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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.

user112495
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4 Answers4

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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...

DonAntonio
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  • Sorry, I don't really understand what you're doing there. Could you try and explain it a bit more? – user112495 Dec 14 '13 at 19:52
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    Well, check the relation you define actually means $$;P_1\sim P_2\iff x\mid(P_1-P_2)\iff P_1(x)-P_2(x)=xQ(x)\iff P_1(0)=P_2(0);$$ Hmm...have you studied already algebraic structures? Groups , rings...? – DonAntonio Dec 14 '13 at 19:54
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    DonAntonio: I think that the term "ring homomorphism" might be more confusing than helpful in this context. – Asaf Karagila Dec 14 '13 at 19:56
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    "Ant Onion", @AsafKaragila ?? Anyway, I think you may be right if the OP hasn't yet covered that, and I think it may be he hasn't. – DonAntonio Dec 14 '13 at 19:57
  • We have studied groups and rings (although in a different module). But we haven't come across the term homomorphism before. – user112495 Dec 14 '13 at 19:58
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    Uh? Then you're beginning to study that thing, or you haven't really studied them...Anywaym Asaf's answer greatly complements my answer and, in fact, shows you the way to prove what you want without rings and homomorphisms. – DonAntonio Dec 14 '13 at 19:59
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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?

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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.

Asaf Karagila
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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 $\,\Bbb R[x]/(x)\cong \Bbb R$

Bill Dubuque
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