Is the limit $C_n[x]/I_n$ exsit when the dimension is infinite？

Question

The question is finding the dimension of $\mathbb C[x]/I$, where $\mathbb C[x]$ is the polynomial ring over $\mathbb C$, and $I=\{f(x)\in \mathbb C[x]\mid f(1)=f'(1)=f^{(k)} (1)=0\}$.

My answer is to compute the dimension for $C[x]$ and $I$ when the order of $f(x)$ is less than or equal to n,

use $dimC_n[x]$ and $dim I_n$ for the dimension of these two rings when degrees is less or equal to n,

Solve $a_n+a_n-1…+a_1+a_0=0,\\ na_n+(n-1)a_n-1+…+a_1=0\\n(n-1)a_n+(n-1)(n-2)a_{n-1}+…2a_2=0\\…\\n(n-1)(n-2)…(n-k+1)an+…k!a_k=0$

It is same as solving

$k!a_k=-n(n-1)…(n-k+1)a_n+…\\k!a_k+(k-1)!a_{k-1}=-n(n-1)…(n-k+2)a_n+…\\…\\a_0+…a_k=-an+(-)a_{n-1}+…+(-)a_{k+1}$

The fundamental solution is $n-k$ dimension. We get $dimC_n[x]=n+1$ and $dimI_n=n-k$

so $dimC_n[x]/I_n$ is $k+1$.

As $n \to \infty$ $dimC_n[x]/I_n \to {k+1} $ $dimC_n[x]/I_n=dimC[x]/I$.

Is the method for sequence also suit for compute the dimension？

$dimC_n[x]/I_n=k+1$ for $\forall n \in N+$.And $dimC_n[x]/I_n=dimC[x]/I(n\to \infty)$ a correct expression?

The primary problem is to compute the dimension when $I=\{f(x)\in \mathbb C[x]\mid f(1)=f'(1)=0\}$.But it seems the repliers do not understand what I asked.I change it into a general case so it is easy to understand why I list two linear equations.

Answer 1

$\newcommand{\N}{\mathbb{N}}$$\newcommand{\C}{\mathbb{C}}$$\newcommand{\Set}[1]{\left\{ #1 \right\}}$You would be in trouble trying to compute the dimension of $\C[x]$.

Moreover, one should state (and prove!) that $I$ is an ideal, so that you want to compute the dimension of the quotient ring. In fact, it is clear that the sum of two elements of $I$ lies in $I$, Now if $f \in I$ and $g \in \C[x]$, we have $fg(1) = f(1) g(1) = 0$, and $$(f g)'(1) = (f' g + f g')(1) = f'(1) g(1) + f(1) g'(1) = 0,$$ so $fg \in I$.

Still, you appear to be arguing in the right direction. First use Taylor to show that all polynomials can be written uniquely as $$ a_{0} + a_{1} (x-1) + \dots + a_{n} (x-1)^{n} \qquad n \in \N, a_{i} \in \C. $$

Then prove that $$ I = \Set{ a_{0} + a_{1} (x-1) + \dots + a_{n} (x-1)^{n} : n \in \N, a_{i} \in \C, a_{0} = a_{1} = 0} $$ and proceed from that.

Answer 2

Let $V = \{a + bX \mid a, b \in \Bbb C\}$ be the vector space of polynomials of degree at most $1$. Let $\varphi : \Bbb C[x] \to V$ be given by $\varphi (f) = f(1) + f'(1) X$. Check for yourself that $\varphi$ is a linear map (it is easy).

Let us show that $\varphi$ is surjective: if $a + bX \in V$ then $\varphi \big( (a-b) + bX \big) = a + bX$, so indeed $\varphi$ is surjective.

Looking at the kernel of $\varphi$ we see that

$$\varphi (f) = 0 \iff f(1) + f'(1)X = 0 \iff f(1) = f'(1) = 0 \iff f \in I ,$$

which shows that $\ker \varphi = I$.

Applying now the first fundamental isomorphism theorem for vector spaces, we get that $\Bbb C [X] / I \simeq V$, so $\dim \Bbb C [X] / I = \dim V$. But $\dim V = 2$ (obviously), so $\dim \Bbb C [X] / I = 2$.