1

Show that $1+X^4$ is reducible over $\mathbb Z_p$ for every prime $p$.

MY ATTEMPT==>I have used Fermat's theorem for this as $X^{p-1}≡1\bmod p$, then this can also be written in the form $1≡X^{p-1}\bmod p$. So, in $\mathbb Z_p$, $$1+X^4 =X^{p-1}+X^4 = X(X^3 + X^{p-2})$$ as $p\ge2$, i.e., $1+X^4 = X(X^3 + X^{p-2})$.

Am I correct?

user26857
  • 52,094
Styles
  • 3,539
  • 1
    but is my approach correct? – Styles Apr 18 '16 at 09:56
  • 1
    No. It is not correct. The polynomial $X^{p-1}$ is not the same polynomial as the constant $1$, Little Fermat notwithstanding. Two polynomials $p(X)$ and $q(X)$ are equal if all the powers of $X$ have the same coefficients in $p$ and $q$. You may be confused by the difference between a polynomial and a polynomial function. You are asked to show that the polynomial $X^4+1$ is a product of two lower degree polynomials. – Jyrki Lahtonen Apr 18 '16 at 10:36
  • @JyrkiLahtonen,you meant to say that we are not sure about the integer nature of X. – Styles Apr 18 '16 at 14:06
  • $X$ is a variable. It is not an integer (nor an element of $\Bbb{Z}_p$). It is a formal variable. You can substitute many a thing in place of $X$, but it is "more than the totality of all the possible substitutions". Mind you, this takes some time getting used to. I didn't get used to thinking about it as a formal variable right away myself. Read more about the difference between (formal) polynomials and polynomial functions. A polynomial with coefficients in $\Bbb{Z}_p$ cannot be equated with a function from $\Bbb{Z}_p$ to itself. There are only finitely many such functions, – Jyrki Lahtonen Apr 18 '16 at 15:09
  • (cont'd) but infinitely many polynomials. For example $1,X,X^2,\ldots$ are all different polynomials. Among other things they have different degrees. – Jyrki Lahtonen Apr 18 '16 at 15:15
  • @JyrkiLahtonen:now i got it.thanks for help and suggestion. – Styles Apr 20 '16 at 07:12
  • Glad to hear that! – Jyrki Lahtonen Apr 20 '16 at 08:20

0 Answers0