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Sep 26, 2016 at 7:35 comment added Wilberd van der Kallen @Pierre MATSUMI.This is not the way it is done. If you agree, you should accept the answer.
Sep 25, 2016 at 12:17 comment added Pierre MATSUMI Dear Wilberd van der Kallen, thanks a lot. Pierre Matsumi
Sep 21, 2016 at 13:18 comment added Wilberd van der Kallen No, that is false. In my answer $g$ is irreducible, but $(g)$ is not prime. Could you rephrase your question about `the relation' ?
Sep 21, 2016 at 13:15 history edited Wilberd van der Kallen CC BY-SA 3.0
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Sep 21, 2016 at 10:11 comment added Pierre MATSUMI Because g is irreducible, this means (g) is prime. Then as you say, the answer is Yes, isn't it? What's the relation of your example of real formal power series ring? Pierre
Sep 21, 2016 at 8:08 history answered Wilberd van der Kallen CC BY-SA 3.0