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Let $\alpha_i$ is a root of $k(t) \in \mathbb Z[t]$. Let $p_0(t)$ be the minimal polynomial of $\alpha_1$ over $\mathbb Z$. After this, it is written in a text that

Then, we have $p_0(t)\mid k(t)$. So, $\alpha_1$ is also a root of $k(t)$.

My question is how we know $p_0(t)\mid k(t)$?

EDIT

Let me rephrase my problem.

  1. the first statement: $\alpha_i$ is a root of $k(t) \in \mathbb Z[t]$, this is according to a previous hypothesis.

  2. the second statement: Let $p_0(t)$ be the minimal polynomial of $\alpha_1$ over $\mathbb Z$, this is true because for $\alpha_1$, there exists a minimal polynomial.

  3. the third statement: We have $p_0(t)\mid k(t)$.

Now every algebraic number $\alpha_1$ has a minimum polynomial, say $p_0$, and if polynomial $k(t)$ has root $\alpha_1$ then $p_0$ divides $k(t)$ (in general every polynomial that$\alpha_1$ satisfies). But at this point we don't know $\alpha_i=\alpha_1$, then how we know $p_0(t)\mid k(t)$?

  1. the fourth statement: "So", $\alpha_1$ is also a root of $k(t)$

Now if we already used the fact that $\alpha_1$ is also a root of $k(t)$ imply $p_0(t)\mid k(t)$ then how can we say $p_0(t)\mid k(t)$ implies $\alpha_1$ is also a root of $k(t)$?

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    $\begingroup$ What was wrong with my answer to your previous version of the question? It appears you have deleted it. Is that supposed to be $\alpha_i$ in the second line, or is it $\alpha_1$ like all the rest? If so, you started by assuming $\alpha_1$ was a root of $k(t)$ $\endgroup$ Aug 9, 2020 at 15:06
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    $\begingroup$ That is exactly what I answered last time. If $\alpha_1$ is a root of $p_0(t)$, it is a root of anything $p_0(t)$ divides. $\endgroup$ Aug 9, 2020 at 15:18
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    $\begingroup$ Finally, we have a good question, which I don't know the answer to. It is similar to unique factorization. The claim is that for every algebraic number $\alpha$ the minimum polynomial that $\alpha$ satisfies divides every polynomial that $\alpha$ satisfies. That implies there are not two incompatible polynomials that $\alpha$ satisfies. Do you know that $\Bbb Z[x]$ is a unique factorization domain? $\endgroup$ Aug 10, 2020 at 1:41
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    $\begingroup$ You have chosen $k(t)$ so that $\alpha$ satisfies it. That is what $\alpha$ being a root of $k(t)$ means. Then the theorem you quote is what you need because $p_0(t)$ is the minimal polynomial that $\alpha$ satisfies. The text is applying this theorem. $\endgroup$ Aug 10, 2020 at 5:06
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    $\begingroup$ I asked about that in a previous comment. Presumably the subscript should stay constant through the paragraph. As it does not, one is likely a typo. I can't tell which without more context. $\endgroup$ Aug 10, 2020 at 5:13

2 Answers 2

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He has constructed $k(t)$ so that $\alpha_1$ is a root of $k(t)$ by showing that $h(g(\alpha_1))=0$ and naming $k(t)=h(g(t))$. Then if $p_0(t)$ is the minimal polynomial of $\alpha_1$, the theorem you quote in the comments "(for) every algebraic number α the minimum polynomial that α satisfies divides every polynomial that α satisfies." gives us $p_0(t)|k(t)$ because $k(\alpha_1)=0$

