Prove that $\mathbb{Q}[x] / (x^{3} - 2)$ is a field and find the inverse of of $[x - 3]$

Question

I'm not sure how to prove the first part other than to go through all the axioms. As for the second, I know I need to find an element such that when it is multiplied to $[x - 3]$, it results in $[1]$

Any help would be much appreciated

Answer 1

Hint $\ $ Just like for integers, since $\,x^3-2\,$ is prime, by Bezout's GCD identity we have

$\qquad x^3\!-2\nmid f\,\Rightarrow\, 1 = \gcd(x^3\!-2,f) = (x^3\!-2)h + fg\ \ {\rm for\ some}\,\ g,h\in\Bbb Q[x]$

Said ${\rm mod}\ x^3-2\!:\,\ f\not\equiv 0\,\Rightarrow\, fg\equiv\,1 \Rightarrow\, g\equiv f^{-1},\,$ so $\,\Bbb Q[x]/(x^3-1)\,$ is a field.

Let $\, y = x\!-\!3,\,$ To calculate $\,y^{-1}$ we can use the extended Euclidean algorithm to get the above Bezout identity, e.g. here, or else change variables $\,x\to y\,$ in the minimal polynomial $\,x^3-2\,$ for $\,x\,$ to get a polynomial having $y$ as root, from which we can read off the inverse. Proceding this way

$$\,2 = x^3 = (y\!+\!3)^3 = y(y^2\!+\!9y\!+\!27) + 27 =: y f(y)\!+\!27\,$$ hence $\,yf(y) = -25\,$ so $\,y\,f(y)/(-25) = 1,\,$ so $\,y^{-1}= f(y)/(-25) = -f(x\!-\!3)/25.$

Answer 2

To find the inverse of $x-3$ use the euclidean algorithm to write $1$ as a polynomial linear combination of $x-3$ and $x^3-2$. Then apply the quotient map and the term with $x^3 -2$ disappears so you have found an inverse for $x-3$.

If you are familiar with maximal ideals then to prove the quotient is a field it suffices to prove $x^3 - 2$ generates a maximal ideal. This is true since $x^3 -2$ is irreducible by, say, Eisenstein's criterion.