4 ms·
"there are no non-fields for which every nonzero element is a power of a fixed nonzero nonunit element". There are enough qualifiers here to go quite a bit fur
by Chinjut 12y ago
"there are no non-fields for which every nonzero element is a power of a fixed nonzero nonunit element".
There are enough qualifiers here to go quite a bit further: Not only are there no non-fields of this sort, but, in fact, there are no rings of this sort whatsoever: if x is our fixed nonzero, nonunit element, we must have x + 1 as some power of x (x + 1 being nonzero, as x is not -1, as x is not a unit). Furthermore, this power cannot be zero (as x + 1 does not equal x^0 = 1, as x is nonzero). Thus, we would have x + 1 = x^(m + 1) for some m. Which is to say, 1 = x * (x^m - 1). But this makes x a unit, contra stipulation. Thus, there is no such ring.
[Of course, another way of looking at this is that, dropping the "nonzero nonunit" stipulation, we prove that the generating element x is either zero or a unit, and thus the rings with this property are, as noted, necessarily fields (or the trivial ring in which 0 = 1)]
- tprice7 12y agoIt's seems a bit absurd to me to go further than non-fields, because fields have no non-zero non-units by definition. That is a nicer argument than the one I had come up with though, which used the observation that all such rings must be quotients of the polynomial ring Z[x].