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Apr 29, 2014 at 8:10 history edited Kevin Ventullo CC BY-SA 3.0
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Apr 28, 2014 at 12:48 comment added Eric Wofsey @SJR: The maximal ideals in that ring are naturally in bijection with ultrafilters on $\mathbb{N}$; your special case of Kevin's construction is the same as the ultrafilter construction.
Apr 28, 2014 at 12:20 comment added Sidney Raffer Nice. Alternatively, we can think of the last displayed item as a ring $R$, every element of which satisfies the equation $x^p=x$. So if $I$ is any maximal ideal of $R$, then the elements of the field $R/I$ all satisfy the same equation, whence $R/I$ is the $p$-element field.
Apr 28, 2014 at 12:04 comment added Eric Wofsey Note that this is a bit weaker than the construction with ultrafilters, because that construction actually can be lifted to a homomorphism $\prod_{n=1}^\infty \mathbb{Z} / \bigoplus_{n=1}^\infty \mathbb{Z}\to\widehat{\mathbb{Z}}$.
Apr 28, 2014 at 7:50 history edited Kevin Ventullo CC BY-SA 3.0
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Apr 28, 2014 at 5:41 history answered Kevin Ventullo CC BY-SA 3.0