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I just read a very short definition of a field where it was said that a field is a set of elements $K$ with two maps from the field into the field itself, such that

$K$ is an abelian group with $+$.

$K \backslash \{0\}$ is an abelian group with $\cdot$. Furthermore, we have the distributive law $x(y+z)=xy+xz$.

Is this already sufficient to show that $0x=0$?

If anything is unclear, please let me know.

Jyrki Lahtonen
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  • BTW,here's an even shorter definition of a field: A field F is a commutative division ring ( i.e. for every x,y in F, if x*y =0, then either x =0 or y=0 or both). – Mathemagician1234 Nov 02 '14 at 07:48
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    Just for the benefit of users who don't have enough rep to see comments on deleted answers: the issue here is that it's only assumed that $xy=yx$ when $x$ and $y$ are nonzero, so that one can't take the reversed distributive law $(y+z)x=yx+zx$ for granted if zero is involved. – Hans Lundmark Nov 02 '14 at 10:39
  • Doesn't a field really have that the multiplication is commutative? Then right and left distributivity imply each other. I think the definition given is not quite right, as it is missing this. – Dom Nov 02 '14 at 13:17
  • @Dom $K\setminus{0}$ is an abelian group. – egreg Nov 02 '14 at 15:45
  • @egreg I don't see how that would give you $0x=x0$ or right distributivity unless you specify that it is also a ring. – Dom Nov 04 '14 at 04:39

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You can show that $0=x-x=x(1-1)=x0$ which is the other way around than what you wanted. To get $0x=0$ I think you will need to add that multiplication is commutative (which I believe you need to properly define a field) or alternatively that multiplication is also right distibutive. Another (short) definition of a field would be a commutative ring so that the non-zero elements form a group under multiplication.

Dom
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