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Let $R$ be an integral domain, $P$ be a prime ideal of $R$. Let $f\in R[x]$, where $f(x):=a_nx^n +\dots + a_1x+a_0$, such that $a_n\not\in P$, $a_i\in P$ ($i<n$), $a_0\in P^2$. Then $f$ is irreducible.

I was trying to prove this by analogy with the proof in the book Contemporary Abstract Algebra by Gallian, but I have no idea what the author means when he says "... so there is a least integer $t$ such that $p\nmid b_t$. Also, where he says "Let $a_t=b_tc_0 + b_{t-1}c_1+\dots+b_0c_0$" - it is not clear where this is coming from and what this means. [Aside: Actually, I have been penalized several times for omitting much more obvious things in my proof. So, if Gallian wrote his proofs in an assignment or an exam in my university, he'd surely not get top marks].

I would appreciate your help in trying to figure out this stuff.

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sequence
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1 Answers1

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In fact he wants to say the smallest $t$ such that $p$ does not divide $b_t$ such a $t$ exists since $p$ does not divide $b_r$ So $p$ divides $b_{t-1},...,b_0$.

The expression of $a_t$ comes from the multiplication of $g$ and $h$, he writes the coefficient $t$ of $gh$ ,if you write $a_t=b_ta_0+b_{t-1}c_1+...+b_0a_t$, you have $p$ divides $b_{t-i}a_i, i=1,...,t$. Since $p$ divides $b_{t-i}$, you deduce that $p$ divides $b_ta_0$ since $p$ divides $a_t$. This is impossible since $p$ is a prime and $p$ does not divide neither $a_0$ nor $b_t$.

Tsemo Aristide
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  • Sorry, but you basically said the same words that the author of the book mentioned above. Unfortunately, this does not add any clarity. So, if $p$ does not divide $b_r$, then why is it supposed to divide some $b_t$. What is $b_t$, in the first place?

    For your second paragraph, it is also not clear what the index $t$ stands for. Is $0 \le t< n$? It is not clear since the author did not define $t$.

     – sequence Mar 18 '17 at 02:02
  • $g(x)=b_0+b_1x+..+b^rx^r$. $t$ is an integer in the set ${0,1,...,r}$. The author says $p$ divides $b_0$ and $p$ does not divide $b_r$, so you can take $t$ to be the smallest integer such that $p$ does not divide $b_t$. – Tsemo Aristide Mar 18 '17 at 02:20
  • "Let us say $p|;b_0$". So $p$ DOES divide $b_t$ for some $ t\in {0,1,...,r}.$ He did not need to define $t.$ He defined ${b_t: 0\leq t\leq r}.$ And since $p\not |;b_r$ but $p|;b_0$ there exists a least non-negative integer $t$ such that $p\not |;b_t$, and we must have $t\leq r$.... It is customary in def'ns such as $g(x)=b_rx^r+...b_0$ to assume that $b_t$ only exists for integer $t\in {0,...,r}$. – DanielWainfleet Mar 18 '17 at 02:28
  • Does the author of the book mean that $t$ is at most $r$? – sequence Mar 18 '17 at 02:46