5

For this exercise, you are supposed to show that the polynomial $$ x^4+1 $$ is irreducible in $\mathbb{Z}_5$. However, I found that $$ (x^2+2)(x^2+3) = x^4+5x^2+6 = x^4+1. $$

Is the question wrong or am I?

Book citation: Pinter, Charles C. "A book of abstract algebra. Reprint of the second (1990) edition." (2010).

J. W. Tanner
  • 60,406
dryoung
  • 71
  • No, I'm pretty sure that $(x^2+2)(x^2+3)$ would be equal to $x^4\color{red}{+5x^2+6}$ and not $x^4+1$. Would you please show how you got that $(x^2+2)(x^2+3)=x^4+1$? – CrSb0001 May 30 '23 at 21:35
  • 7
    @CrSb0001 The OP is doing this over $\mathbb Z_5$ – balddraz May 30 '23 at 21:38
  • @0XLR Ah sorry, I must have been thinking of a different set there :\ – CrSb0001 May 30 '23 at 21:39
  • 3
    Well, we can't argue with direct computations. This equality shows the polynomial is reducible over $\mathbb{Z_5}$. – Mark May 30 '23 at 21:43
  • @Mark That's about what I thought but it seemed too simple a problem to not only be wrong but I believe you are also meant to use this to prove a part of a later question in the chapter which makes it worse somehow. – dryoung May 30 '23 at 21:46
  • 3
    In fact, $x^4 + 1$ is 'famous' for beeing reducible over every finite field, but irreducible over $\mathbb Q$. – Zag May 30 '23 at 21:53
  • 1
    Yes, there are many sources proving that $x^4 + 1$ is reducible over finite fields check Wikipedia itelf and examples on this site e.g.

    https://math.stackexchange.com/questions/77155/irreducible-polynomial-which-is-reducible-modulo-every-prime

     – balddraz May 30 '23 at 22:01
  • 3
    This seems to be an instance of the common student error that a polynomial with no root must be irreducible. By the way, write $+1=-4$, and the factorization is obvious. – Ted Shifrin May 30 '23 at 22:10

0 Answers0