Bijection from set of equivalence classes to $\mathbb R$

Question

In $\mathbb R[X]$, define an equivalence relation ~ by $P_1$~$P_2$ if $P_1-P_2$ is divisible by X.

I have shown that $X$ is an equivalence relation.

Let $\mathscr Q$ denote the set of equivalence classes of ~ in $\mathbb R[X]$.

I now have to find an explicit bijection $\mathscr Q \to \mathbb R$. Any help would be appreciated. Thanks.

Answer 1

Hint $ \ \rm x\mid p\!-\!q\!\iff\! p\equiv q\pmod{\! x}.\, $ $\rm\, p(x)\equiv p(0)\pmod{\! x},\,$ so $\rm\ p\sim q\!\iff\! p(0) = q(0).$

So the class of $\rm\,p\,$ consists of every polynomial with the same constant term $\rm\,p(0).\,$ A natural choice of representative is the "simplest" (lowest degree) element, i.e. the constant polynomial $\rm\,p(0).$

Remark $ $ If perchance you already know about quotient rings and evaluation homomorphisms of polynomial rings then you should examine this problem from that perspective, noting that $\rm\,p(0)\,$ is the result of applying the evaluation hom $\rm\,x\mapsto 0\,$ to $\rm\,p(x).$ Then, applying the First Isomorphism Theorem lifts our set-theoretic isomorphism to a ring-theoretic isomorphism $\,\Bbb R[x]/(x)\cong \Bbb R$

Answer 2

Hint:

Define a map

$$\phi:\Bbb R[x]\to\Bbb R\;,\;\;\phi(p(x)):=p(0)$$

The nicest thing of the above is that $\;\phi\;$ is a ring homomorphism...

Answer 3

Here is a hint for DonAntonio's hint:

Note that if $f\sim g$ then $f(0)=g(0)$. Use this to prove that the map $[f]/\sim\mapsto f(0)$ is well-defined, and it is a bijection.

Answer 4

We have $P_1\sim P_2\iff x|(P_1-P_2)\iff \exists Q$ such that $P_1(x)=xQ(x)+P_2$ and by the Euclidean division we have $$P_1(x)=xQ(x)+P_1(0)$$ hence $$P_1\sim P_1(0)$$ Can you see now the bijection?