1. ## Rings

1. $R = \{ a +bi : a,b \in Z \}$

is a subring of a ring $C$ (complex)

Find an element of R different from the identity element which has a multiplicative inverse in R.

2. Let R,S be rings and suppose $f:R \rightarrow S$ is a ring homomorphism. Suppose $im(f) = S$

(a) Show that if R has an identity element then S has an identity element.
(b) Show that if R is commutative then S is commutative
(C) Is it true that if R is an integral domain then S is an integral domain? Justify your assertion.
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I'm not 100% sure on some of this.

2. For #1. Is it true that $\left\{ {i, - i} \right\} \subseteq R$?

For #2b. If $\left\{ {p,q} \right\} \subseteq S$ then because the mapping is onto we get $\left( {\exists r_p } \right)\left( {\exists r_q } \right)\left[ {f(r_p ) = p\,\& \,f(r_q ) = q} \right]$.
This means because R is commutative, $\left( {r_p * r_q } \right) = \left( {r_q * r_p } \right)$. Which means $p \cdot q = f\left( {r_p } \right) \cdot f\left( {r_q } \right) = f\left( {r_p * r_q } \right) = f\left( {r_q * r_p } \right) = f\left( {r_q } \right) \cdot f\left( {r_p } \right) = q \cdot p$

3. Originally Posted by Jason Bourne
2. Let R,S be rings and suppose $f:R \rightarrow S$ is a ring homomorphism. Suppose $im(f) = S$
(a) Show that if R has an identity element then S has an identity element.
Suppose that there exists an element $1$ such that $1x=x1=x$ for all $x\in R$. Then we claim that $f(1)$ is the identity element for $S$. We need to show $sf(1)=f(1)s=s$ for all $s\in S$. Since the function is onto it means $s=f(r)$. That means $sf(1) = f(r)f(1)=f(r1)=f(r)=s$ and similarly $f(1)s=s$.

(C) Is it true that if R is an integral domain then S is an integral domain? Justify your assertion.
No, it is not true. Let $R = \mathbb{Z}$ and $S = \mathbb{Z}_4$. For $x\in R$ let $f(x) = [x]_4$ to be the congruence class of $x$ mod 4. Then $f$ is a ring homomorphism and $\text{Im}(f) = S$ but $R$ is an integral domain and $S$ is not.