Question

Problem 1 . In the lecture, a ring isomorphism from T to R is a map θ : T → R satisfying 4 properties: (1) θ preserves addition. (2) θ(h) IR, (3) θ preserves multiplication, and (4) θ is bijective. However, in Chapter 3.1 in the textbook, the author only uses properties (1), (3), and (4) in the definition and says that property (2) will follow from them. We verify it here. Forget the above motivation, here's the problem: let θ : T → R be a map satisfying the following 2 properties: (a) θ preserves multiplication: 6(t1t2) (b) θ is surjective -6(h)6(t2) Yh,t2ET. Prove that θ(h)-1 R. (3pts)

Answer 1

$\dpi{100} \theta(t_1t_2)=\theta(t_1)\theta(t_2)$ for all $\dpi{100} t_1,t_2\in T$

$\dpi{100} \theta$ is surjective

$\dpi{100} t_1=t_2=1_T$ means $\dpi{100} \theta(1_T1_T)=\theta(1_T)\theta(1_T)$

So $\dpi{100} \theta(1_T)=\theta(1_T)\theta(1_T)$

And $\dpi{100} \theta(1_T)=1_R\theta(1_T)$ as $\dpi{100} \theta(1_T)\in R$

So we have $\dpi{100} 1_R\theta(1_T)=\theta(1_T)\theta(1_T)$

Thus, we must have $\dpi{100} \theta(1_T)=1_R$

$\dpi{100} \blacksquare$

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