Prove multiplication in fields is commutative This is Problem $16$ from Halmos' Linear Algebra Problem Book.
The problem asks whether or not multiplication must be commutative in a field.
The solution uses the distributive properties $a(b+c)=ab+ac$ and $(a+b)c=ac+bc$ with $(0+1)x$ and $x(0+1)$ to show that both $0x=x0=0$.
From here it states that this implies that multiplication is both commutative and associative. I can't seem to grasp the underlying logic of this. How does this imply that $xy=yx$ for every $x$ and $y$ in the field?
Thanks for any help!
-Tusike
 A: From the pages preceeding the problem:

The author is intending to ask if all pairs multiplicatively commute, since we've been given that all nonzero products commute.
Thus the only thing left to prove is that $0$ commutes with everything.

As an aside, keep in mind that some older books, especially ones translated from French abstract algebra, use the word field to mean division ring (a "noncommutative field"). Nowadays fields are mostly assumed to be commutative rings, though.
A: Start here:
a*0 + a
= a*0 + a*1   (Because it's a group, it has a multiplicative identity element 1)
= a*(0 + 1)   (Using the distributive law)
= a*1         (0 + 1 = 1)
= a + 0     (0 is the additive identity element, which exists because it's a group)
So a*0 + a = a + 0   (Now add the inverses of a to both sides)
=> a*0 + a - a = a - a + 0  
=> a*0 = 0 (Because he element + inverse equals the additive inverse, which is 0; and then an element + the additive inverse, is the element)
Do the same thing, except use 0 * a first. You've proved multiplication with 0 is commutative, which is all that matters, because we know that multiplication with between every other element is commutative
