D+F 13.2.3

Determine the minimal polynomial over $\mathbb{Q}$ for the element $1+i$.

$\textit{Solution}$ : We want a monic, irreducible polynomial of minimal degree over $\mathbb{Q}$ s.t. $m(1+i)=0$. $$(1+i)(1+i)=2i$$ $$(2i)^2=-4$$ So, $$f(x)=x^4+4$$ satisfies $$f(1+i)=0$$ However, $x^4+4$ actually factors partially over $\mathbb{Q}$ as follows: $$x^4+4=(x^2+2)^2-4x^2=(x^2+2)^2-(2x)^2$$ $$=((x^2+2)-(2x))((x^2+2)+(2x))=(x^2-2x+2)(x^2+2x+2)$$ The first of the above quadratics has the property that $$(1+i)^2-2(1+i)+2=2i-2-2i+2=0$$ ...a smaller degree polynomial with $1+i$ as a root. This is consistent with the observation that $$\mathbb{Q}[1+i]=\{a+b(1+i):a,b\in{\mathbb{Q}}\}=\{a+b+bi:a,b\in{\mathbb{Q}}\}$$ $$=\{c+bi:c,b\in{\mathbb{Q}}=\mathbb{Q}[i]$$ ...a degree 2 extension.
By Prop. 11 p. 521, we know that the degree of our field extension should be equal to the degree of our minimal polynomial. So, we may safely conclude that the minimal polynomial for the element $1+i$ with $\mathbb{Q}$ as our base field is $m(x)=x^2-2x+2$.