# Homomorphisms from a unital ring to a ring with no zero divisors preserve unity?

I'm having a little problem with a trouble from Hungerford is *Algebra * worrying ring homomorphisms.

Allow $f\colon R\to S$ be a homomorphism of rings such that $f(r)\neq 0$ for some nonzero $r\in R$. If $R$ has an identification and also $S$ has no absolutely no divisors, after that $S$ is a ring with identification $f(1_R)$.

I've exercised a little an argument. Take $r$ to be as in the trouble, so $$ f(r)=f(1_R)f(r)=f(r)f(1_R)\implies f(r)-f(1_R)f(r)=0_S. $$ This reveals that $f(1_R)$ is the identification of $f(R)$, yet I'm not exactly sure if that is much usage. Nonetheless, if $1_S$ exists, after that from $f(r)-f(1_R)f(r)=1_Sf(r)-f(1_R)f(r)=0_S$ and also the distributive regulation, I would certainly have $$ \bigl(1_S-f(1_R)\bigr)f(r)=0_S\implies 1_S=f(1_R) $$ given that $S$ has no absolutely no divisors, and also $f(r)\neq 0_S$. Yet I do not see a means to confirm $1_S$ exists, so if it is no trouble, I'm wanting to get a tip on just how to show that, or probably a prod in the appropriate instructions if I'm off track. Thanks.

Since the solution was not noticeable to me quickly, I simply intend to upload a solution to make sure that the inquiry has a complete solution.

Placing it with each other, $f(1_R)s=f(1_R)^2s\implies f(1_R)[s-f(1_R)s]=0_S$. Yet $f(1_R)\neq 0$ given that $f(1_R)f(r)=f(r)\neq 0$, so $s=f(1_R)s$ given that $S$ has no absolutely no divisors. Given that $sf(1_R)=sf(1_R)^2$ too, the very same argument reveals $s=sf(1_R)$, so $f(1_R)=1_S$.

Hint: If $s$ remains in $S$, after that $f(1_R)s=f(1_R)^2s$.

I left the above quick tip based upon the demand at the end of your inquiry, not intending to "give way too much away, " today I'm satisfied to specify a little bit.

You intend to show that $f(1_R)$ is an identification for $S$. As you stated in a comment, it adheres to from the reality that $f$ is a homomorphism that $f(1_R)=f(1_R)^2$. You additionally recognize that $f(1_R)\neq 0$, due to the fact that $f(1_R)f(r)=f(r)\neq 0$, and also hence it is alluring to intend to "cancel" $f(1_R)$ in the formula of the previous sentence to get $1_S=f(1_R)$. Yet certainly, we still need to show that $1_S$ exists. In order to terminate $f(1_R)$, we need to be increasing it with something, and also this encourages increasing both sides of the formula by some $s\in S$ to get the formula on top of this solution. Currently you can do some cancelling.

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