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I am reading Fraleigh in which Fraleigh proves that $f(x) = x^3 + 3x+ 2$ is irreducible over $\mathbb{Z}[5]$. He explains that the if $f(x) = x^3 + 3x+ 2$ were reducible over $\mathbb{Z}[5]$, then there would exist ar least one linear factor of $f(x)$ of the form $x-a$ for some $a \in \mathbb{Z}[5]$. Then, $f(a)$ would be $0$. However, $f(0) = 2, f(1) = 1, f(2) = 1, f(-1) = -2,f(-2) = -2$.

I don't understand why Fraleigh didn't analyze $f(4)$ and $f(3)$ instead of $f(-1)$ and $f(-2)$ respectively. Can someone please explain? Thanks!

Ricky_Nelson
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    $4 \equiv -1 \pmod 5$ and $3 \equiv -2 \pmod 5$. – Alex Nolte Apr 26 '20 at 02:01
  • $3=-2$ and $4=-1$. – Angina Seng Apr 26 '20 at 02:01
  • I meant to ask why he chose to analyze $f(-1)$ and $f(-2)$ in the first place. Did he do this for the sake of computational ease? – Ricky_Nelson Apr 26 '20 at 02:03
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    Analyzing $-2$ and $-1$ is equivalent to analyzing $3$ and $4$, so most likely. – Ryan Shesler Apr 26 '20 at 02:05
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    Side note: $\Bbb Z[5]$ is an odd notation, since $R[\alpha]$ usually indicates adjoining an element to a ring. Are you referring to $\Bbb Z/5\Bbb Z$? – Greg Martin Apr 26 '20 at 02:14
  • note: $f(-x)+f(x)=4, $ so $f(-x)=4-f(x)$ – J. W. Tanner Apr 26 '20 at 02:14
  • He was being either lazy or efficient, depending on how you like to describe it. Cubing $-1$ is easier than cubing $4$, though just a little. He might also be doing it for a didactic reason, to shake up the student, and to drive home the fact that even though we are using familiar symbols, like $3$ and $4$, they don't mean exactly the same thing as the student is used to. – JonathanZ Apr 26 '20 at 03:02
  • Oh, and @GregMartin's point is right. You should write it as either $\mathbb{Z}/5\mathbb{Z}$ or $\mathbb{Z}_5$. The notiation $\mathbb{Z}[5]$ is at best uncommon and at worst nonsensical, and I've certainly never seen it before. – JonathanZ Apr 26 '20 at 03:06
  • This is the standard (balanced) residue system (see the linked dupe). Likely he chose it for the same reason many do, viz. the smaller numbers simplify calculations (or may yield symmetry based simplifications, being closed under negation) – Bill Dubuque Apr 26 '20 at 04:16
  • I had copied the $\mathbb{Z}[5]$ notation straight from Fraleigh. – Ricky_Nelson Apr 26 '20 at 17:03
  • @Ricky_Nelson: Interesting. I learned from Fraleigh (a long time ago), and his notation looks strange to me now. – JonathanZ Apr 28 '20 at 21:27

3 Answers3

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The set $\{0,1,2,-1,-2\}$ forms a complete residue system modulo $5$, just like $\{0,1,2,3,4\}$. This means all possibilities for the linear factor of $f$ are still covered with the former set as they are with the latter set.

It is just a matter of convention to adopt one system or the other. The more appropriate choice may also depend on context – calculations related to quadratic reciprocity, for example, heavily rely on manipulating negative numbers (e.g. is $-1$ a square modulo $n$ or not).

Parcly Taxel
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Just check the polynomial's value $\pmod 5$ at each of $\{0,1,2,3,4\}=\Bbb Z_5$.

Oh, and in answer to your question, we have $-1\equiv 4\pmod5$ and $ -2\equiv 3\pmod5$.

user26857
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When multiplying modulo $n$ for relatively small $n$ if you choose to use negative numbers instead of higher positive numbers you keep the numbers small at the expense of having to keep track of the sign. This is often easier when using brute force to check for irreducibly of a polynomial and other such computations. For example it can make proofs of quadratic reciprocity easier to follow.

CyclotomicField
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