The divisors of $X^n$

Question

Consider the ring $\mathbb{R}[X]$

Show that the only divisor of $X^n$ is $X^i$ with $ i=0,...,n$.

$My~~ work :$

Let $P\in\mathbb{R}[X]$ : if $P|X^n$, then 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 $i+j=n$.

My problem is : how do I show that there is an $i$ such that $P = a_iX_i$.

Answer 1

The thing becomes obvious once you know that $K[X]$ is a Principal Ideal Domain (PID), which is a consequence of the existence of an euclidean division $$ A(X)=Q(X)B(X)+R(X)\qquad\text{with $\deg(R)<\deg(B)$}. $$ Since PIDs are UFDs (UFD = Unique Factorization Domain) you have unique factorization in $K[X]$. The element $X^n$ has an obvious decomposition: $$ X^n=\underbrace{X\cdot\cdots\cdot X}_{\text{$n$ factors}} $$ thus by unicity the only way to get a divisor is to aggregate a few of those factors together. But that's $X^k$ for some $k\in\{0,...,n\}$.

Answer 2

Assume $P(X)\mid X^n$, so $P(X)Q(X)=X^n$ for some $Q(X)$. Then $P(X)$ is not the zero polynomial and hence we can write $P(X)=X^pP_1(X)$ where $p\ge0$ and $P_1(0)\ne0$. Similarly, we can write $Q(X)=X^qQ_1(X)$. Thus we have $X^{p+q}P_1(X)Q_1(X)=X^n$.

• If $p+q< n$, we divide both sides by $X^{p+q}$ and arrive at $P_1(X)Q_1(X)=X^{n-p-q}$ which is false at $X=0$
• If $p+q> n$, we divide both sides by $X^n$ and arrive at $X^{p+q-n}P_1(X)Q_1(X)=1$ which again is false at $X=0$

We conclude $p+q=n$ (and in particular, $0\le p\le n$). Then $P_1(X)Q_1(X)=1$ and by comparing degrees, we find $\deg P_1(X)=\deg Q_1(X)=0$, i.e., $P_1(X)$ is a non-zero constant. Hence $$ P(X)=cX^i$$ with $c\ne 0$ and $0\le i\le n$. If you additionally require that the leading term of $P$, then of course $c=1$ and $P(X)=X^i$.