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Field is a set $F$ with two operations: $\cdot, +;$ stated as multiplication and addition respectively.
It is a commutative group under addition, and commutative group under multiplication for non-zero elements; and has the distributive property that links the two operations.

It has additive identity, also shown as: $0,$ and the multiplicative identity, also shown as: $1.$ Though the sense for the two identities, can be quite different than the usual algebraic meaning (like, say in $Z$) of $0, 1;$ based on the domain of the problem at hand.

Let the field $F$ have more than one elements, and let $a\in F$ be not an additive identity.

Then, the property $a\cdot 0 = 0$ seems to be not derived by any of the $11$ axioms (five axioms, each for the addition & multiplication operations, & one for the distributive property for the two operations).
If not, please show using which of the axioms of the field, is the given property (also stated as: 'multiplicative property of the additive identity') derived.

jiten
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  • Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed. – Shaun Sep 30 '24 at 12:19
  • See the linked dupe. More conceptually, as here, $0$ is the unique additive idempotent in an additive group. But $0x$ remains idempotent, being the image of $0$ under a group hom $,r\mapsto rx.\ \ $ Thus $,0x=0,$ by uniqueness. $\ $ Please search for answers before posting questions. $\ \ $ – Bill Dubuque Sep 30 '24 at 16:52
  • @BillDubuque Please note that it is a heavily edited question, based on multiple comments it recd. Though question has remained the same. At the time of framing, saw no link, that linked to it. Might be the earlier awkward framing caused so. But, then there seems no option during Edit, to see link(s) of similar earlier posts; which think must be added. Also, no comment (till edit) pointed to it being a duplicate; though there were many comments. Also, think that the best post, for this is: https://math.stackexchange.com/q/556593/424260. Though this awareness came after reading all comments. – jiten Oct 01 '24 at 04:45

1 Answers1

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The formula of your question follows from the following axioms of fields:

  1. Associativity of the sum: $\forall x,y,z\in F$, $(x+y)+z=x+(y+z)$.
  2. Existence of the additive identity: $\forall x\in F$, $0+x=x+0=x$.
  3. Existence of additive inverse: $\forall x\in F$, $\exists y\in F$ (denoted as $-x$) such that $x+(-x)=0$.
  4. Distributive property: $\forall x,y,z\in F$, $(x+y)\cdot z=x\cdot z+y\cdot z$ and $x\cdot(y+z)=x\cdot y +x\cdot z$.

Now, take any element $a\in F$. By Axiom 2 taking $x=0$ we obtain the equality $0+0=0$. Hence, $$a\cdot0=a\cdot(0+0).$$ By Axiom 4, we can distribute this sum as follows: $$a(0+0)=a\cdot0+a\cdot0.$$ Hence, we get that $a\cdot0=a\cdot0+a\cdot0$. By Axiom 3, there exists the additive inverse $-a\cdot 0\in F$ satisfying $a\cdot0+(-a\cdot 0)=0$. Adding up in both sides of the equiality, we get $$0=a\cdot0+(-a\cdot 0)=(a\cdot0+a\cdot0)+(-a\cdot 0).$$ Finally, by Axiom 1, we can move the parenthesis conveniently, so that $$(a\cdot0+a\cdot0)+(-a\cdot 0)=a\cdot0+(a\cdot0+(-a\cdot 0))=a\cdot0+0=a\cdot0.$$ This proves that $a\cdot0=0$ using only four of the eleven axioms of fields.

Remark: On the one hand, since we did not use any commutative axiom for the product, this proves that the equality is satisfied for non-commutative rings. On the other hand, the commutativity of the sum is neither useful for this proof. To emphasize this idea, and also as a fun fact, the commutative property of addition can also be derived from some of the other axioms (in particular, the existence of identity element for the product, together with the four axioms introduced in this answer).

Ikeroy
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