# The Unapologetic Mathematician

## The Complex Numbers

Yesterday we defined a field to be algebraically closed if any polynomial over the field always has exactly as many roots (counting multiplicities) as we expect from its degree. But we don’t know a single example of an algebraically complete field. Today we’ll (partially) remedy that problem.

First, remember the example we used for a polynomial with no roots over the real numbers. That is $p=X^2+1$. The problem is that we have no field element whose square is $-1$. So let’s just postulate one! This, of course, has the same advantages as those of theft over honest toil. We write our new element as $i$ for “imaginary”, and throw it in with the rest of the real numbers $\mathbb{R}$.

Okay, just like when we threw in $X$ as a new element, we can build up sums and products involving real numbers and this new element $i$. But there’s one big difference here: we have a relation that $i$ must satisfy. When we use the evaluation map we must find $\mathrm{ev}(X^2+1,i)=0$. And, of course, any polynomial which includes $X^2+1$ as a factor must evaluate to ${0}$ as well. But this is telling us that the kernel of the evaluation homomorphism for $i$ contains the principal ideal $(X^2+1)$.

Can it contain anything else? If $q\in\mathbb{R}[X]$ is a polynomial in the kernel, but $q$ is not divisible by $X^2+1$, then Euclid’s algorithm gives us a greatest common divisor of $q$ and $X^2+1$, which is a linear combination of these two, and must have degree either ${0}$ or ${1}$. In the former case, we would find that the evaluation map would have to send everything — even the constant polynomials — to zero. In the latter case, we’d have a linear factor of $X^2+1$, which would be a root. Clearly neither of these situations can occur, so the kernel of the evaluation homomorphism at $i$ is exactly the principal ideal $(X^2+1)$.

Now the first isomorphism theorem for rings tells us that we can impose our relation by taking the quotient ring $\mathbb{R}[X]/(X^2+1)$. But what we just discussed above further goes to show that $(X^2+1)$ is a maximal ideal, and the quotient of a ring by a maximal ideal is a field! Thus when we take the real numbers and adjoin a square root of $-1$ to get a ring we might call $\mathbb{R}[i]$, the result is a field. This is the field of “complex numbers”, which is more usually written $\mathbb{C}$.

Now we’ve gone through a lot of work to just add one little extra element to our field, but it turns out this is all we need. Luckily enough, the complex numbers are already algebraically complete! This is very much not the case if we were to try to algebraically complete other fields (like the rational numbers). Unfortunately, the proof really is essentially analytic. It seems to be a completely algebraic statement, but remember all the messy analysis and topology that went into defining the real numbers.

Don’t worry, though. We’ll come back and prove this fact once we’ve got a bit more analysis under our belts. We’ll also talk a lot more about how to think about complex numbers. But for now all we need to know is that they’re the “algebraic closure” of the real numbers, we get them by adding a square root of $-1$ that we call $i$, and we can use them as an example of an algebraically closed field.

One thing we can point out now, though, is the inherent duality of our situation. You see, we didn’t just add one square root of $-1$. Indeed, once we have complex numbers to work with we can factor $X^2+1$ as $(X-i)(X+i)$ (test this by multiplying it out and imposing the relation). Then we have another root: $-i$. This is just as much a square root of $-1$ as $i$ was, and anything we can do with $i$ we can do with $-i$. That is, there’s a symmetry in play that exchanges $i$ and $-i$. We can pick one and work with it, but we must keep in mind that whenever we do we’re making a non-canonical choice.

## 10 Comments	»

1. Of course, the extent to which singling out $i$ is non-canonical in $\mathbb{C}$, constructed this way, is no more and no less than the extent to which singling out $X$ is non-canonical in $\mathbb{R}[X]$.

Comment by Sridhar Ramesh | August 8, 2008 | Reply

2. This is true: there are some great examples of situations where it’s useful to view certain algebras of polynomials (subalgebras of the full polynomial algebra) are actually isomorphic to full polynomial algebras in other collections of variables.

Comment by | August 8, 2008 | Reply

3. [...] Properties of Complex Numbers Today I’ll collect a few basic properties of complex numbers. [...]

Pingback by | August 8, 2008 | Reply

4. In your first sentence, “Yesterday we defined a field to be algebraically closed if it always has exactly as many roots (counting multiplicities) as we expect from its degree,” the first use of the word “it” should perhaps be replaced by something like “any polynomial over it”.

Comment by John Palmieri | August 8, 2008 | Reply

5. Sorry, you’re right. distractions abound.

Comment by | August 8, 2008 | Reply

6. [...] Okay, we know that we can factor any complex number into linear factors because the complex numbers are algebraically closed. But we also know that real polynomials can have too few roots. Now, there are a lot of fields out [...]

Pingback by | August 14, 2008 | Reply

7. [...] of last time, we can write this as and find two solutions — and — by taking the two complex square roots of . But the equation doesn’t use any complex numbers. Surely we can find real-valued [...]

Pingback by | October 13, 2008 | Reply

8. [...] Until further notice, I’ll be assuming that the base field is algebraically closed, like the complex numbers [...]

Pingback by | February 2, 2009 | Reply

9. [...] of finite dimension over an algebraically closed field . If you want to be specific, use the complex numbers [...]

Pingback by | March 4, 2009 | Reply

10. [...] Numbers and the Unit Circle When I first talked about complex numbers there was one perspective I put off, and now need to come back to. It makes deep use of [...]

Pingback by | May 26, 2009 | Reply

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