# Prove that if $f:R\rightarrow S$ is a semiring homomorphism and $R$ is a ring, then $S$ is a ring as well.

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.

• Are you sure the homomorphism $f$ is not required to be onto? Feb 3 '18 at 14:46
• Oh I think it is required to be onto. That would be the reason why I know $-f(a)$ is in $(S,*_0)$. Feb 3 '18 at 14:50
• Without onto, $R=\Bbb Z$, $S=\Bbb Z\times \Bbb N_0$, $x\mapsto (x,0)$ would be a counterexample Feb 3 '18 at 14:51
• Precisely, you can conclude from $0 = f(a) + f(-a)$ that $f(-a)$ is the additive inverse of $f(a)$ but without onto this implies that the inverses of elements in $S$ exists only if they are taken from the image of $f$. On the other hand, if $f$ is onto then we know that every $x \in S$ has $a \in R$ such that $f(a) = x$ and $-x$ can be found as you did it (using the fact that $R$ contains $-a$ and $f$ is homo, i.e. using the above equation). Feb 3 '18 at 14:57

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 $s+f(-r)=f(r)+f(-r)=f(r-r)=f(0)=0$.

• So I can use the absorption property to show that inverses exist in $(S,*_0)$. I think the professor showed the class how we don't need to explicitly state the absorption property because we get it from the distributive property. Feb 3 '18 at 15:07
• @dan I'm not sure how you can, unless the monoid $(S,*_0)$ is assumed to be cancellative. Feb 3 '18 at 15:12
• The example he showed us was $a0=a(0+0)=a0+a0$ then we subtract $a0$ from both sides to get $a0=a0-a0=0$. Feb 3 '18 at 15:20
• @dan You can't subtract in a monoid. Feb 3 '18 at 18:10
• But since $R$ is a ring, $(R,*_0)$ is a group. Feb 3 '18 at 23:19