Given a finite dimensional commutative $\mathbb{C}$-algebra, generated by one element $a \in A$. Write $(X^i)$ for the ideal generated by the polynomial $X^i \in \mathbb{C}[X]$ and $\frac{\mathbb{C}[X]}{(X^i)}^{\times n_i}= \underbrace{\frac{\mathbb{C}[X]}{(X^i)}\times \frac{\mathbb{C}[X]}{(X^i)} \times ... \times \frac{\mathbb{C}[X]}{(X^i)}}_\text{$n_i$ times}$

I have to proof that $A$ is isomorphic to $\mathbb{C}^{n_1} \times \frac{\mathbb{C}[X]}{(X^2)}^{\times n_2} \times \frac{\mathbb{C}[X]}{(X^3)}^{\times n_3} \times ... \times \frac{\mathbb{C}[X]}{(X^m)}^{\times n_m}$, for certain $n_1, n_2, ..., n_m \in \mathbb{N}$.

As a hint, they tell you to show that there exists a surjection $\mathbb{C}[X] \to A$ and then look at the kernel. I've made a $\mathbb{C}$-lineair ringepimorphism between the two $\mathbb{C}$-algebras by mapping a polynomial $P \in \mathbb{C}[X]$ to $P(a) \in A$. The kernel is equal to the ideal $(X-a)$, so by using the first ring isomorphism theorem, I get that $\mathbb{C} \cong \frac{\mathbb{C}[X]}{(X-a)} \cong A$. But how will this help me with my proof? If I pick $n_1 = 1 $ and $n_i=0$ $\forall m \geq i >1$, then I'm done? Or am I wrong with my surjection?

(note: I work with algebras over a ring, not over a field)

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    $\begingroup$ The kernel is not equal to the ideal $(X - a)$ [unless $A = \mathbb{C}$], because $a \notin \mathbb{C}$, so $X - a \notin \mathbb{C}[X]$. The kernel is generated by the minimal polynomial of $a$. $\endgroup$ – Daniel Fischer Nov 22 '16 at 20:55
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    $\begingroup$ The minimal polynomial of $a$ is the polynomial $P \in \mathbb{C}[X]$ of least degree such that $P(a)=0 \in A$, right? (I'm not familiar with the concept) $\endgroup$ – J.Bosser Nov 22 '16 at 21:07
  • $\begingroup$ Basically, yes. That doesn't determine it uniquely yet, only up to scalar multiples, so to get a unique one, we take the monic polynomial of lowest degree such that $P(a) = 0$. $\endgroup$ – Daniel Fischer Nov 22 '16 at 21:24

Considering the epimorphism $\mathbb{C}[X]\to A$, $p(X)\mapsto p(a)$, as you did, use the fact that $\mathbb{C}[X]$ is a PID to find $p_a(X)\in \mathbb{C}[X]$ with $\ker=(p_a(X))$, the ideal generated by $p_a(X)$ (this is precisely the minimal polynomial of $a$). Then decompose $p_a(X)=\prod_{i=1}^p(X-\lambda_i)^{k_i}$, where $\lambda_i\in\mathbb{C}$, which can be done since $\mathbb{C}$ is algebraically closed. We then have an isomorphism $$A\simeq\frac{\mathbb{C}[X]}{(p_a(X))}\simeq \prod_{i=1}^p\frac{\mathbb{C}[X]}{(X-\lambda_i)^{k_i}}$$ (this is the elementary divisor decomposition of $A$: Elementary divisors decomposition, see Dummit and Foote Theorem 12.6).

The automorphism $\mathbb{C}[X]\to\mathbb{C}[X]$, $p(X)\mapsto p(X-\lambda_i)$ maps $(X^{k_i})$ to $((X-\lambda_i)^{k_i})$, so $\mathbb{C}[X]/((X-\lambda_i)^{k_i})\simeq\mathbb{C}[X]/X^{k_i}$.

If we let $m$ be the largest of the $k_i$, we can simply reorganize terms and obtain an isomorphism $$A\simeq\prod_{i=1}^p\frac{\mathbb{C}[X]}{X^{k_i}}\simeq\prod_{j=1}^m\left(\prod_{k_i=j}\frac{\mathbb{C}[X]}{(X^{k_i})}\right)=\prod_{j=1}^m\left(\frac{\mathbb{C}[X]}{(X^j)}\right)^{n_j},$$ where $n_j=\#\left\{i:k_i=j\right\}$.


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