Question

Give an example of a polynomial $\mathit{f}\left(x\right)\mathbf{\in }\mathit{Z}\left[x\right]$and a prime p such that $\mathit{f}\left(x\right)$is reducible inrole="math" localid="1649243074209" $\mathit{Q}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$ but $\mathit{f}\left(x\right)$is irreducible in$\mathit{f}\left(x\right)\mathbf{\in }\mathit{Z}\left[x\right]$. Does this contradict Theorem 4.25?

Short answer

The example is$2x^4+x^2+x+1=\left(x^2-x+1\right)\left(2x^2+2x+1\right)$

Step-by-step solution

Write the theorem

According to Theorem 4.25, assume that$f\left(x\right)=a_k{x}^k+\dots +a_{1}x+a_0$ is a polynomial with integer coefficients, and consider p as a positive prime. It does not divide a_k . This implies if$f\left(x\right)$ is irreducible in $Z_p\left[x\right]$, then $f\left(x\right)$is irreducible in$Q\left[x\right]$ .

Write the example

Consider.$2x^4+x^2+x+1=\left(x^2-x+1\right)\left(2x^2+2x+1\right)$

Rewrite the equation in Z₂asfollows:

$2x^4+x^2+x+1=x^2+x+1$

The obtained polynomial is irreducible. This example does not contradictTheorem4.25becauseone of the assumptions is that the prime does not divide the leading term of the polynomial.

Hence, the example is $2x^4+x^2+x+1=\left(x^2-x+1\right)\left(2x^2+2x+1\right)$ , and in Z₂ , we have $2x^4+x^2+x+1=x^2+x+1$, that is irreducible.