Question

Show that${\mathit{x}}^{\mathbf{3}}\mathbf{-}\mathbf{6}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathit{i}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{+}\mathbf{3}\mathit{i}$ is irreducible in$\left(\mathrm{\mathbb{Z}}\left[i\right]\right)\left[\mathit{x}\right]$ .[ Hint :Exercise 11 .]

Short answer

It is shown that$x^3-6x^2+4ix+1+3i$ is irreducible in$\left(\mathrm{\mathbb{Z}}\left[i\right]\right)\left[x\right]$ .

Step-by-step solution

Referring to Theorem 4.24 (Eisenstein’s Criterion)

Theorem 4.24

Let $\mathit{f}\left(x\right)\mathbf{=}{\mathit{a}}_{\mathbf{n}}{\mathit{x}}^{\mathbf{n}}\mathbf{+}\mathbf{\dots }\mathbf{+}{\mathit{a}}_{\mathbf{1}}\mathit{x}\mathbf{+}{\mathit{a}}_{\mathbf{0}}$, be a non-constant polynomial with integer coefficients. If there is a prime p such that p divides each of ${\mathit{a}}_{\mathbf{0}}\mathbf{,}{\mathit{a}}_{\mathbf{1}}\mathbf{,}\mathbf{\dots }\mathbf{,}{\mathit{a}}_{\mathbf{n}\mathbf{-}\mathbf{1}}$but p does not divide ${\mathit{a}}_{\mathbf{n}}$ and ${\mathit{p}}^{\mathbf{2}}$does not divide ${\mathit{a}}_{\mathbf{0}}$ then, $\mathit{f}\left(x\right)$ is irreducible in $\mathrm{\mathbb{Q}}\left[\mathit{x}\right]$.

Showing that x3-6x2+4ix+1+3i is irreducible in ℤix

Factorizing the coefficient of$x^3-6x^2+4ix+1+3i$as:

role="math" localid="1653730815594" $\begin{array}{rcl}1+3i& =& \left(1+i\right)\left(2+i\right)\\ 4i& =& {\left(1+i\right)}^2{\left(1-i\right)}^2·2\\ 6& =& \left(1+i\right)\left(1-i\right)·3\end{array}$

From the factorization of coefficients, we can see that the irreducible element$\left(1+i\right)$ divides all the coefficients of$x^3-6x^2+4ix+1+3i$ .

But${\left(1+i\right)}^2=2i$ does not divide the coefficient$1+3i$ .

Therefore, according toExercise 11and Theorem 4.24, the polynomial$x^3-6x^2+4ix+1+3i$is irreducible.

Hence, it is proved that $x^3-6x^2+4ix+1+3i$is irreducible in$\left(\mathrm{\mathbb{Z}}\left[i\right]\right)\left[x\right]$ .