# Rings

• Jan 9th 2008, 02:17 PM
Jason Bourne
Rings
1. $\displaystyle R = \{ a +bi : a,b \in Z \}$

is a subring of a ring $\displaystyle C$ (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 $\displaystyle f:R \rightarrow S$ is a ring homomorphism. Suppose $\displaystyle im(f) = S$

(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.
• Jan 9th 2008, 03:35 PM
Plato
For #1. Is it true that $\displaystyle \left\{ {i, - i} \right\} \subseteq R$?

For #2b. If $\displaystyle \left\{ {p,q} \right\} \subseteq S$ then because the mapping is onto we get $\displaystyle \left( {\exists r_p } \right)\left( {\exists r_q } \right)\left[ {f(r_p ) = p\,\& \,f(r_q ) = q} \right]$.
This means because R is commutative, $\displaystyle \left( {r_p * r_q } \right) = \left( {r_q * r_p } \right)$. Which means $\displaystyle p \cdot q = f\left( {r_p } \right) \cdot f\left( {r_q } \right) = f\left( {r_p * r_q } \right) = f\left( {r_q * r_p } \right) = f\left( {r_q } \right) \cdot f\left( {r_p } \right) = q \cdot p$
• Jan 9th 2008, 05:52 PM
ThePerfectHacker
Quote:

Originally Posted by Jason Bourne
2. Let R,S be rings and suppose $\displaystyle f:R \rightarrow S$ is a ring homomorphism. Suppose $\displaystyle im(f) = S$
(a) Show that if R has an identity element then S has an identity element.

Suppose that there exists an element $\displaystyle 1$ such that $\displaystyle 1x=x1=x$ for all $\displaystyle x\in R$. Then we claim that $\displaystyle f(1)$ is the identity element for $\displaystyle S$. We need to show $\displaystyle sf(1)=f(1)s=s$ for all $\displaystyle s\in S$. Since the function is onto it means $\displaystyle s=f(r)$. That means $\displaystyle sf(1) = f(r)f(1)=f(r1)=f(r)=s$ and similarly $\displaystyle f(1)s=s$.

Quote:

(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 $\displaystyle R = \mathbb{Z}$ and $\displaystyle S = \mathbb{Z}_4$. For $\displaystyle x\in R$ let $\displaystyle f(x) = [x]_4$ to be the congruence class of $\displaystyle x$ mod 4. Then $\displaystyle f$ is a ring homomorphism and $\displaystyle \text{Im}(f) = S$ but $\displaystyle R$ is an integral domain and $\displaystyle S$ is not.