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Given $R=\mathbb{C}[x]/\langle x^n\rangle$, each element in $R$ may be represented as $a_0+a_1x+\cdots+a_{n-1}x^{n-1}$. I'm guessing that $P=\langle x^i\rangle$ for $i=\{1,\ldots,(n-1)\}$ represents all possible primes in $R$. Am I correct or are there aditional prime ideals that I'm missing.

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Those are not prime, unless $i=1$. In fact, $k[X]/(X^n)$ is a local ring with maximal nilpotent principal ideal $(\bar X)$, hence with finitely many ideals. Since the maximal ideal is nilpotent, it is the only prime ideal: $(\bar X)^n=0\subseteq \mathfrak p$ implies $(\bar X)\subseteq \mathfrak p$ implies $(\bar X)=\mathfrak p$ by maximality.

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  • $\begingroup$ I notice a bar above your $X$ variables, is that intentional to represent the complex conjugate of the variable? $\endgroup$ – Andrew Brick Feb 2 '15 at 8:12
  • $\begingroup$ @AndrewBrick No. I use it to denote the class of $X$ in the quotient. I thought it might have been misleading to denote the "$X$" in $k[X]/(X^n)$ which has $X^n=0$ by the same $X$ as in $k[X]$. $\endgroup$ – Pedro Tamaroff Feb 2 '15 at 8:52
  • $\begingroup$ Seeing this example, I thought I would get a more interesting result. Would swapping $\mathbb {C}$ with the reals $\mathbb{R}$ make a difference to the set of Prime ideals? $\endgroup$ – Andrew Brick Feb 2 '15 at 16:00
  • $\begingroup$ Also, with this question would you prove maximality of P by assuming an intermediary proper ideal exists between $R$ and $P$, and end up with a contradiction that this intermediate field must either be $R$ or $P$. What otger possible methods could I use? $\endgroup$ – Andrew Brick Feb 2 '15 at 16:03
  • $\begingroup$ @AndrewBrick You can show that the set of non units is an ideal. Every element of $k[X]/(X^n)$ can be written as $\mu+\bar Xp(\bar X)$ for $\mu$ in $k$. Since $\bar Xp(\bar X)$ is nilpotent, this is a unit whenever $\mu\neq 0$. Thus the set of non units is precisely $(\bar X)$. $\endgroup$ – Pedro Tamaroff Feb 2 '15 at 16:07

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