On pp.2-3 of Curtis, Abstract Linear Algebra, he gives a definition of a field which seems to fail to exclude a pathological example. He says a field is a set k with two operations (a+b) and (ab) such that:
$k$ is an abelian group with +
$k - {0}$ (where 0 is the additive identity of the above group) is an abelian group over multiplication
for all a,b,c, $a(b + c) = ab + ac$.
It is easy to prove, using distributivity, that for any $a$, $a0 = 0$. But proving that $0a = 0$ seems to require knowing that $(b + c)a = ba + ca$. However, proving that this is the case seems to require knowing that $0a = 0$.
To be clear, if for nonzero $a, b$, $0a = b$, we get something quite pathological. $0b = 0(0a) = (00)a = 0a = b$, so $01 = 0(bb^{-1}) = (0b)b^{-1} = bb^{-1} = 1$. Now for any nonzero a, $a = a1 = a(01) = (a0)1 = 01 = 1$, so the only possible nonzero element is 1. But there's no contradiction here that I see.
Define:
0+0 = 0 = 1+1 0+1 = 1 = 1+0 00 = 0 = 10 11 = 1 = 01
Addition clearly gets you an abelian group. k - {0} is the trivial group, and distributivity is satisfied:
0(0+0) = 00 = 0 = 0 + 0 = 00 + 00 0(0+1) = 01 = 0 + 01 = 00 + 01 0(1+1) = 00 = 0 = 1 + 1 = 01 + 01 1(0+0) = 10 = 0 = 0 + 0 = 10 + 10 1(0+1) = 11 = 1 = 0 + 1 = 10 + 11 1(1+1) = 10 = 0 = 1 + 1 = 11 + 11
I see the same definition for field on p. 34 of Dummit and Foote, 3ed. So have I made a mistake? Or is this definition lacking an additional axiom to exclude out the above case:
- for any a, b, c, $(a + b)c = ac + bc$
EDIT, Will Jagy: Corrected in Errata for the third edition of Dummit and Foote, see ERRATA PDF