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Let $K$ be a field and consider the ring $A=K[x_1,x_2,\cdots]$ in countably infinite indeterminates and its ideal $\alpha=(x_1,x_2^2,\cdots,x_n^n,\cdots)$. Then Atiya and MacDonald in their "Introduction to Commutative Algebra", top of page 91, mention that the only prime ideal of $A/ \alpha$ is the image of $(x_1,x_2,\cdots)$. I can see that the image of this ideal is prime, however how can we see that it is the only prime?

Thanks.

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    @AsafKaragila: Hahahaha, i see now...Good for me, i have nothing to do with this film :-)2012-09-03

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As Matt mentions in the comments:

Let $B=A/\alpha$ as in the question. If $x \in B$ satisfies $x^n = 0$ and $P$ is a prime ideal, then since $o \in P$, either $x$ or $x^{n-1}$ in $P$. By induction, $x \in P$. In particular each $x_i + \alpha \in B$ is contained in each prime ideal $P$. Hence the image $P$ of $(x_1,x_2,\ldots)$ is the unique minimal prime ideal. The quotient ring is isomorphic to the field $K$, so $P$ is also a maximal ideal. Hence it is the only prime ideal of $B$. $B$ is not artinian, since it contains the descending chain of ideals $P_i = (x_i, x_{i+1}, \ldots) \leq P$.