Let $I = \langle x^2-3\rangle \subset \Bbb Q[x]$ and let $f=x-1 \in \Bbb Q[x]$. Show that $f+I$ is invertible in $\Bbb Q[x]/I$ and find the inverse.

Question

Let $I = \langle x^2-3\rangle \subset \Bbb Q[x]$ and let $f=x-1 \in \Bbb Q[x]$. Show that $f+I$ is invertible in $\Bbb Q[x]/I$ and find the inverse.

In $\Bbb Q[x]/I$ we have that $x^2-3=0$ so the identity is $x^2-2=1$. So in order to show that $f+I$ inverts I need to find $g+I$ such that $$(f+I)(g+I)=fg+I=1+I$$ that is I need to find a polynomial $g$ for which $(x-1)g(x)=x^2-2$.

I cannot figure this $g$ out. I don't think that $x^2-2$ has a nice factorization with $x-1$ as one of the factors even in $\Bbb Q[x]/I$?

Answer 1

In this ring, $x = \pm \sqrt 3$. If $x=\sqrt 3$, then we want $\dfrac{1}{\sqrt 3 -1}$. Rationalizing the denominator yields $\dfrac {\sqrt 3 + 1}{2}$, so try $g(x)=\dfrac {x+1}{2}$.

$$(x-1)\frac{x+1}{2}=\frac{x^2-1}{2}\equiv \frac{3-1}{2}\pmod{x^2-3}=1.$$

Answer 2

(Repeating comment here): You need to find some $g \in \mathbb{Q}[x]$ such that $(x-1) \, g(x) \equiv 1 \pmod{x^2 - 3}$, i.e. $$ (x-1) \, g(x) - 1 = (x^2 - 3) h(x) $$ for some $h \in \mathbb{Q}[x]$. You are implicitly assuming that $h = 1$, and that doesn't work.

If you rearrange this equation to the form $$ (x-1) \, g(x) - (x^2 - 3) h(x) = 1 $$ it looks like Bézout's identity (in $\mathbb{Q}[x]$ rather than in $\mathbb{Z}$). You can find such $g$ and $h$ if and only if $$ \gcd(x-1, x^2-3) = 1, $$ which you can probably see is true. Now, the Extended Euclidean Algorithm gets you those coefficient polynomials.