Consider the quotient ring $M:=\mathbb{R}[x]/\langle x^2+x+1\rangle$ as a module over $\mathbb{R}$ and let $\overline{f(x)}\in M$ nonzero. Is it true that $\{\overline{f(x)},\overline{xf(x)}\}$ is linearly independent?
My attempt:
We let
$$k_1f(x)+k_2xf(x)=g(x)(x^2+x+1)$$
for some $k_1,k_2\in \mathbb{R},g(x)\in \mathbb{R}[x]$. Suppose LHS is nonzero.
The LHS is equal to $f(x)(k_1+k_2x)$ which must be divisible by $h(x):=x^2+x+1$. As $h(x)$ does not factor in $\mathbb{R}[x]$, the nonzero $k_1+k_2x$ is relatively prime to $h(x)$, and hence $f(x)$ is divisible by $h(x)$, contradicting that $\overline{f(x)}$ is nonzero in $M$. Thus, LHS must be zero and as $f(x)$ again cannot be zero, $k_1=k_2=0$. So, yes, our set is linearly indepentend. Q.E.D.
- Is the above reasoning correct?
- I am not sure whether I can use the term relatively prime in polynomial. Even if it is OK, are there other ways to say what I mean above?
Thank you so much for all suggestions.