Let $V$ be a $n$-dimensional $\mathbb{C}$-vector space, so $V\cong \mathbb{C}^n$. Let further $T:\mathbb{C}\to \mathbb{C}$ be a $\mathbb{C}$-linear transformation. We consider $V$ as a $\mathbb{C}[X]$ module by $$p(X)\cdot v=p(T)(v).$$ The classification theorem for finitely generated modules over PID tells us that there exist $a_i\in\mathbb{C}[X]$ such that $$V\cong \mathbb{C}[X]^n\oplus_i^m \mathbb{C}[X]/a_i\mathbb{C}[X]\tag{1}.$$ Now, since $V$ is finite dimensional over $\mathbb{C}$ we have $n=0$.
Here comes my question:
In my lecture notes it's written that we can show that (1) collapses into $$ V\cong\bigoplus_{i=1}^m\bigoplus_{k=1}^{r_i}\mathbb{C}[X]/(X-\lambda_{i,k})^{d_{i,k}}\mathbb{C}[X] \tag{2}$$ and so we conclude that $\mathbb{C}[X]$ is the direct sum of Jordan Blocks and have proven the existence of the Jordan Normal Form.
Could someone explain how (2) follows form (1)?
Let $a_i=\prod_{k=1}^{r_i}(X-\lambda_{i,k})^{d_{i,k}}$. Then by the Chinese Remainder Theorem we have that $$\mathbb{C}[X]/a_i\mathbb{C}[X]\cong\bigoplus_{k=1}^{r_i}\mathbb{C}[X]/(X-\lambda_{i,k})^{d_{i,k}}\mathbb{C}[X]$$ then $$V\cong\bigoplus_{i=1}^m\bigoplus_{k=1}^{r_i}\mathbb{C}[X]/(X-\lambda_{i,k})^{d_{i,k}}\mathbb{C}[X]$$
Alternatively, if you're allowed to use the primary decomposition form of the classification (given as the second statement here), we have a neater proof.
Over a PID, a primary ideal must be generated by an element of the form $p^s$ for some prime $p$ (see here for an explanation). Since $\mathbb{C}$ is algebraically closed, primes in $\mathbb{C}[X]$ are of the form $(X-\lambda)$, so $$V\cong\bigoplus_{i=1}^r\mathbb{C}[X]/(X-\lambda_i)^{s_i}\mathbb{C}[X]$$