Factor ring of polynomials by an ideal that corresponds to multiple roots

Question

Let's consider the polynomial ring $\mathbb C[x]$ and the ideal $$I = ((x-c_1)^{m_1}\cdot (x-c_2)^{m_2}\cdots(x-c_n)^{m_n}),$$ where $c_1,\ldots,c_n\in \mathbb C$ are fixed numbers. I need to find some structure that is isomorphic to the factor ring $\mathbb C[x]/I$.

For example, if there aren't multiple roots and $I = ((x-c_1)\cdot (x-c_2)\cdots(x-c_n))$ then we can consider homomorphism $\alpha: \mathbb C[x] \rightarrow \mathbb C^n$ where each polynomial $f \in \mathbb C[x]$ corresponds to a vector $\vec a$ (with component-wise operations $+$ and $\cdot$) where $a_1 = f(c_1),a_2 = f(c_2),\ldots,a_n = f(c_n)$.

Since $\ker(\alpha)$ is equal to $I$ then from the homomorphism theorem we can infer that $\mathbb C[x]/I \cong \mathbb C^n$ and each element of the factor ring corresponds to $\vec a \in \mathbb C^n$ on the following way $[\vec a] = \{f\mid f(c_1)=a_1,f(c_2)=a_2,\ldots,f(c_n)=a_n\}$, here $[\vec a]$ denotes equivalence class of factor ring.

I stuck with performing the same procedure in case of multiple roots. I understand that root of $k$-th degree corresponds to zero of $(k-1)$-th derivative in the point of the root, but what exactly this structure could look like?

Answer 1

The Chinese remainder theorem for general commutative rings says the following: for a commutative ring $ R $ and pairwise comaximal ideals $ I_1, I_2, \ldots, I_n $, there is an isomorphism

$$ R/(\cap_{i=1}^n I_i) \cong \prod_{i=1}^n R/I_i $$

In a principal ideal domain such as $ K[X] $ over any field $ K $, comaximal ideals are the ideals generated by coprime elements, and the intersection of two ideals is their least common multiple. Since powers of distinct primes are always coprime, the Chinese remainder theorem reduces the task of understanding the quotients of a principal ideal domain (or, more generally, a Dedekind domain) to the task of understanding the quotients by prime-power ideals. Since the primes in $ \mathbb C[x] $ are first degree polynomials, we only need to understand the isomorphism type of $ \mathbb C[x]/(x^n) $ for $ n \geq 1 $. Your intuition was correct: this quotient ring is isomorphic to the ring $ R $ of entire functions $ \mathbb C \to \mathbb C $, quotiented out by the ideal $ I = \{ f \in R : f(0) = f'(0) = \ldots = f^{(n-1)}(0) = 0 \} $.

Answer 2

Let $m$ denote the sum of the $m_i$'s. Define an $m\times m$ complex matrix $A$ as the companion matrix (see https://en.wikipedia.org/wiki/Companion_matrix ) of the polynomial $f(x)=\prod_i (x-c_i)^{m_i}$. This subalgebra of $m\times m$ matrix algebra generated by this $A$ will be isomorphic to $\mathbf{C}[X]/I$. (It is in fact the linear span of all powers of $A$, upto the $m$th power)