To show that the real numbers $a+b\alpha+c\alpha^2$ with $a,b,c$ in $\mathbb{Q}$ form a field. (where $\alpha$ is the real cube root of 2)

Question

If we define the law of composition for multiplication as follows:

$(a+b\alpha+c\alpha^2)\times (x+y\alpha+z\alpha^2)=(ax+by\alpha+cz\alpha^2)$

And use addition of real numbers as the law of composition of addition then it is easy to verify it forms a field.

But if we use the law of composition of multiplication as real multiplication we have:

$(a+b\alpha+c\alpha^2)\times (x+y\alpha+z\alpha^2)=((ax+2bz+2cy)+(bx+ay+2cz)\alpha+(cx+by+az)\alpha^2)$

But with this composition, I am unable to find a multiplicative inverse for a given element.

ie I am unable to show the following:For $(a,b,c) \in \mathbb{Q}^3\setminus\{0\}$, there exists a $(3\times 1)$-matrix $\textbf{X}$, such that, if $$ \textbf{M}=\begin{pmatrix} a&2c&2b\\\ b&a&2c\\\ c&b&a\end{pmatrix},$$

then $\textbf{MX}=\begin{pmatrix} \ 1\\\ 0\\\ 0\end{pmatrix}.$

Answer 1

Your first assertion is patently false. If you define the product by $$(a+b\alpha+c\alpha^2)(r+s\alpha+t\alpha^2) = ar + bs\alpha + ct\alpha^2$$ then your structure is most definitely, not a field; it’s isomorphic to $\mathbb{Q}^3$, and when you multiply, for example, $\alpha$ by $\alpha^2$, you get $0$ even though neither factor is $0$. That cannot happen in a field.

If you do the multiplication correctly, as $$(a+b\alpha+c\alpha^2)(r+s\alpha+t\alpha^2) = (ar+2bt+2cs) + (as+br + 2ct)\alpha + (at+bs+cr)\alpha^2$$ then you do get a field.

To find the inverse of $a+b\alpha+c\alpha^2$, note that $a+bx+cx^2$ is relatively prime to $x^3-2$. So find polynomials $p(x)$ and $q(x)$ such that $$1 = p(x)(x^3-2) + q(x)(a+bx+cx^2).$$ What happens when you evaluate at $\alpha$?

Answer 2

If you do the multiplication in the "usual" way, then you get $\Bbb Q(\alpha)\cong\Bbb Q[x]/(x^3-2)$, which is a field because it is an integral domain modulo an irreducible polynomial (which is hence an integral domain modulo a maximal ideal).