Consider the ideal $\mathfrak{a} = \langle x^3 \rangle \subseteq \Bbbk[x]$. The ideal $\langle x + \mathfrak{a} \rangle$ is a prime ideal in $ \Bbbk[x]/\mathfrak{a}$. Then why is the affine algebra, $ \Bbbk[x]/\mathfrak{a}$ zerodimensional?
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3To have nonzero Krull dimension, you need to have a chain $\mathfrak{p}_0 \subsetneq \mathfrak{p}_1$ of distinct prime ideals. The prime ideal $\langle x + \mathfrak{a}\rangle$ is neither contained in nor contains any other prime ideals. – Rolf Hoyer Jun 17 '15 at 05:01

isn't $\langle 0 + \mathfrak{a}\rangle $ a strict subset of $\langle x + \mathfrak{a} \rangle$? – Zoey Jun 17 '15 at 05:31

3that is an ideal, but it's not prime, since your ring isn't an integral domain. – Rolf Hoyer Jun 17 '15 at 05:41

thank you! what happens when i quotient it with a prime ideal like $\langle x^2 + 1\mathfrak{a}\rangle$? Even then the quotient ring is zerodimensional. But the zero ideal is prime, right? – Zoey Jun 17 '15 at 06:22

Note that $\langle x^2+1\rangle$ may or may not be a prime ideal in $\Bbb k[x]$. It is if $\Bbb k = \Bbb R$, but not if $\Bbb k = \Bbb C$. If $x^2+1$ is in fact irreducible, then the given quotient will in fact be a field isomorphic to the field extension formally adjoining a square root of $1$. – Rolf Hoyer Jun 17 '15 at 06:31

If I consider the ring $\mathbb{R}[x,y]/\mathfrak{a}$ where $\mathfrak{a}=\langle x^2 + 1, y^2 + 1 \rangle$. Isn't this chain a valid chain of prime ideals? where am i getting it wrong? $\langle 0 + \mathfrak{a}\rangle \subsetneq \langle x + \mathfrak{a}\rangle \subsetneq \langle x + \mathfrak{a}, y + \mathfrak{a} \rangle$ – Zoey Jun 17 '15 at 06:47

In this case $(x+y)(xy) = x^2y^2 = x^2+1(y^2+1) \in \mathfrak{a}$. Your ring is isomorphic to $\Bbb C[y]/(y^2+1)$, which is not a domain as $y^2+1$ is reducible in $\Bbb C$. – Rolf Hoyer Jun 17 '15 at 06:54

thank you! but is there a general way of showing that the chain of prime ideals is always going to be zero? I have asked that general question at http://math.stackexchange.com/questions/1328532/howcanoneshowthattheascendingchainofprimeidealsintheaffinebbbk – Zoey Jun 17 '15 at 07:29

Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – Fly by Night Jul 22 '15 at 22:51
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The prime ideals of $k[x]/(x^3)$ are of the form $\mathfrak p/(x^3)$ where $\mathfrak p$ is a prime ideal in $k[x]$ containing $(x^3)$. But $k[x]$ is a principal ideal domain, so $\mathfrak p=(f)$ with $f\in k[x]$ irreducible. From $(x^3)\subseteq\mathfrak p$ we deduce $x\in\mathfrak p$, so $f\mid x$. It follows that $f=ax$ for some $a\in k$, $a\ne0$. Then $(f)=(x)$, and thus $(x)/(x^3)$ is the only prime ideal of $k[x]/(x^3)$.
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