After trying to recall some fundamental field theory, I got very confused at the notion of field extensions. For example, when we make $\mathbb{C}$ out of $\mathbb{R}$, we can simply think of it as adding $i$, which is a symbol for which $i^2 = -1$. So: $\mathbb{C} = \mathbb{R}(i)$.
However, $\mathbb{C}$ is algebraically closed, since each polynomial with complex coefficients has a root in $\mathbb{C}$. I suppose what confuses me is why it follows from here that there exists no further field extension of $\mathbb{C}$. This is always stated as obvious, but I can not make a formal argument.
Why can't we always simply add a new symbol $z$ to a given field $k$, which squares to any element of $k(z) = \{ a + b z \; | \; a,b \in k \}$. For simplicity, let's say that $z^2 = 1$. I know that we already have $1^2 = 1$ and $(-1)^2 = 1$, but why is this an issue? Why can't we have two distinct elements square to the same element?
By defining addition as
$$ (a+bz) + (c+dz) = (a+c) + (b+d)z $$
and multiplication as
$$ (a+bz)\cdot(c+dz) = ac + adz + bcz + bdz^2 = (ac + bd) + (ad + bc)z,$$
all the properties for being a field are easily verified. So, what's the issue here?
Edit: Thank you everyone for very insightful comments. As a follow-up, since we can do extensions of any fields (although they are not themselves necessarily fields), how can we conclude that an algebraically closed field (say $\mathbb{C}$) has no finite field extension, i.e. none of such extension by some symbol is a field?
Best Answer
Take your example of adding to $\mathbb{C}$ a new element $z \notin \mathbb{C}$ such that $z^2=1$. The resulting structure would not be a field, because it would have zero divisors. $$(z-1)(z+1) = z^2 - 1 = 0,$$ while $$z \neq 1 \implies z-1 \neq 0, \text{ and } z \neq -1 \implies z+1 \neq 0.$$ The same happens whenever you add a new root to a polynomial that already splits into linear factors.
Also, it is not correct to say:
You CAN in fact extend any field by adding a new element, as long as it is transcendental, i.e. not a root of any polynomial. It just will be a transcendental extension instead of an algebraic one.