Minimal Polynomial of $1+i$ over $\Bbb Q$

Question

What is minimal polynomial of $1+i$ over $\mathbb{Q}$?

My attempt:

Let $x=1+i$, hence $(x-1)^2= i^2 = -1$, which is not possible in $\mathbb{Q}$, hence squaring further we obtain the minimal polynomial as $ x^4-4x^3+4x^2-4$.

Doubt:

1. Is this a required polynomial?

2. Are the justifications for the steps enough? ( particularly where I have written that this is not possible in $\mathbb{Q}$ )

Answer 1

You already show that for $x=1+i$ you have $$(x-1)^2=-1.$$ This means that $x$ is a root of $$(X-1)^2+1=0.$$ In particular, the minimal polynomial divides this quadratic polynomial. From here it is not difficult to conclude that this is the minimal polynomial.

Answer 2

The conjugate of $1 + i$ is $1 - i$, so try:

$\begin{align*} (x - (1 + i)) (x - (1 - i)) &= x^2 - 2 x + 2 \end{align*}$

No linear polynomial over $\mathbb{Q}$ can have complex zeros, this is minimal.