# polynomial cohomology

Hope this finds you all well.

I want to make sure of one thing : Do we usually have polynomial cohomology only in case the cohomology modules are free of rank 1 at most in each degree?

PS:I don't know who is the person who voted negatively for this question but it seems like some one rude is around

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No, certainly not. The product of two spaces with polynomial cohomology rings (I guess that's what you mean) again has a polynomial cohomology ring -- at least if the number of polynomial generators in each degree is finite. This should be discussed in Hatcher's book, for example. Think about an $n$-fold product $X = (\mathbb{CP}^{\infty})^n$, for example, which has $H^{\ast}(X) \cong \mathbb{Z}[x_1,\ldots,x_n]$ where the degree of each $x_{i}$ is two. –  t.b. May 21 '11 at 10:36
By the way: Googling for polynomial cohomology yields Hatcher's book (page 226) as one of the first hits, containing precisely the example I gave in the previous comment (and further discussion). –  t.b. May 21 '11 at 10:42
@EL Moro: th efact that googling for 'polynomial cohomology' takes you to Hatcher's excat discussion of this is a not a great indication that you put much effort in answering this yourself. If you keep asking questions like this, people will get annoyed. –  Mariano Suárez-Alvarez May 21 '11 at 13:04
@Mariano: I agree wholeheartedly with that. @El Moro: Honestly, I don't understand your question in the comment. To follow up on Mariano's comment and your reaction, let me quote what I found to be the essential part of Alex B's comment to one of my answers: "(...) as a general rule of thumb for the future (...), I recommend this: to display ignorance in mathematics is great. To display laziness is very bad." I'd guess (as Mariano did) the downvote is probably linked to this sound piece of advice. –  t.b. May 21 '11 at 13:46
@el moro: I understand that, surely, and I did not at all mean to suggest you stop asking questions, even if you think they might be silly. But nothing beats answering your questions yourself at being instructive! This does not mean you must come up with the answers magically or anything like that: I don't know Terry Tao and I've surely not seen him at work... but I bet he also googles for stuff all the time :) –  Mariano Suárez-Alvarez May 21 '11 at 16:05