I know that invertible elements of $\Bbb{Q}[x]$ are constants, so $\Bbb{Q}$. But in $\Bbb{Q}[x]/(x^{600})$, I suppose there are more invertible elements. How to find all of them?
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3Can you find the invertible elements of $\mathbb{Q}[X]/(X^2)$? I think that's much easier and should help. – Daniel Fischer Jul 14 '13 at 20:48
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HINT:
Show first that if $R$ is a commutative ring with identity with a unique maximal ideal $\frak m$ (such rings are called local) the invertible elements of $R$ are exactly the elements in $R\setminus\frak m$.
Now show that $\Bbb Q[X]/(X^N)$ has a unique maximal ideal for all $N\geq1$, namely the ideal generated by (the class of) $X$.

Andrea Mori
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does that mean that the maximal ideal is (x) so that invertible elements are precisely Q[x]/(x) which is isomorphic to Q? – stella Jul 14 '13 at 21:27
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@stella : no, $R\setminus\frak m$ is the difference set, the elements in $R$ not in $\frak m$, not the quotient $R/\frak m$. – Andrea Mori Jul 14 '13 at 22:36
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@AndeaMori, I just came back to this question, and I am not sure how to describe these invertible elements. Are they all the polynomials in Q[x] excluding classes of polynomials with zero constant term? – stella Jul 20 '13 at 21:53
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@stella : exactly. The unique maximal ideal is that generated by $X$. An element in the quotient is represented by a unique polynomial of degree $\leq N-1$ which is invertible if and only if its constant term is not $0$. – Andrea Mori Jul 21 '13 at 13:41
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