Show that the ideal $(x+1, x^2 + 1)$ in $\mathbb{Q}[x]$ is equal to $\mathbb{Q}[x]$.
I've never dealt with ideals in spaces of polynomials, I know the requirements for it to be an ideal, but I don't know how to apply these to $\mathbb{Q}[x]$?
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$\begingroup$This is easiest done by showing that $1$ is an element of the ideal, i.e. that it is a linear combination of the two generators. For instance, $$ 1 = \frac12(x^2 + 1) - \frac12(x-1)(x+1) $$
$\endgroup$ 1 $\begingroup$We have the ideal $(x^2+1)$ is maximal in $\mathbb Q[x]$ because $\mathbb Q[x]/(x^2+1) \cong \mathbb Q(i)$ and $(x^2+1)\subseteq(x+1,x^2+1)$ then by definition of maximal ideal we have $(x+1,x^2+1)=\mathbb Q[x]$ and $(x+1,x^2+1) \nsubseteq (x^2+1) $ because $x+1\notin (x^2+1) $
$\endgroup$ 3 $\begingroup$Hint: first show $x\in (x+1,x^2+1)$. Then use that if $x$ and $x+1$ are in the ideal, so is $x+1-x=1$.
Finally, use/prove that if an ideal contains a unit, it is the whole ring.
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