Is this mapping a well-defined mapping?

Question

Let $E = \mathbb{R}[x]_{\leq 3}$.

Let $G = \{ a + b · x + c · x^2 + d · x^3 ∈ E : a, b, c, d \in \mathbb{R}$ such that $b + c = 0 \}$.

Let $f(a + b·x + c·x^2 + d·x^3) = (a+b+c+d) + (b + c)·x - (b + c)·x^2 + (a - b - c + d)·x^3$.

Let $g: E/\langle 1 - x^3 \rangle \to G$ such that $[a + b · x + c · x^2 + d · x^3] \to f(a + b · x + c · x^2 + d · x^3)$. Being $[p(x)]$ a class of $E/\langle 1 - x^3 \rangle$ that admits $p(x)$ as its representative.

I have to show that $g$ is a well-defined mapping. I've tried to show it using that if $p(x) \sim q(x)$ then $p(x) - q(x) \in \langle 1 - x^3 \rangle$. Then, $p(x) = q(x) + \lambda_1 - \lambda_2 · x^3$, for some $\lambda_1, \lambda_2 \in \mathbb{R}$. But I can't seem to work it out. I think I've gotten myself in a wrong way to prove it.

Answer 1

Key Point: Since $E$ is a real vector space, $\langle 1-x^3\rangle$ is a (linear) subspace of $E$ generated by one vector $1-x^3$. In other words, $$ \langle 1-x^3\rangle = \{ \lambda(1-x^3) = \lambda - \lambda x^3 \mid \lambda\in\mathbb{R} \} $$

However, you thought that $\langle 1-x^3\rangle=\{\lambda_1-\lambda_2 x^3\mid \lambda_1,\lambda_2\in\mathbb{R}\}$, which is not a line but a plane.

Following your proof:

If $p(x)\sim q(x)$ then $p(x)−q(x)\in\langle 1−x^3\rangle$. Then $p(x)=q(x)+\lambda−\lambda x^3$ for some $\lambda\in\mathbb{R}$.

Set $p(x)=a+bx+cx^2+dx^3$. Then $$ q(x)=(a+\lambda)+bx+cx^2+(d-\lambda)x^3 $$ and we have $$ \begin{align*} f(q(x)) &= (a+\lambda+b+c+d-\lambda) + (b+c)x - (b+c)x^2 \\ &\qquad + (a+\lambda-b-c+d-\lambda)x^3 \\ &= (a+b+c+d) + (b+c)x - (b+c)x^2 + (a-b-c+d)x^3 \\ &= f(p(x)) \end{align*} $$ Therefore, $g([p(x)])=f(p(x))=f(q(x))=g([q(x)])$ which you wanted.