ideals of ring $A $

Question

Let $A=\mathbb{R}[X]/\langle X^n\rangle$. Show that $A$ admits $n-1$ proper ideals.

I know that:

$$ A = \{a_0+a_1x+\dots+a_{n-1}x^{n-1} : a_i \in\mathbb{R}\} $$ and $$I_i=\langle x^i\rangle $$ for $i=1,\dots,n-1$ are proper ideals of $A $. But how do I show that she is are unique?

Answer 1

Hint 1 If $I$ is an ideal in $\mathbb{R}[X]/\langle X^n\rangle $ then there exists an ideal $J$ in $\mathbb{R}[X]$ such that $\langle X^n\rangle \subseteq J$ such that $I=J/ \langle X^n\rangle $.

Hint 2 $\mathbb{R}[X]$ is a PID.

Hint 3: For polynomials $P,Q \in \mathbb{R}[X]$ you have $\langle P \rangle \subseteq \langle Q\rangle $ if and only if $Q$ divides $P$.

Answer 2

An ideal of $\mathbb{R}[X]/(X^n)$ corresponds to an ideal $I=(P)$ of $\mathbb{R}[X]$ such that $(X^n)\subseteq I$ where $P$ is a polynomial ($I$ is principal because $\mathbb{R}[X]$ is a PID). Therefore there is polynomial $Q$ such that $X^n=PQ$. Writing $P=\sum a_iX^i$ and $Q=\sum b_jX^j$ the equation $X^n=PQ$ says that the term of degree $0$ of the product $PQ$ is zero, that is, $a_0b_0=0$. Thus $a_0=b_0=0$. There are $i,j$ such that $a_i\neq0, b_j\neq0$ and $i+j=n$. By induction one can check that $a_l=0$ for $l<i$ and $a_i=1$. With a similar argument one can verify that $a_p=b_q=0$, where $p$ and $q$ are the degrees of the $P$ and $Q$ respectively, and that $a_l=0$ for $l>i$. This argument implies that $P=X^i$.