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I have seen an assertion that the "commutativity of addition" axiom in a field is redundant because it can be proved from the other axioms. There are relevant posts here and here.

However, both of these proofs use the two cancellation laws $$a+c=b+c\ \Rightarrow\ a=b\qquad\hbox{and}\qquad c+a=c+b\ \Rightarrow\ a=b\ .$$ In my source, cancellation is not specifically mentioned, and only the one-sided identity and negative axioms are assumed: $$a+0=a\qquad\hbox{and}\qquad a+(-a)=0\ .$$ Can commutativity of addition be proved with these axioms (and the other usual field axioms)? Or is it an oversight on the part of the author?

To be completely clear: it is not assumed that $0+a=a$, nor that $(-a)+a=0$.

The source is Howard Eves, Great Moments in Mathematics after 1650, chapter 28.

David
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Yes, these are sufficient assumptions. $0$ is also a left additive identity. Indeed, let $a\in F$, and denote $b=0+a$. Then:

$b+(-a)=(0+a)+(-a)=0+(a+(-a))=0+0=0=a+(-a) \ \ \ \ (*)$

The element $-a$ also has a right additive inverse. To simplify notation, denote $c=-(-a)$. Then adding $c$ to both sides of $(*)$ from the right gives $b=a$, as desired.

Similarly, it can be proved that $(-a)+a=0$ for any $a\in F$. Indeed, let $b=(-a)+a$. Then:

$b+(-a)=((-a)+a)+(-a)=(-a)+(a+(-a))=(-a)+0=-a=0+(-a)$

In the last step we use the fact that we already know that $0$ is a two sided additive identity. And again, adding $c=-(-a)$ to both sides from the right gives $b=0$, i.e $(-a)+a=0$.

Note that we never used multiplication here. This proof shows that the axioms of a group can be replaced by only having a left (or right) identity, and a left (respectively right) inverse for each element.

Mark
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