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  • $\begingroup$ "gives us $p_0(t)|k(t)$ because $k(\alpha_1)=0$ " ...Oh god! not again!! Ross, my friend, this is not what the thesis says, thesis says, "$p_0(t)|k(t)$ So,(THEN) $k(\alpha_1)=0$", also note that the construction u r talking is made upon $\alpha_i$, there is no specification in the thesis that $i=1$ or the $\alpha_i$ is root for all $i$, it says for some $\alpha_i$ .. this is the source of my confusion which i could not express properly!! :( $\endgroup$ Aug 10, 2020 at 15:28
  • $\begingroup$ Also if we know already $k(\alpha_1)=0$, then why we need $p_0(t)|k(t)$? Are we saying $p_0(t)|k(t)$ to show $k(\alpha_1)=0$ or $k(\alpha_1)=0$ implies $p_0(t)|k(t)$? Which one? We don't need to show $k(\alpha_1)=0$ implies $p_0(t)|k(t)$ because our objective is to show that $k(\alpha_1)=0$ not to show $p_0(t)|k(t)$. $\endgroup$ Aug 10, 2020 at 15:37
  • $\begingroup$ He wants to show that $k(t)$ is a power of $p_0(t)$. First he says that $p_0(t)|k(t)$ because of the theorem. Then he says let $k(t)=p_0^n(t)q(t)$ with $q(t)$ coprime to $p_0(t)$ and shows that $q(t)$ is constant. $\endgroup$ Aug 10, 2020 at 17:42
  • $\begingroup$ The theorem u r referencing does not apply here. This is the key problem, I am sorry to say but u failed to answer at least clarify. $\endgroup$ Aug 12, 2020 at 8:57
  • $\begingroup$ @Andrew Forgive me, but you're being pretty demanding of strangers' kindness and generosity while showing very little gratefulness for their time and effort. Answering you is likely to result in a lengthy and possibly stressful back-and-forth, which greatly reduces the desire of most people to answer. $\endgroup$ Aug 16, 2020 at 23:48
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I think the result that you require is this one. Since $\alpha_i$ is a root of $k(t)\in Z[t]$, we can apply the embedding that carries to $\alpha_i$ to $\alpha_1$, say $\sigma$, to the polynomial $k(t)$. If $\alpha_i$ is a root of $k(t)$, then $\sigma(\alpha_i)=\alpha_1$ is a root of $\sigma(k(t))=k(t)$. This happens because $0=\sigma(0)=\sigma(k(\alpha_i))=k(\sigma(\alpha_i))=k(\alpha_1)$.

Therefore, your $\alpha_1$ also satisfies the same polynomial $k(t)$ and is a root. So, the minimal polynomial of $\alpha_1$ must divide it. Can you take it from here?

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  • $\begingroup$ Hi Umesh. thanks for ur answer, but I have 2 issues: 1. the embedding you are refereeing is not mentioned in the paragraph (last paragraph on page 25), it does not look like the author used that embedding to say $p_0(t)|k(t)$ 2. The author first proved that $p_0(t)|k(t)$ then implies $\alpha_1$ is root (look the first line on page 26), not that the author proved $\alpha_1$ is a root then implied $p_o(t)|k(t)$, plz reconsider. Link: scribd.com/document/471950968/… $\endgroup$ Aug 19, 2020 at 11:17
  • $\begingroup$ I read the proof in the link before I gave the answer. The embedding is mentioned in same proof right before equation 4.1. I'm aware that he doesn't make use of the arguement explicitly. But he has indeed used the arguement. This arguement, however, is one where my professors just believed that we would know. If this argument was applied in my courses, it was almost never explicitly stated. $\endgroup$ Aug 19, 2020 at 11:32
  • $\begingroup$ The problem is it looks like there is another reasoning, the way it is written, I mean my second point, The author first proved that $p_0(t)|k(t)$ then implies $α_1$ is root (look the first line on page 26), not that the author proved $α_1$ is a root then implied $po(t)|k(t)$, so I cant accept your answer, if I accept $\alpha_1$ is a root then I dont need to post this question, i hope you understand. $\endgroup$ Aug 19, 2020 at 11:42
  • $\begingroup$ If this link doesn't clear this up, I don't know what will. math.stackexchange.com/questions/1160606/… $\endgroup$ Aug 19, 2020 at 12:10
  • $\begingroup$ I understand what you say, but could u plz explain what is the point of saying $p_0(t)|k(t)$ in text if we already know (by using embedding) that $\alpha_1$ is a root? $\endgroup$ Aug 19, 2020 at 12:14

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