# Equivalence class of polynomials

$X$ is the set of all polynomials over $\mathbb{R}$. We define an equivalence relation on $X$ such that $p$~$q$ iff $p(0)=q(0)$.

($1$) What is the equivalence class of $p(x)=x$?

($2$) Give a description of $X$/~ by showing a set bijective to $X$/~.

For ($1$), I think the equivalence class is something like all polynomials with zero constants, so $p(x)$ s.t. $\forall p(x)$, $p(0) = 0$.

For ($2$), I've started by defining a function $f:X \to$ $X$/~ s.t. $p(x) = x$ for $\forall x \in X$ but am pretty sure this isn't right.

• What is the equivalence relation in question? – Git Gud Mar 3 '13 at 22:00
• Sorry -- the equivalence relation on X s.t. p~q iff p(0)=q(0). – Juniper Leaf Mar 3 '13 at 22:03
• Edit it into the question. – Git Gud Mar 3 '13 at 22:04
• @JuniperLeaf As for $(1)$, the equivalence relation of $x$ is the set of polynomials $q(x)$ such that $q(0)=p(0)=0$, that is $[x\textbf{]}=\{xg(x): g(x)\in \Bbb R[x\textbf{]}\}$. – Git Gud Mar 3 '13 at 22:09
• Do you know any abstract algebra? In particular, do you know about rings, and quotient rings? – Math Gems Mar 3 '13 at 22:37

Based on your comments, we see that this is an exercise in set theory, not in ring theory.

Relations like yours $\rm\ p\sim q \smash[t]{\overset{\ def}{\iff}} f(p) = f(q),\:$ for $\rm\:f(p) = p(0),\:$ are always equivalence relations.

Generally, suppose $\rm\,\ u\sim v\ \smash[t]{\overset{\ def}{\iff}}\, f(u) \approx f(v)\$ for some function $\rm\,f\,$ and equivalence relation $\,\approx.\, \$ Then the equivalence relation properties of $\,\approx\,$ transport (pullback) to $\,\sim\,$ along $\rm\,f\,$ as follows:

• reflexive $\rm\quad\ f(v) \approx f(v)\:\Rightarrow\:v\sim v$

• symmetric $\rm\,\ u\sim v\:\Rightarrow\ f(u) \approx f(v)\:\Rightarrow\:f(v)\approx f(u)\:\Rightarrow\:v\sim u$

• transitive $\rm\ \ \ u\sim v,\, v\sim w\:\Rightarrow\: f(u)\approx f(v),\,f(v)\approx f(w)\:\Rightarrow\:f(u)\approx f(w)\:\Rightarrow u\sim w$

Such relations are called (equivalence) kernels. One calls $\, \sim\,$ the $\,(\approx)\,$ kernel of $\rm\,f.$

For $(2)$ you are correct, the equivalence class of $\rm\:p(x)=x\:$ is precisely the set of all polynomials $\rm\:q\:$ such that $\rm\:q(0) = p(0) = 0.\:$

Hint for $(3)\!:\:$ choose one "simple" representative of each equivalence class, e.g. one of least degree.

• Thank you that really cleared things up. – Juniper Leaf Mar 3 '13 at 23:19

Your solution for (1) is correct. For (2) an idea: define a function

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

(i) Check $\,\phi\,$ is a surjective ring homomorphism

(ii) Show that $\,\ker\phi=\langle x\rangle\,$ = the ideal generated by the identity polynomial.

(iii) Use the first isomorphism theorem.

Added by OP's comment: A simpler (I think) approach could be to define a map of set

$$\phi: X/\sim\;\longrightarrow\Bbb R\,\;\;,\;\;\phi[f(x)]:=f(0)$$

a) Check the map above is well defined, i.e.: $\,[f(x)]=[g(x)]\Longrightarrow f(0)=g(0)\,$

b) Check the map $\,\phi\,$ is injective and surjective.

• Thank you for your help. I have to admit though that your solution is a little beyond me as I haven't learned any of that yet. – Juniper Leaf Mar 3 '13 at 22:17
• I see, @JuniperLeaf...it would perhaps be a good idea to add something in your question/profile about your mathematics level. – DonAntonio Mar 3 '13 at 22:20
• @Juniper, I added a second idea to my answer. Check this. – DonAntonio Mar 3 '13 at 22:24
• Thanks! Will work on this. Have also edited my profile. – Juniper Leaf Mar 3 '13 at 22:27
• @GitGud? It's just the same, isn't it? $\,\phi[f(x)]=f(0)\,$ so showing the identity by means of $\,\phi\,$ or of its definition on each equivalence class is the same... – DonAntonio Mar 3 '13 at 22:36

Hint: If $p = a_0 + a_1 x + a_2 x^2 + \ldots + a_n x^n$, then $p(0) = a_0$. So two polynomials are equivalent iff ...