Length of the $\mathbb{C}$-module $\mathbb{C}[x]/((x^2+1)^2)$

Question

I want to prove that $l(\mathbb{C}[x]/((x^2+1)^2))=4$ but I am stuck. For the $\mathbb{R}$ -module, $\mathbb{R}[x]/((x^2+1)^2)$ we knew that the ideal $(x^2+1)/(x^2+1)^2$ was unique maximal ideal so it was easy to show that the length equals to 2.

In my case this is not maximal. I know that the maximal ideals of $\mathbb{C}$ are of the form: $(x-a)$ so the maximal ideals of $\mathbb{C}[x]/((x^2+1)^2)$ must have the form $(x-a)/((x^2+1)^2)$. Doesn't that mean that the maximal ideals are infinite?

Edit. Now I i understand that $(x-i), (x+i)$ are maximal ideals of the module. So we can think of this chain: $$0 \subseteq \frac{(x^2+1)}{((x^2+1)^2)} \subseteq \frac{(x-i)}{((x^2+1)^2)}\subseteq \frac{(x+i)}{((x^2+1)^2)}\subseteq \mathbb{C}[x]/((x^2+1)^2)$$

Is it a composition series? I do not know how to prove that the quotients of the 2nd/3rd term and 3rd/4th are simple modules ($0$ is the 1st term).

Any help would be appreciated.

Answer 1

No, it is not a composition series. For instance, the first term $\frac{(x^2+1)}{((x^2+1)^2)}$ is isomorphic to $\mathbb C[x]/(x^2+1)$ which is not simple.

In the following it is useful to notice that if $R$ is an integral domain and $a,b\in R$, $a,b\ne0$, then $(a)/(ab)\simeq R/(b)$.

It is not difficult to see now that $$ (0) \subset \frac{(x^2+1)(x-i)}{((x^2+1)^2)} \subset \frac{(x^2+1)}{((x^2+1)^2)} \subset \frac{(x+i)}{((x^2+1)^2)}\subset \frac{\mathbb{C}[x]}{((x^2+1)^2)}$$ is a composition series.

Answer 2

Let $I=aR$ be a principal ideal in the domain $R$. Consider the map $$ f\colon R \to I/I^2,\qquad f(r)=ar+I^2 $$ This is a homomorphism of $R$-modules and its kernel is $I$ (prove it), so you have that $I/I^2\cong R/I$.

If $R$ is an $k$-algebra, for $k$ a field, then the isomorphism is also of $k$-vector spaces. In your case you have the exact sequence $$ 0 \to \frac{(x^2+1)}{(x^2+1)^2} \to \frac{\mathbb{C}[x]}{(x^2+1)^2} \to \frac{\mathbb{C}[x]}{(x^2+1)}\to 0 $$ so you just have to prove that $$ \dim_{\mathbb{C}}(\mathbb{C}[x]/(x^2+1))=2 $$