求证：若半环同态f:R→S的原像R是环，则像S必为环

Let's walk through this proof step by step, starting with clear restatements of the definitions we'll use to avoid confusion.

First, here's the semiring definition provided in your course:

半指环R配备两个二元运算，满足：
(1) $(R,_0)$是带有单位元$e_0$的交换幺半群；
(2) $(R,_1)$（第二个运算）存在单位元$e_1$；
(3) 第二个运算对第一个运算满足分配律，即对任意$a,b,c∈R$，有$a*_1(b*_0c)=(a*_1b)_0(a_1c)$和$(b*_0c)_1a=(b_1a)_0(c_1a)$。

For context, a ring (matching this semiring structure) requires:

• The first operation ($*_0$) forms an abelian group (a commutative monoid where every element has an inverse)
• The second operation ($*_1$) forms a monoid (associative with an identity element)
• The distributive laws between the two operations hold

We already know S is a semiring (semiring homomorphisms preserve semiring structure). The only gap we need to fill is proving that $(S, *_0)$ is an abelian group—all other ring conditions are already satisfied by S being a semiring.

Step 1: $(S, *_0)$ is a commutative monoid #

Since f is a semiring homomorphism, it carries over the commutative monoid properties from R to S:

• $f(e_0)$ acts as the identity element for $*_0$ in S (homomorphisms always map identity elements to identities)
• Commutativity holds: for any $s_1, s_2 ∈ S$, $s_1 *_0 s_2 = s_2 *_0 s_1$ (preserved directly from R's commutative structure)
• Associativity holds for $*_0$ (inherited from R via the homomorphism)

Step 2: Every element in S has a $*_0$-inverse #

Note: For this to hold, we assume f is surjective (otherwise the statement isn't true—for example, take R as the ring of integers, S as the semiring of non-negative integers, and f as the inclusion map: this is a semiring homomorphism, but S isn't a ring). This is a standard assumption for this type of proof.

Take any element $s ∈ S$. Since f is surjective, there exists some $r ∈ R$ where $f(r) = s$. Because R is a ring, r has an inverse $r'$ under the $*_0$ operation—meaning $r *_0 r' = e_0 = r' *_0 r$.

Using the homomorphism property (preservation of the $*_0$ operation):
$$f(r *_0 r') = f(r) *_0 f(r')$$
Since $r _0 r' = e_0$, the left side equals $f(e_0)$ (the identity for $_0$ in S). So:
$$f(e_0) = s *_0 f(r')$$
Similarly, $f(r' *_0 r) = f(r') _0 f(r) = f(e_0)$, so:
$$f(r') _0 s = f(e_0)$$
This means $f(r')$ is the inverse of s under $_0$. Since s was arbitrary, every element in S has an inverse for the $_0$ operation.

Step 3: Confirm the remaining ring conditions #

These are already covered by S being a semiring and the homomorphism:

• $(S, _1)$ has an identity element: $f(e_1)$ (preserved directly from R's $_1$ identity)
• Associativity for $*_1$: For any $s_1, s_2, s_3 ∈ S$, pick $r_1, r_2, r_3 ∈ R$ with $f(r_i) = s_i$. Then:
  $$s_1 *_1 (s_2 *_1 s_3) = f(r_1) *_1 f(r_2 *_1 r_3) = f(r_1 *_1 (r_2 *_1 r_3)) = f((r_1 *_1 r_2) *_1 r_3) = (s_1 *_1 s_2) _1 s_3$$
  This works because associativity holds in R's $_1$ operation, and homomorphisms preserve associativity.
• Distributive laws: These are part of the semiring definition, and the homomorphism ensures they carry over to S (we can verify this the same way we did associativity, using the homomorphism to pull operations back to R where we know distributivity holds).

Conclusion #

Since $(S, *_0)$ is an abelian group, $(S, *_1)$ is a monoid, and the distributive laws hold, S satisfies all the conditions of a ring.

内容的提问来源于stack exchange，提问作者daniel