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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?

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2 Answers 2

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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: $$\mathbb{C}[X]/f(x)=\mathbb{C}[X]/ \prod_{i=1}^n (x-\lambda_i)=\prod_{i=1}^n \mathbb{C}[X]/ (x-\lambda_i)=\mathbb{C}^n$$

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  • $\begingroup$ Thanks! great and simple idea! $\endgroup$
    – Extremal
    Dec 25, 2015 at 5:57
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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).

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  • $\begingroup$ But I need a surjective homomorphism. $\endgroup$
    – Extremal
    Dec 25, 2015 at 5:50
  • $\begingroup$ @EpsilonDelta It is surjective. I've added a hint. $\endgroup$
    – Slade
    Dec 25, 2015 at 5:52
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    $\begingroup$ @NoahOlander I'm confused too, especially since an "explicit" homomorphism was requested. $\endgroup$
    – Slade
    Dec 25, 2015 at 6:00
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    $\begingroup$ @NoahOlander I think using Lagrange interpolation prevents it from being self-contained. The whole point of the CRT is that such a homomorphism is onto. (I have nothing to do with the downvote.) $\endgroup$
    – rschwieb
    Dec 27, 2015 at 3:40
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    $\begingroup$ @rschwieb I suppose that the most natural thing to do, for an elementary approach, is to calculate the kernel and then deduce surjectivity by counting dimension. $\endgroup$
    – Slade
    Dec 27, 2015 at 4:11

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