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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.

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    $\begingroup$ Are you sure the homomorphism $f$ is not required to be onto? $\endgroup$
    – egreg
    Feb 3 '18 at 14:46
  • $\begingroup$ Oh I think it is required to be onto. That would be the reason why I know $-f(a)$ is in $(S,*_0)$. $\endgroup$
    – dnem41
    Feb 3 '18 at 14:50
  • $\begingroup$ Without onto, $R=\Bbb Z$, $S=\Bbb Z\times \Bbb N_0$, $x\mapsto (x,0)$ would be a counterexample $\endgroup$ Feb 3 '18 at 14:51
  • $\begingroup$ 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). $\endgroup$ Feb 3 '18 at 14:57
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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 $s+f(-r)=f(r)+f(-r)=f(r-r)=f(0)=0$.

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  • $\begingroup$ 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. $\endgroup$
    – dnem41
    Feb 3 '18 at 15:07
  • $\begingroup$ @dan I'm not sure how you can, unless the monoid $(S,*_0)$ is assumed to be cancellative. $\endgroup$
    – egreg
    Feb 3 '18 at 15:12
  • $\begingroup$ The example he showed us was $a0=a(0+0)=a0+a0$ then we subtract $a0$ from both sides to get $a0=a0-a0=0$. $\endgroup$
    – dnem41
    Feb 3 '18 at 15:20
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    $\begingroup$ @dan You can't subtract in a monoid. $\endgroup$
    – egreg
    Feb 3 '18 at 18:10
  • $\begingroup$ But since $R$ is a ring, $(R,*_0)$ is a group. $\endgroup$
    – dnem41
    Feb 3 '18 at 23:19

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