11 '22 at 20:12
  • Note: Let $v = \sqrt[3]{3}$, then $(a+bv)\left(a-bv+\frac{b^2}{a}v^2\right) = a^2 + 3\frac{b^3}{a} \in \mathbb{Q}$. ETA: Oops, I misinterpreted your question. – Brian Tung Apr 11 '22 at 20:15
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    ${a+b\sqrt[3]{3}\mid a,b\in\mathbb{Q}}$ is not even closed under mutliplication: $(\sqrt[3]{3})^2 = \sqrt[3]{9}$ cannot be expressed as $a+b\sqrt[3]{3}$ with $a$ and $b$ rationals. – Arturo Magidin Apr 11 '22 at 21:20

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You may have seen $$ x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - yz - zx - xy) $$

We are going to introduce $ d^3 = 3$ then make changes to the identity using $ x \mapsto x, y \mapsto d y, z \mapsto d^2 z$ to arrive at $$ x^3 + d^3y^3 + d^6z^3 - 3d^3xyz = (x+dy+d^2z)(x^2 + d^2y^2 + d^4z^2 - d^3yz - d^2zx - dxy) $$ for $$ \frac{1}{x+dy+d^2z \; } \; = \; \frac{ \; x^2 + d^2y^2 + d^4z^2 - d^3yz - d^2zx - dxy \; }{x^3 + d^3y^3 + d^6z^3 - 3d^3xyz} $$

You need $d^3 = 3$

Will Jagy
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