I need to show the uniqueness of a maximal ideal in the ring $S_{n}=\mathbb{C}[[z]]/z^{n}\mathbb{C}[[z]]$. The ideal in question is the ideal $I/I^{n}$ where $I^{n}=z^{n}\mathbb{C}[[z]]$. Now I know that $S_{n} \cong \mathbb{C}[z]/(z^n)$ as rings so I guess the question can be asked their in analogue and I know that $S_{n}$ is a PID. However I can't find a good description of the ideal $I/I^{n}$ or whether it is even an ideal. To prove it is maximal I guess I can determine this by finding its generator when considered in $\mathbb{C}[z]/(z^n)$ and prove it is irreducible as a polynomial, but how do I go about proving uniqueness?
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Do you know that $\mathbb C[z]$ and $\mathbb C[[z]]$ are PIDs? – user26857 Nov 12 '15 at 12:28
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They are because $\mathbb{C}$ is a field I guess, atleast that works for $\mathbb{C}[z]$, since it is a polynomial ring over a field and therefore Euclidean. For $\mathbb{C}[[z]]$ I don't know whether any general argument shows that it is a PID or not? Is it possible to prove without using that fact? – Bartuc Nov 12 '15 at 15:18
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For $\mathbb C[[x]]$ the argument is trivial: a power series writes $x^tg(x)$ with $g$ invertible, and a power series with $t$ minimal in an ideal $I$ is a generator of $I$. (I wouldn't continue to study such rings without knowing their basic properties.) – user26857 Nov 12 '15 at 15:58
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The most important thing to know about the ring of power series over a field ($F[[z]]$) is that its nonzero ideals are exactly of the form $(z^n)$ (counting the whole ring as $(z^0)$.) Consequently its ideals are all linearly ordered, and therefore it has a unique maximal ideal.
By the correspondence theorem for quotient rings, every possible quotient of a ring with a unique maximal ideal has a unique maximal ideal (the image of the original maximal ideal.)
So you see it is not all that important that the ideals are principal or even finitely generated since this is true of all local rings with unique maximal ideals (they are called local rings.)
The method you've chosen will work, too. In a PID, the ideals containing $(x)$ are principal ideals generated by divisors of x, and hopefully you know the divisors of $z^n$. By correspondence, again, your quotient is a ring with linearly ordered ideals, and so there is a unique maximal ideal.
and I know that $S_{n}$ is a PID
Er, it's not a PID since z is usually a zero divisor in it. But yeah, it is at least a principal ideal ring.
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I have never seen the isomorphism theorem like that before, but I guess it isn't all that hard to prove. The maximal ideal of $\mathbb{C}[[z]]$ should be of the form $(z^{n}$ for some $n$, since all ideals are of this form, but exactly what $n$ will do, since if the correponding ideal is to be maximal it must divide the exponent $m$ of every other ideal $(z^{m})$, so it should be $z^{1}$ right?, and that surely gets sent to $I/I^{n}$ by the isomorphism theorem, right? I guess this works only because $\mathbb{C}[[z]]$ is a PID, right? – Bartuc Nov 13 '15 at 09:09
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@Bartuc the proof that the ideals are linearly ordered can be based in the fact that it is a domain ( being principal is something you get simultaneously with the conclusion about the ideal structure.) the correspondence theorem is true for all rings and you will encounter it in any chapter introducing rings and ideals http://math.stackexchange.com/q/39370/29335 – rschwieb Nov 13 '15 at 11:04
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Funny, in the book I have used they never speak of correspondences like that, although it is rather simple as a follow up from the isomorphism theorem for rings. Thank you! – Bartuc Nov 14 '15 at 10:15