Question #146090
Show that( R [x] /<x²+1>) isomorphic To C as fields.
1
Expert's answer
2020-11-25T17:55:57-0500

We give a map f:R[X]C.f:\mathbb{R}[X]\longrightarrow \mathbb{C}. xi.x\mapsto i. In general, g(x)g(i).g(x)\mapsto g(i). Hence a+bXa+ib.a+bX\mapsto a+ib. Hence surjective. Homomorphism is obvious since gh(i)=g(i)h(i)gh(i)=g(i)h(i) for polynomials. Also (g+h)(i)=g(i)+h(i).(g+h)(i)=g(i)+h(i). Now for kernel. g(x)0g(i)=0.g(x)\mapsto 0\Rightarrow g(i)=0. Now by division algorithm, g(x)=aX+b+(X2+1)(q(x))g(x)=aX+b+(X^2+1)(q(x)) . Hence g(i)=0ai+b=0a=b=0.g(i)=0\Rightarrow ai+b=0\Rightarrow a=b=0. Hence

kernel is generated by X2+1.X^2+1. Hence by first isomorphism theorem, the result follows.


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