I'll rewrite the question here: Prove that if $f:R \rightarrow S$ is a semiring homomorphism and $R$ is a ring, then $S$ is a ring as well.
So here are some quick definitions since the use of "ring" is not universal yet. In my class we define "semiring" as a set $R$ with two binary operations where (1) $(R,*_0)$ is a commutative monoid with identity, $e_0$, (2) $(R,*_1)$ (the second operation) requires the identity element, $e_1$, and (3) the second operation is distributable over the first. Then a "ring" satisfies the above requirements and has the additional requirement that $(R,*_0)$ is a group.
Here is my attempt at the proof. The part I'm not sure about is how I know that $-f(a)$ exists in $(S,*_0)$. Do I get that automatically because $R$ is a ring? If so, I don't understand how.
Proof: Let $f: R \rightarrow S$ be a semiring homomorphism and $R$ be a ring. We want to see that (1) $(S, *_0)$ is an abelian group with identity $e_0$, (2) $(S, *_1)$ is a monoid with identity $e_1$, and (3) the operation $*_1$ distributes over $*_0$. We get (2) and (3) because $f$ is a semiring homomorphism. We want to show that (1) $(S, *_0)$ is a group from the fact that $(R,*_0)$ is a group. For convenience, we will denote $ *_0 $ using additive notation and $ *_1 $ using multiplicative notation.
At this point, we're saying that $f:R \rightarrow S$ is only a semiring homomorphism. From that, we know that for any $a,b \in R$, we have $f(a+b)=f(a)+f(b)$, $f(0)=0$, $f(ab)=f(a)f(b)$, and $f(1)=1$. We don't know yet if $S$ is a ring, but we're assuming $R$ is a ring. Since $R$ is a ring, we know that for any $a \in R$, we must have $-a \in R$ too. By definition, $a + (-a) = 0$. Because $f: R \rightarrow S$ is a homomorphism, $f(a + (-a))=f(a)+f(-a)$. So we have $f(a + (-a)) = f(0) = 0 =f(a)+f(-a)$. Then, $0=f(a)+f(-a)$ implies that $-f(a)=f(-a)$. This shows that $S$ contains inverses and is thus a group.
Therefore $S$ is also a ring.
Thanks for any help.
Forgive me if I use more common notation instead of $*_0$ and $*_1$.
The operations on $R$ and $S$ are denoted by addition and multiplication. In addition to the properties you list, I assume that $S$ also satisfies $0s=s0=0$.
Since $R$ is a ring, there is $-1$ so that $(-1)+1=0$. Set $n=f(-1)$. Then $$ n+1=f(-1)+f(1)=f((-1)+1)=f(0)=0 $$
Let $s\in S$; then $0=0s=(n+1)s=ns+s$, which yields the required negative for $s$.
Without the assumption $0s=s0=0$, I believe you need the assumption that $f$ is surjective (onto): in this case $$ s=f(r) $$ for some $r\in R$ and $s+f(-r)=f(r)+f(-r)=f(r-r)=f(0)=0$.