A non-trivial mapping $\theta : S \to R$

Question

Let $(R,+,\times)$ be a ring with additive identity $0 \in R$. On the set $S = \{(a,b) :\ a,b \in R\}$ the binary operators $\oplus$ and $\otimes$ are defined by:

$$(a, b)\oplus(c, d) = (a+c, b+d),$$ $$ (a, b)\otimes(c, d) = (a\times c+b\times d, a\times d+b\times c).$$

I am currently doing self study of abstract algebra and need some guidance. I need to state a non-trivial mapping $\theta : S \to R$ and show that it is a homomorphism.

Firstly I done (& please correct me if I am wrong):

let $θ : S \to R$ defined by $\theta(a,b) = a$.

Then

$$\theta((a,b) + (c,d)) = (a,b) + (c,d) = (a+c, b+d)$$ $$\theta((a,b)) + θ((c,d)) = (a,b) + (c,d) = (a+c, b+d)$$

$$\theta((a,b) \times (c,d)) = (a,b) \times a (c,d) = (ac + bd, ad + bc)$$ $$\theta((a,b)) \times θ((c,d)) = (a,b) \times (c,d) = (ac+ bd, ad + bc)$$

Therefore it is a homomorphism

Does this then give me what I am looking for? Please feel free to state anything that I have done incorrectly.

Thank you

Answer 1

You are not following your own definition, nor using the right operations on $S$. What you want to do is: $$ \theta((a,b)\oplus(c,d))=\theta(a+c,b+d)=a+c=\theta(a,b)+\theta(c,d) $$ and $$ \theta(a,b)\otimes(c,d))=\theta(ac+bd,ad+bc)=a\times c+b\times d. $$ This last term is not necessarily equal to $a\times b=\theta(a,b)\times\theta(b,c)$, so your $\theta$ is not multiplicative (unless the product on $R$ is trivial).

One map that is a homomorphism is $\theta(a,b)=a+b$.