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 . The problem is that we have no field element whose square is
. 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
for “imaginary”, and throw it in with the rest of the real numbers
.
Okay, just like when we threw in as a new element, we can build up sums and products involving real numbers and this new element
. But there’s one big difference here: we have a relation that
must satisfy. When we use the evaluation map we must find
. And, of course, any polynomial which includes
as a factor must evaluate to
as well. But this is telling us that the kernel of the evaluation homomorphism for
contains the principal ideal
.
Can it contain anything else? If is a polynomial in the kernel, but
is not divisible by
, then Euclid’s algorithm gives us a greatest common divisor of
and
, which is a linear combination of these two, and must have degree either
or
. 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
, which would be a root. Clearly neither of these situations can occur, so the kernel of the evaluation homomorphism at
is exactly the principal ideal
.
Now the first isomorphism theorem for rings tells us that we can impose our relation by taking the quotient ring . But what we just discussed above further goes to show that
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
to get a ring we might call
, the result is a field. This is the field of “complex numbers”, which is more usually written
.
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 that we call
, 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 . Indeed, once we have complex numbers to work with we can factor
as
(test this by multiplying it out and imposing the relation). Then we have another root:
. This is just as much a square root of
as
was, and anything we can do with
we can do with
. That is, there’s a symmetry in play that exchanges
and
. 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.
Of course, the extent to which singling out
is non-canonical in
, constructed this way, is no more and no less than the extent to which singling out
is non-canonical in
.
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.
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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”.
Sorry, you’re right. distractions abound.
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