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.


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.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.