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This's about of exercise 6.8 in Atiyah. I think not: to take $A=k[x_1, \cdots, x_n, \cdots ]$ but I had have some problems for proof that $\text{Spec}(A)$ is Noetherian.

Tianlalu
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Tom Ryddle
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    A polynomial ring with infinitely many variables (like your $A$) isn't noetherian. So it's not surprising you're having trouble showing that $Spec(A)$ is noetherian. – Arthur Oct 11 '18 at 05:26
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    You're right. I want to show a counterexample. – Tom Ryddle Oct 11 '18 at 05:28
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    @Tom Ryddle- there can be a non-Noetherian ring $A$ such that $ \text{Spec}(A)$ is Noetherian. See this one-https://math.stackexchange.com/questions/7392/a-non-noetherian-ring-with-noetherian-spectrum?rq=1 – MAS Oct 11 '18 at 05:29
  • I see. I misunderstood the purpose of your post then. – Arthur Oct 11 '18 at 05:38

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