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 S be a semigroup and let R be a ring. Let R[S] = \{ r : S \rightarrow R\ |\ r_s = 0\ \mathrm{for\ all\ but\ finitely\ many}\ s \in S\}. That is, R[S] is the set of all functions S \rightarrow R which vanish everywhere except for a finite subset of S.

Given f,g \in R[S], we define f+g and f \cdot g as follows.

  1. (f+g)(s) = f(s) + g(s)
  2. (f \cdot g)(s) = \sum_{tu = s} f(t)g(u)

In the definition of \cdot, if the summation is empty, then (f \cdot g)(s) = 0 as usual.

We claim that (R[S], +, \cdot) is a ring.

  1. (+ is associative) For all s \in S, we have ((f+g)+h)(s) = (f+g)(s) + h(s) = f(s) + g(s) + h(s) = f(s) + (g+h)(s) = (f+(g+h))(s). So (f+g)+h = f+(g+h).
  2. (A +-identity exists) Define 0 : S \rightarrow R by 0(s) = 0_R for all s \in S. Now (f+0)(s) = f(s) + 0(s) = f(s) + 0 = f(s), so that f+0 = f. Similarly, 0 + f = f.
  3. (Every element has a +-inverse) Given f, define \overline{f} : S \rightarrow R by \overline{f}(s) = -f(s). Then (f + \overline{f})(s) = f(s) + \overline{f}(s) = f(s) - f(s) = 0 = 0(s). So f + \overline{f} = 0. Similarly, \overline{f} + f = 0. We will denote the additive inverse of f by -f.
  4. (+ is commutative) We have (f+g)(s) = f(s) + g(s) = g(s) + f(s) = (g+f)(s).
  5. (\cdot is associative)
    ((f \cdot g) \cdot h)(s)  =  \displaystyle\sum_{tu = s} (f \cdot g)(t) h(u)
     =  \displaystyle\sum_{tu = s} \left(\displaystyle\sum_{vw = t} f(v)g(w)\right) h(u)
     =  \displaystyle\sum_{tu = s} \displaystyle\sum_{vw = t} f(v)g(w)h(u)
     =  \displaystyle\sum_{vwu = s} f(v)g(w)h(u)
     =  \displaystyle\sum_{vt = s} \displaystyle\sum_{wu = t} f(v) g(w) h(u)
     =  \displaystyle\sum_{vt = s} f(v) \left( \displaystyle\sum_{wu = t} g(w) h(u) \right)
     =  \displaystyle\sum_{vt = s} f(v) (g \cdot h)(t)
     =  (f \cdot (g \cdot h))(s)

    So f \cdot (g \cdot h) = (f \cdot g) \cdot h.

  6. (\cdot distributes over + from the left)
    (f \cdot (g + h))(s)  =  \displaystyle\sum_{tu = s} f(t) (g+h)(u)
     =  \displaystyle\sum_{tu = s} f(t)(g(u) + h(u))
     =  \displaystyle\sum_{tu = s} \left( f(t) g(u) + f(t) h(u) \right)
     =  \displaystyle\sum_{tu = s} f(t) g(u) + \displaystyle\sum_{tu = s} f(t)h(u)
     =  (f \cdot g)(s) + (f + h)(s)
     =  ((f \cdot g) + (f \cdot h))(s)

    So f \cdot (g + h) = (f \cdot g) + (f \cdot h)

  7. (\cdot distributes over + from the right) Similar to the proof for left distributivity.
  8. (If S has a left identity, then R[S] has a left identity) If e \in S is a left identity, define \overline{e} : S \rightarrow R by \overline{e}(e) = 1 and \overline{e}(s) = 0 if s \neq e. Now (\overline{e} \cdot f)(s) = \sum_{tu = s} \overline{e}(t) f(u). If t \neq e, then \overline{e}(t) = 0, and if t = e, then u = s. (The pair (e,s) is certainly in the index set of the summation.) So this is equal to f(s). Hence \overline{e} \cdot f = f.
  9. (If S has a right identity, then R[S] has a right identity) Similar to the proof for left identity.

So R[S] is a ring. Rings of this type are called semigroup rings, and their elements are typically written as ‘polynomials’ in the elements of S with coefficients from R. 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 s \in S, we define \varphi_s : S \rightarrow R by \varphi_s(t) = 1 if t = s and 0 otherwise. We claim that the map \Phi : S \rightarrow R[S] given by s \mapsto \varphi_s is a semigroup homomorphism. Indeed, \Phi(st)(a) = \varphi_{st}(a) = 1 if a = st and 0 otherwise. Now (\varphi_s \cdot \varphi_t)(a) = \sum_{uv = a} \varphi_s(u) \varphi_t(v). If u \neq s, then \varphi_s(u) = 0, and if v \neq t, then \varphi_t(v) = 0. So (\varphi_s \cdot \varphi_t)(a) = 1 if a = st and 0 otherwise. Hence \Phi(st) = \Phi(s) \cdot \Phi(t).

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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