I would like to know if what I'm saying is correct when I state the following: suppose $z = 4 - i$ and $w = 1 - i$ are two complex numbers, and say I want to adjoin them to $\mathbb{Q}$. The first thing I do is find irreducible polynomials with rational coefficients such that $z$ is a root of the first polynomial and $w$ is the root of the second polynomial respectively.
For $z$ I used the polynomial $(x - (4-i))(x - (4+i)) = x^2 - 8x + 17$. Now, I take the quotient of $\mathbb{Q}[x]$ over the ideal generated by this polynomial, and since it is irreducible in $\mathbb{Q}$, then the quotient ring is a field. By some theorem whose name I forgot, this field is isomorphic to $\mathbb{Q}(4 - i)$.
So by the above construction, we can write the field $\mathbb{Q}(4 - i)$ in set notation like tihs:
$$ \mathbb{Q}(4 - i) = \{a + b(4 - i): a,b\in\mathbb{Q}\}$$
Now, suppose I want to do the same thing for $1 - i$; i.e I want to adjoin $1 - i$ to $\mathbb{Q}$. If I do the same thing with the polynomial $q = (x - (1 - i))(x - (1 + i))$, I will end up with the definition of $\mathbb{Q}(1 - i)$ in set notation like this:
$$ \mathbb{Q}(1 - i) = \{a + b(1 - i): a, b\in\mathbb{Q}\}$$
Now, I have two questions. Are the above two constructions I did correct? Can I just use any arbitrary polynomial with $4 - i$ as a root that is irreducible over $\mathbb{Q}$? Also, if I am correct, am I also correct in saying that the above two fields are isomorphic? They look pretty much the same after expanding out the terms.
Thanks!
If you take the adjoined elements form $\mathbf C$ (or any algebraically closed field containing $\mathbf Q$), your fields are are both equal (not only isomorphic) to $\mathbf Q(i)$ since, say, $\mathrm i=4-(4-\mathrm i)$.
Of course if you make an abstract construction, they're only isomorphic: $\mathbf Q[x]/(x^2-8x+17)$ vs $\mathbf Q[x]/(x^2+1)$.