Why can't we extend any field by simply adding a new symbol to it?

Question

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?

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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?

Answer 1

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:

[...] there exists no further field extension of $\mathbb{C}$.

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.

Answer 2

It's not true that there are no field extensions of $\mathbb{C}$. For example, there is $\mathbb{C}(x)$, the field of rational functions of $x$ over $\mathbb{C}$. However, there is no finite-dimensional field extension of $\mathbb{C}$, or any algebraically closed field $F$.

First, a lemma. If $F$ is algebraically closed, then by definition every polynomial $p(x)$ over $F$ has a root; but a stronger fact is true - that all roots of $p(x)$ must be in $F$. This is true by induction on degree, since if $p(x)$ has a single root in $\alpha \in F$ then we can write $p(x) = (x-\alpha) q(x)$ for $q(x) \in F[x]$, and inductively $q(x)$ must have all its roots in $F$.

Now suppose $K: F$ is a proper, finite-dimensional field extension of $F$ where $F$ is algebraically closed. Choose $\alpha \in K \backslash F$. Then $\alpha^k$ cannot be in $F$ for any $k > 0$, because then $\alpha$ would be a root of $x^k - \alpha^k$ and so would be in $F$, since $F$ is algebraically closed. Then the set $\{1, \alpha, \alpha^2, \alpha^3, \cdots\}$ cannot be linearly independent over $F$, because if so, this would mean $K$ is infinite-dimensional over $F$. So, there are coefficients $c_0, \ldots, c_n\in F$ with $c_0 + c_1 \alpha + \cdots + c_n \alpha^n = 0$, ie, $\alpha$ is a root of $p(x) = c_0 + c_1x + \ldots + c_nx^n \in F[x]$. But since $F$ is algebraically closed, $\alpha \in F$ (by the lemma), a contradiction.

Answer 3

Suppose we add an element $z$ to out field $\mathbb{F}$ that is algebraically complete (all polynomials can be factored linearly). And there is some polynomial $p(x)$ such that $p(z) = 0$.

We factor $p(x)$ in $\mathbb{F}$ into linear terms -- $(x - a_0)...(x - a_n) = t_0 .. t_n$, with $a_n \in \mathbb{F}$, where $t_i = (x-a_i)$. Now, if $t_i(z)$ is zero then $z-a_i = 0$, which means $z = a_i$ and $z \in \mathbb{F}$. No extension.

If $z-a_i \ne 0$ for all $i$, then $0 = p(z) = (z-a_0)...(z-a_n)$ is a product of non-zero terms whose result is 0 - we have a zero divisor. So we don't have a field.

Suppose $\mathbb{F}[z]$ has dimension K. Then any set of K+1 terms are linearly dependent over $\mathbb{F}$, such as ${1, z, z^2, z^3, ..., z^K}$. This produces a K-degree polynomial $p(x)$ over $\mathbb{F}$ for which $p(z) = 0$, running into the above problem.

Basically once a field is algebraically complete, you run out of room to add algebraic extensions to it - any attempt to do so introduces a zero divisor.

If your field isn't algebraically complete you can add algebraic extensions, until you hit the problem of completing your field.

This doesn't block infinite dimensional or non-algebraic extensions of your field, however.