I have a field extension $\mathbb Q (2^{1/3}) = a + b2^{1/3} + c2^{2/3}$ where $a,b,c\in \mathbb Q$. I want an elementary proof it indeed is a field. How to go about proving it contains its inverse $(\forall x \in \mathbb Q(2^{1/3}))(x^{-1} \in \mathbb Q(2^{1/3}))$?
Thank you very much!
On
Let $K=\mathbb{Q}[\sqrt[3]{2}]$. Take $\alpha \in K$, $\alpha\ne0$ and consider $\phi: K \to K$ given by $\phi(x)=\alpha x$. Then $\phi$ is an injective $\mathbb{Q}$-linear transformation because $K\subset \mathbb R$, which has no zero divisors. Since $K$ is a finite-dimensional vector space over $\mathbb{Q}$, the injectivity of $\phi$ implies its surjectivity. In particular, $1$ is in the image of $\phi$, which proves that $\alpha$ is invertible in $K$.
On
I'm not sure what you mean by an elementary proof.
Note that $p(x)=x^3-2$ is irreducible over $\mathbb Q[x]$ (e.g. by Eisenstein).
Hence given any polynomial $q(x)$ either $q(x)=r(x)p(x)$ for some polynomial r(x), or we have $$q(x)r(x)+p(x)s(x)=1$$ for some polynomials $r(x)$ and $s(x)$ which can be determined using the Euclidean algorithm.
If we then substitute $x=\sqrt[3] 2$ we get $p(\sqrt[3] 2)=0$ and therefore $q(\sqrt[3] 2)r(\sqrt[3] 2)=1$
Note that in the first case $q(\sqrt[3] 2)=0$ and we are only looking for inverses for non-zero elements.
On
Your statement of the existence of inverse is incorrect, because the notation $x^{-1}$ already assumes the existence of an inverse in the field already. Rather it should be:
$\forall x \in \mathbb{Q}(2^{\frac{1}{3}}) \left( \exists y \in \mathbb{Q}(2^{\frac{1}{3}}) \left( xy=1 \right) \right)$
The simplest way is to show that this set corresponds to the set of polynomials over $\mathbb{Q}$ modulo the irreducible $(x^3-2)$, which gives a field. In a Galois theory course it is usually derived as a consequence of the theorem that the quotient ring of a unital commutative ring over a maximal ideal is a field. But without using any ring theory at all, it can be seen by observing that we can define $\gcd$ for polynomials over $\mathbb{Q}$ using the Euclidean algorithm, and so irreducibility automatically implies that any non-multiple has an inverse modulo it.
On
Here is another answer.
Let $K=\mathbb{Q}[\sqrt[3]{2}]$. Take $\alpha \in K$, $\alpha\ne0$. Because $K$ has dimension $3$ over $\mathbb{Q}$, the four numbers $1$, $\alpha$, $\alpha^2$, $\alpha^3$ must be linearly dependent and so there are $a_3$, $a_2$, $a_1$, $a_0$ in $\mathbb Q$, not all zero, such that $a_3 \alpha^3 + a_2 \alpha^2 + a_1 \alpha + a_0=0$.
If $a_0\ne0$, then $(-\frac{a_3}{a_0}\alpha^2-\frac{a_2}{a_0}\alpha-\frac{a_1}{a_0})\alpha=1$ and you have found the inverse of $\alpha$.
If $a_0=0$, then cancel one $\alpha$ and repeat.
Write the system of equations corresponding to $(a+br+cr^2)(x+yr+zr^2)=1$, where $r=\sqrt[3]{2}$: \begin{cases} ax+2cy+2bz=1\\ bx+ay+2cz=0\\ cx+by+az=0 \end{cases} The determinant of the system is $a^3+2b^3+4c^3-6abc$, which is non zero provided at least one of the coefficients is non zero.
How to see this? We can assume $a$, $b$ and $c$ are integers, because if we're able to find the inverse of $a+br+cr^2$ in this case, then also the inverse of $$ \frac{a}{d}+\frac{b}{d}r+\frac{c}{d}r^2 $$ can easily be found, multiplying by a rational number. We can also assume that $a$, $b$ and $c$ have no common prime factor, for a similar reason.
So, suppose $a^3+2b^3+4c^3=6abc$.
Note that $a$ must be even, so we can write $a=2A$; therefore $$ 8A^3+2b^3+4c^3=12Abc $$ and dividing by $2$ we get $4A^3+b^3+2c^3=6Abc$, which means $b$ is even; write $b=2B$ so that $$ 4A^3+8B^3+2c^3=12ABc $$ which, like before, has the consequence that $c$ is even. Contradiction.
By Cramer's theorem, the above system has a solution for every non zero number of the form $a+br+cr^2$.