# Is Complex Numbers the biggest field? If yes, is there any easy proof to understand it?

1. Is the Complex Numbers the biggest field?
2. If yes, does anyone have a "simple"/"easy to understand" proof?
• How do you define "size"? Commented Nov 22, 2013 at 19:43
• The field of rational functions $\mathbb{C}(t)$ contains $\mathbb{C}$. Commented Nov 22, 2013 at 19:44

It depends on your notion of "biggest". The complex numbers aren't the biggest field in the sense that you could adjoin a new variable to them, say $\mathbb{C}(x)$, and obtain a new field into which $\mathbb{C}$ sits embedded as a subfield. They also aren't the biggest field in terms of cardinality, since you could probably put field structures on sets of arbitrarily large cardinalities.

But they are biggest, in the sense they are algebraically closed. This means that if you try to adjoin a new element to them that satisfies some polynomial relation, you... can't. The variable $x$ in $\mathbb{C}[x]$ doesn't satisfy any relations, so this extension isn't algebraic. But something like $\mathbb{C}[x]/{(3x^2 + 1)}$, where we add in a "new" thing $\alpha$ satisfying the polynomial equation $3\alpha^2 + 1 = 0$, is an algebraic extension. But you know that every complex polynomial has a root, so we can't add in this $\alpha$ to $\mathbb{C}$ - it's already there. On the other hand, $\mathbb{R}[x]/{(3x^2 + 1)}$ is "bigger" than $\mathbb{R}$ since we added in a new thing - a solution to $3\alpha^2 + 1 = 0$. This didn't already live in $\mathbb{R}$, so adding this thing actually extends the field.

• Perhaps you might like to explain the importance of polynomials and polynomial relations. Why can't the OP consider solving non-polynomial equations, and why can't the OP consider quotients like $\mathbb{C}[x]/(\operatorname{e}^x)$? In short, why do we confine ourselves to polynomials when working with fields? Commented Nov 22, 2013 at 19:47

It depends on the interpretation of the word 'biggest'. There are certainly fields of larger cardinality than $\mathbb C$ that contain $\mathbb C$ (for instance, take any set $S$ and consider the field $\mathbb C(S)$ where the elements of $S$ are adjoined as independent transcendental elements). So, if you measure size by cardinality, then for every field $\mathbb F$ there is a larger field containing it, and thus, in particular, $\mathbb C$ is not largest.

However, if by biggest you mean that every polynomial over it has a root in it, in other words being algebraically closed, then $\mathbb C$ is well-known to be algebraic closed and thus in that sense $\mathbb C$ is the largest algebraic extension of $\mathbb R$.

• Perhaps you could use the chain icon to include some hyperlinks to help the OP understand this very detailed reply. There are many technical terms used here, and I think that hyperlinks would make this answer more accessible. Commented Nov 22, 2013 at 19:53
• @FlybyNight I don't find that your comments (here and to the other answer) are constructive. Any answer would have to omit some details since this is not a textbook. Both Zach and myself in our answers used very standard terminology and our answers are correct and to the point. Your comments feel argumentative, as if you were looking for some detail that we left out and said "oh, but it would be even better to include that detail in the answer". If you find a missing detail a problem, then you can add short comment filling the gap yourself. Commented Nov 22, 2013 at 21:18