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The question is :

Construct a bijection $\mathrm{Hom}_{\mathbb C}(\mathbb C[x,y]/(xy-1),\mathbb C) \to \mathbb C\setminus \{0\}$.

Here $\text{Hom}_{\mathbb C}(\mathbb C[x,y]/(xy-1),\mathbb C)$ is the set of all homomorphisms $$\phi :\mathbb C[x,y]/(xy-1) \to \mathbb C\;\; \text{s.t.}\;\; \phi|_{\mathbb C}=\mathrm{id}|_{\mathbb C}.$$

Please someone give some hints how can I do this?

Thank you.

user26857
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MADmind
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2 Answers2

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Hint: $~$ A $\mathbb{C}$-algebra homomorphism $\mathbb{C}[x,x^{-1}]\to\mathbb{C}$ is determined by where $x$ is sent.

anon
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  • I didn't studied $\mathbb C$-algebra yet so if there is any other process please provide – MADmind Aug 17 '17 at 20:13
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    @nisha "$\Bbb C$-algebra homomorphism" = "$\Bbb C$-linear ring homomorphism," which turns out to be exactly what you are studying in this problem. – anon Aug 17 '17 at 20:14
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A homomorphism from $R=\Bbb C[x,y]/(xy-1)$ to $\Bbb C$ is a homomorphism $\Phi$ from $\Bbb C[x,y]$ to $\Bbb C$ with $\Phi(xy-1)=0$. Each homomorphism $\Phi:\Bbb C[x,y]\to \Bbb C$ has has the form $\Phi_{a,b}:f(x,y)\to f(a,b)$ where $a$, $b\in\Bbb C$. Then $\Phi_{a,b}(xy-1)=ab-1$. So $\Phi_{a,b}$ defines a homomorphism on $R$ iff $ab=1$, etc.

Angina Seng
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