## Every semigroup is isomorphic to a multiplicative subsemigroup of some ring

Show that every semigroup is isomorphic to a multiplicative subsemigroup of some ring. Exhibit a semigroup which is not isomorphic to the (entire) multiplicative semigroup of any ring.

Let be a semigroup and let be a ring. Let . That is, is the set of all functions which vanish everywhere except for a finite subset of .

Given , we define and as follows.

In the definition of , if the summation is empty, then as usual.

We claim that is a ring.

- ( is associative) For all , we have . So .
- (A -identity exists) Define by for all . Now , so that . Similarly, .
- (Every element has a -inverse) Given , define by . Then . So . Similarly, . We will denote the additive inverse of by .
- ( is commutative) We have .
- ( is associative)

So .

- ( distributes over from the left)

So

- ( distributes over from the right) Similar to the proof for left distributivity.
- (If has a left identity, then has a left identity) If is a left identity, define by and if . Now . If , then , and if , then . (The pair is certainly in the index set of the summation.) So this is equal to . Hence .
- (If has a right identity, then has a right identity) Similar to the proof for left identity.

So is a ring. Rings of this type are called *semigroup rings*, and their elements are typically written as ‘polynomials’ in the elements of with coefficients from . Addition and multiplication are then carried out as usual for polynomials – multiply the semigroup elements (preserving order) with their coefficients, and then collect like terms.

Given , we define by if and 0 otherwise. We claim that the map given by is a semigroup homomorphism. Indeed, if and otherwise. Now . If , then , and if , then . So if and otherwise. Hence .

So every semigroup can be embedded in the multiplicative semigroup of a ring. However, not every semigroup is isomorphic to the full multiplicative semigroup of some ring, as we argue. Note that every multiplicative semigroup of a ring has a zero. So any semigroup without a zero cannot be the multiplicative semigroup of a ring.

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