In $Q[x]/(x^2-2)$ find inverses of $[3x-2]$

Question

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In $\mathbb{Q}[x]/(x^2-2)$

• Find inverses of $[3x-2]$

Answer 1

You do to complicate:

$$(3x-2)(ax+b)=3ax^2+(3b-2a)x-2b$$ and thus, $$(3x-2)(ax+b)=k(x^2-2)+1\iff(3a-k)x^2+(3b-2a)x-2(k-b)-1=0\iff...$$

Answer 2

It's easier if you write $[3x-2]=3r-2$, where $r^2=2$. Then $$ \frac{1}{3r-2}=\frac{3r+2}{(3r-2)(3r+2)}=\frac{3r+2}{18-4}= \frac{3}{14}r+\frac{1}{7} $$

More generally, any element of your field can be written in a unique way as $a+br$, with rational $a$ and $b$, not both zero. Then \begin{align} (a+br)^{-1} &=((a+br)(a-br))^{-1}(a+br)\\[6px] &=(a^2-2b^2)^{-1}(a+br)\\[3px] &=\frac{a}{a^2-2b^2}a+\frac{b}{a^2-2b^2}r \end{align} and you know that $a^2-2b^2\ne0$ (why?).