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The question is related to ring polynoms. But to what else?

Given:

  1. $P(X) = X^k + \cdots + a_1 X + a_0$ in $\mathbb{Q}[X]$ is normed and irreducible.
  2. Let $w,v \in \mathbb{C}$ be roots of $P$, such that $q:= v-w$ is in $\mathbb{Q}$.
  3. Define $P_n(X) := P(X-nq)$, with $n \in \mathbb{N}$.

Prove that: $P_n = P$, for all $n \in \mathbb{N}$.

JohnBlue
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  • I don't understand what you mean by "I could already that $P_n \in \mathbb{Q}[X]$ and that $P_n$ is irreducible too in $\mathbb{Q}[X]$". Do you mean, you can see that it is the case? – Pastudent Dec 19 '22 at 14:55
  • Welcome to MSE! If you format your question using MathJax the probability that someone is going to answer your question is higher. For some basic information about writing mathematics at this site see, e.g., here, here, here and here. – linkja Dec 19 '22 at 14:57
  • The n in the definition of your polynomial is probably different from the one in $P_n$? And $P_n = P $ probably means that they are the same polynomial. – linkja Dec 19 '22 at 15:00
  • @pastudent: the exercise consists actually out of three parts, and I finished part 1, and are stuck in part 2. The sentence in "I could already..." summarizes what I successfully showed in part 1, and maybe helpful to answer the question. – JohnBlue Dec 19 '22 at 15:33
  • @linkja: you are probably right, the "n" appears in the definition of the Polynomial P(X)= X^n ... and in the definition of P_n(X). I suspect they can be different variables actually, but as I don't know what this exercise is really about, I am not 100% sure. But thanks for showing me this, I actually didn't notice it and assumed they were different. – JohnBlue Dec 19 '22 at 15:35
  • @linkja you are probably right concerning the meaning of "Pn(X) = P": it is to be shown, that they are the same polynomial. But I can't imagine, how P(X) and P(X-nq) can be the same thing. – JohnBlue Dec 19 '22 at 15:53
  • Hint: see Why is the difference of distinct roots of irreducible $f(x)\in\mathbb{Q}[x]$ never rational?. – Bill Dubuque Dec 19 '22 at 17:54
  • After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. – linkja Dec 20 '22 at 18:56
  • I would say that this isn't a duplicate of https://math.stackexchange.com/questions/87324/why-is-the-difference-of-distinct-roots-of-irreducible-fx-in-mathbbqx-ne, as this might be the final step in the solution, but it definitely isn't obvious to see how it does. Therefor it is on a very similar topic, but the angle and phrasing of the question is different enough, that this justifies it as its own question. – linkja Dec 20 '22 at 19:02

1 Answers1

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It is enough to show the claim for $n = 1$, because if $P = P_1$ then $P_n(X) = P(X-nq)$ $ = P(X-q-(n-1)q)$ $ = P_1(X-(n-1)q) = P(X-(n-1)q) = P_{n-1}(X)$ and by induction $P_n = P$. To Clarify $P_n = P$, means that for all $x \in \mathbb Q$, $P_n(x) = P(x)$. This implies that for all $x\in Q$ and $z \in \mathbb Z$, $P(x+zq) = P(x)$.

Now we take a look at any $P\in \mathbb Q[X]$ and any $q\in\mathbb Q$. Assume $P(X-q) = P(X)$, so $P(x-q) = P(x)$ for all $x\in\mathbb Q$. This is clearly true if $q = 0$, so assume $q \not = 0$. Then we get, like above, that $P(x-nq) = P(x)$, so define the Polynomial $Q_{x_0}(X) = P(X)-P(x_0)$ for any $x_0\not = 0 \in \mathbb Q$. This polynomial is zero for all $x_0+nq$. Those are infinitely many different values for which $Q_{x_0}$ is zero, so $Q_{x_0}$ is the zero-polynomial $0$ and thereby $P = P(x_0)$ a constant polynomial.

With this we can see that, as constant polynomials are not irreducible, $q$ has to equal zero, so the claim now translates to: "If $v,w$ roots of an irreducible polynomial $P\in\mathbb Q[X]$ and $v-w \in \mathbb Q$, then v = w"

Assume $v-w = q \not = 0$ then the polynomial $Q(X) = P(X)+P(X-q)$ has root $v$, but $deg(Q)<deg(P)$, this is a contradiction, as $P$ is normed and irreducible with root $v$, it has to be the minimal polynomial of $v$, but $Q$ also has $v$ as a root. Therefor $v-w = 0 \implies v = w$.

EDIT: The comment by Bill leads to an interesting post on the topic, where I got the last paragraph.

linkja
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    For proofs of that basic claim see here – Bill Dubuque Dec 19 '22 at 17:49
  • I am still struggling to understand your answer, but this is a huge step forward, I had no clue where to start and now there is some light at the end of the tunnel! Thank you very much for your efforts!!!! :) I am delighted! :-) – JohnBlue Dec 19 '22 at 19:00
  • @JohnBlue Where are you struggling? – linkja Dec 19 '22 at 19:03
  • With "Polynomial Qx0(X)=P(X−nq)−P(x0) for any x0≠0∈Q. This polynomial is zero for all nx0. " I don't get, why "P(X-nq) - P(X)" is the zero polynom, maybe I lack some basic theorems. – JohnBlue Dec 19 '22 at 19:09
  • @JohnBlue I changed some things that were a bit messy, I hope that it is easier to understand now, and it shouldn't have been $nx_0$, but $x_0+nq$. – linkja Dec 19 '22 at 20:34
  • Concernig the polynomials $Q_{x_0} = P(X)-P(x_0)$, notice that we have $-P(x_0)$ the evaluation of $P$ at $x_0$ is the zero polynomial because it has infinitely many roots, which only the zero polynomial has. Alternatively you can just say that $P$ has the same value at all $x_0+nq$ which means that it has to be the constant polynomial, in characteristic 0, as any polynomial that is not constant can only take a value finitely many times. see here: https://math.stackexchange.com/questions/1137190/is-there-a-polynomial-that-has-infinitely-many-roots – linkja Dec 19 '22 at 20:34
  • thank you again!! I think I understood it now better! :-) – JohnBlue Dec 20 '22 at 13:26