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If all of the below is true, does it follow directly from the definitions and the canonical way every abelian group can be considered a $\mathbb{Z}$-module?

EDIT: In the "definitions" below, it is necessary to include left and right distributivity; the two distributive axioms do not follow from the other structures. (For commutative rings technically one of the two will always follow from the other so it is only necessary to assume one -- for non-commutative rings one needs to assume both.)

Question: If we define:

  • ring: abelian group under $+$, semigroup under $\times$,

  • ring with identity: abelian group under $+$, monoid under $\times$,

  • commutative ring: abelian group under $+$, commutative semigroup under $\times$,

  • commutative ring with identity: abelian group under $+$, commutative monoid under $\times$,

then are the following equivalences true? (Yes/no will suffice for an answer.)

  • $R$ ring $\iff$ $R$ associative $\mathbb{Z}$-algebra
  • $R$ ring with identity $\iff$ $R$ unital, associative $\mathbb{Z}$-algebra
  • $R$ commutative ring $\iff$ $R$ commutative, associative $\mathbb{Z}$-algebra
  • $R$ commutative ring with identity $\iff$ $R$ unital, commutative, associative $\mathbb{Z}$-algebra

In particular, no ring is a non-associative $\mathbb{Z}$-algebra?

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    $\begingroup$ All rings have associative multiplication! $\endgroup$
    – Crostul
    Commented Dec 24, 2016 at 13:19
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    $\begingroup$ In the last item, you forgot "with identity". Also, the multiplication must distribute over addition in rings. Yes$^4$ then. $\endgroup$
    – Daniel Fischer
    Commented Dec 24, 2016 at 13:20
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    $\begingroup$ In addition to everything else that has already been said, there is noting wrong with defining nonassociative Z algebras (they're just not rings, is all) $\endgroup$
    – rschwieb
    Commented Dec 24, 2016 at 13:24
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    $\begingroup$ @DanielFischer Ah, ${\rm Yes}^4=$ Yes Yes Yes Yes. First I thought your comment got $4$ votes, on the wrong side. $\endgroup$
    – Dietrich Burde
    Commented Dec 24, 2016 at 14:23
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    $\begingroup$ IIRC an algebra has the structure of a vector space, so necessarily a field is involved and $\bb Z$ is not at field. $\endgroup$
    – Marc Bogaerts
    Commented Dec 25, 2016 at 5:33

1 Answer 1

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If $A$ is a commutative unital ring, then there are two equivalent definitions of an unital associative $A$-algebra:

(1) A unital ring $R$ with a unital ring homomorphism $f : A \rightarrow Z(R)$,

(2) An $A$-module $R$ together with an $A$-bilinear product that is associate and unital.

Given (1), then one defines the scalar multplication by $a \cdot r := f(a)r$, and given (2), one defines the ring homomorphism by $a \mapsto a \cdot 1_R$.

Therefore, there seems to be two candidate definitions for a (non-unital) associative algebra:

(1') A ring $R$ with a ring homomorphism $f : A \rightarrow Z(R)$,

(2') An $A$-module $R$ together with an associative $A$-bilinear product.

These are not equivalent, and option (2') is the standard definition.

If by associative algebra we mean (2') (the standard definition), then the answer is

Yes, Yes, Yes, Yes.

If by associative algebra we mean (1') (the non-standard definition), then the answer is

No, Yes, No, Yes.

For the "No"s:

Then need not be unique ring homomorphism $\mathbb{Z} \rightarrow Z(R)$:

Take any unital ring $R$ (commutative or not) and consider it as a ring. Then there is a unique unital ring homomorphism $\mathbb{Z} \rightarrow Z(R)$, but there can be other ring homomorphisms, such as $$ \mathbb{Z} \xrightarrow{0} Z(R). $$

In particular this shows that (1') and (2') are not equivalent notions.

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