Prove that $R\cong \mathbb{C}^n$

Question

Let $R=\mathbb{C}[x]/(f(x))$ where $f(x)$ is a polynomial of degree $n>0$ which has $n$ distinct complex roots. Prove that $R\cong \mathbb{C}^n$.

I tried to define a homomorphism $\phi$ from $\mathbb{C}[x]$ to $\mathbb{C}^n$ such that $\phi$ is onto and ker$\phi=(f(x))$. Then by first isomorphism theorem I am done. But the hard part is defining such homomorphism explicitely. First I tried to divide a given polynomial $p(x)$ by $f(x)$ and take the remainder polynomial. The remainder is a polynomial of degree at most $n-1$. So it has $n$ coefficients. Then my try was mapping $p(x)$ to the $n$ tuple of those coefficients. But then I was failed proving $\phi$ is a homomorphism. So how do I determine a homomorphism explicitely? Can somebody please help me?

Answer 1

Let $f(x)=\prod_{i=1}^n (x-\lambda_i)$. The ideals $(x-\lambda_i)$ are pairwise coprime, since they are maximal and distinct. Hence, by the Chinese remainder theorem, we have: $$\mathbb{C}[X]/f(x)=\mathbb{C}[X]/ \prod_{i=1}^n (x-\lambda_i)=\prod_{i=1}^n \mathbb{C}[X]/ (x-\lambda_i)=\mathbb{C}^n$$

Answer 2

If the roots are $z_1, \ldots , z_n$, then one simple approach is sending $f\in \mathbb{C}[X]$ to the tuple $(f(z_1), \ldots , f(z_n))$.

It is a homomorphism on each component, therefore a homomorphism. Using distinctness of the roots, we can calculate the kernel without too much fuss.

To show that it is surjective, count dimensions. Alternatively, we can use Lagrange interpolation (using again the distinctness of the roots).