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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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What's the relation between the primes of $B$ and the primes of $B/\mathcal{N}(B)$ where $\mathcal{N}(B)$ is the nilradical? –  Matt Sep 2 '12 at 15:48
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Manos, are you the one with the Hands of Fate? –  Asaf Karagila Sep 2 '12 at 15:49
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@AsafKaragila: Sorry, i don't see what you mean... –  Manos Sep 2 '12 at 17:23
    
@Matt: In your notation, the image of every prime ideal of $B$ is a prime ideal of $B/\mathcal{N}(B)$ and conversely... –  Manos Sep 2 '12 at 17:33
    
Isn't the example $\alpha=(x_1^2,x_2^2,\dots)$ easier? –  i. m. soloveichik Sep 2 '12 at 18:00
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1 Answer

up vote 2 down vote accepted

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 $P_i = (x_i, x_{i+1}, \ldots) \leq P$.

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