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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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    $\begingroup$ A function space? I don't see any functions here. I just see (formal) polynomials. $\endgroup$ – Jyrki Lahtonen Jan 25 '17 at 13:48
  • $\begingroup$ For more discussion about the difference between a polynomial and a polynomial function see here, here and here. $\endgroup$ – Jyrki Lahtonen Jan 25 '17 at 13:58
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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, $$ 1 = \frac12(x^2 + 1) - \frac12(x-1)(x+1) $$

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  • $\begingroup$ @Ar Pet pick: watch the signs: you get $\;-1\;$ on the right side...+1 $\endgroup$ – DonAntonio Jan 25 '17 at 13:37
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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) $

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  • $\begingroup$ Well, we still could, in principle, have equality there: $\;(x^2+1)=(x+1,x^2+1)\;$ , couldn't we? (Of course not, but why?) $\endgroup$ – DonAntonio Jan 25 '17 at 13:46
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    $\begingroup$ if $x+1 \in (x^2+1)$ then $x^2+1|x+1$ that impossible in $\mathbb Q[x]$ $\endgroup$ – Mustafa Jan 25 '17 at 13:58
  • $\begingroup$ thanks @ Jyrki Lahtonen and @ DonAntonio $\endgroup$ – Mustafa Jan 25 '17 at 14:00
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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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