Answer to Question #144997 in Linear Algebra for Sourav Mondal

Question

Prove that C⁴/C ~_ C³

Answer 1

Let $\phi:\mathbb{C}^4\rightarrow \mathbb{C}^3$ be defined as $\phi(z_1,z_2,z_3,z_4)=(z_1,z_2,z_3).$ This projection map is clearly a homomorphism over $\mathbb{C}$ , since $\phi(z_1+z_1',z_2+z_2',z_3+z_3',z_4+z_4')=(z_1+z_1',z_2+z_2',z_3+z_3')=(z_1,z_2,z_3)+(z_1'+z_2'+z_3')=\phi(z_1,z_2,z_3,z_4)+\phi(z_1',z_2',z_3',z_4')$. Also its clearly, linear since, $\phi(\lambda(z_1,z_2,z_3,z_4))=\phi(\lambda z_1,\lambda z_2,\lambda z_3,\lambda z_4)$ $=(\lambda z_1,\lambda z_2,\lambda z_3)=\lambda(z_1,z_2,z_3)=\lambda\phi(z_1,z_2,z_3,z_4).$

Surjectivity is immediate since $(z_1,z_2,z_3)$ has $(z_1,z_2,z_3,z_4)$ as preimage.

Kernel= $\{(z_1,z_2,z_3,z_4)|\phi(z_1,z_2,z_3,z_4)=(0,0,0,0)\}$ Hence Kernel=$\{(0,0,0,z_4)\}.$

Now let $f:$ Kernel$\rightarrow \mathbb{C}$ be given by, $f(0,0,0,z_4)=z_4.$ Clearly, this is an isomorphism. Hence by first isomorphism theorem, $\mathbb{C}^4/Ker \ \phi\cong \mathbb{C}^3\Rightarrow \mathbb{C}^4/Ker \ \phi\cong \mathbb{C}^3.$