Show that the set of polynomials divisible by $x-1$ cannot span $P$, the set of all polynomials.

Question

Proof: (by contradiction)

Let $D$ be the set of polynomials divisible by $x-1$; $D=\{\sigma_i(x)=q_i(x)(x-1) \mid \sigma_i(x),q_i(x)\in P\}$. Considering the set of polynomials $p(x)\in P$ such that $(x-1)\nmid p(x)$. If $D$ spans $P$ then $p(x)=\sum_{i=1}^{n}c_i\sigma_i(x)$,for some $\sigma_i(x)\in D$ and $c_i\in\Bbb{F}$, then \begin{align} p(x)&=c_1q_1(x)(x-1)+...+c_nq_n(x)(x-1)\\ &=(x-1)(c_1q_1(x)+...+c_nq_n(x))\\ \Rightarrow (x-1)\mid p(x)\\ \end{align} Thus a contradiction.

Answer 1

It is correct, exceept that you did not prove that there is a polynomial $p(x)$ such that $(x-1)\nmid p(x)$. Without this, your proof doesn't work.

The simplest way of doing this consists in proving that there is some polynomial outside $D$. Take $p(x)=1$, for instance. If you could write $p(x)$ as $a_1p_1(x)+\cdots+a_mp_m(x)$, with $p_1(x),\ldots,p_m(x)$ all multiples of $x-1$, then you would have $(x-1)\mid1$, which is clearly false.

Answer 2

Any polynomial $p(x)$ multiple of $x-1$ is such that $p(1)=0$ and this property is inherited by any linear combination. So you can't cover $q(1)\ne0$.

Answer 3

What ever you take to be your coefficient ring, will consist of constant polynomials not divisible by $(x-1)$, and will therefore be an example of a set of polynomials that aren't in $\langle x-1 \rangle $.

Answer 4

The set of all polynomials divisible by $(x−1)$ is a (maximal) ideal of $P$ considered as a ring. Thus cannot be $P$ itself.