0

I'm studying analysis and notice a conflict when some authors write about them. Sometimes is uniqueness of the neutral elements are in the axioms, sometimes is a corollary of the axioms which talk about neutral elements of $ + $ and $*$.

Are the uniqueness part of axioms or not?

J. W. Tanner
  • 60,406
Emerson
  • 3
  • By uniqueness, you mean the axiom that $0\neq 1$? If the multiplicative identity is the same as the additive identity, we would get that $0=1$ and so $0 = 0\cdot x = 1\cdot x = x$ implying $x=0$ for all $x$ and so there is only one element in the field. Some authors will specifically avoid calling the field of one element a "field" in order to make phrasing theorems and other results easier so as to not have to continually write and put disclaimers like "except for the trivial field" on everything. – JMoravitz Sep 20 '23 at 15:15
  • Please list the texts you know of which A) include the uniqueness as axioms, and B) Derive the uniqueness from other axioms. – coffeemath Sep 20 '23 at 15:16
  • 1
    This really depends on the philosophy of the author. Some prefer that axioms be as minimalist as possible, including no more than the minimum necessary requirements to define a field, so they derive uniqueness of the identities as corollaries to the definition. Others probably want to avoid unnecessary arguments and include the uniqueness in the definition. I always preferred to derive uniqueness. It's an opportunity to see some general arguments in a concrete situation as shown in the answers. Uniqueness of negative and multiplicative inverse offer the same opportunity. – Chris Leary Sep 20 '23 at 15:25
  • One is free to include redundant axioms if one so desires (a common example is commutativity of addition in rings). Search on "redundant axiom" for many more examples. See here for some ways to discover such uniqueness proofs – Bill Dubuque Sep 20 '23 at 15:38

2 Answers2

1

Identity elements are unique for any binary operation:

Suppose $e,e'$ are both identity elements for our operation $*$. Then $e=e*e'=e'$.

Thus, there is no need for uniqueness to be a part of your axioms.

Adam Z
  • 230
0

Suppose $e_1$ and $e_2$ are both identity elements of $(S, \circ)$.

Then by the definition of identity element: $$\forall s \in S: s \circ e_1 = s = e_2 \circ s$$

Then: $$e_1 = e_2 \circ e_1 = e_2$$

So: $$e_1 = e_2$$ and there is only one identity element after all.

Prime Mover
  • 5,005