The Sum of Two Algebraic Numbers Is Algebraic

I have this mathematician friend who used to make up crazy stuff all the time, just to see how much of it I would believe. He eventually went away to grad school at Texas, because there were “tons of badass topologists there.” I wondered why, if they’re so badass, they aren’t done yet.

He once told me that there’s a fundamental distinction between the irrational numbers \sqrt{2} and \pi. \sqrt{2} is a solution to the equation

x^2 - 2 = 0,

but \pi, so he claimed, is not the solution to any equation

\sum_{k=0}^{n}a_kx^k = 0

with all the a_k integers. (Restricting a_k to integers is no stronger than restricting to rationals. If all the a_k were rationals, you could multiply the whole equation by the a common denominator of all the a_k and get a new equation using integer coefficients, so you might as well stipulate that a_k are integers.)

Turns out he was (as best I can tell) giving me the truth that time. At least, if it’s a joke, a lot of people are in on it.

There’s a nice thing about math, which is that, in principle, no one’s tricky jokes should work. If they say such-and-such is true, you can check it for yourself, just by thinking about it. But you can only do that if you’re sufficiently clever. I do not understand the proof that \pi is not algebraic. (Okay, to be honest, I have never even tried to understand any such proofs. All I’ve done is seen that such-and-such a book says, “it’s hard, here’s a reference if you’re interested”. Well, I’m interested, but not that interested.)

I can check some simpler things, though. For example, I’ve seen it stated that the algebraic numbers form a field, which implies (among other things), that if you take two of them and add them, you get another algebraic number. Curiously, the books I was looking at (and maybe they’re the wrong ones) gave no proof and no reference on this. But hey, it’s math. Just do it.

Let a and b be algebraic numbers, so that

\sum_{k=0}^n\alpha_ka^k = 0,

\sum_{j=0}^m\beta_jb^j = 0

for some set of integer coefficients \alpha_k, \beta_j and two integers n and m.

We’d like to know if a + b is also algebraic.

Looking at the equations that have a and b as roots, they look very similar to the equations defining linear dependence of a set of vectors. In the language of linear algebra, those equations say about a and b:

The set \{ 1 , a , a^2 , ... , a^n \} is linearly dependent.

Likewise for b. I have to admit this is not strictly accurate. Linear dependence is a concept between vectors, but the a^k are not vectors because vectors must be defined over some field. The a^k are only defined over a ring (the integers). But recall that defining algebraic numbers as solutions to polynomials with integer coefficients is no different than letting the polynomials have rational coefficients. The rationals form a proper field, so let’s let all our coefficients \alpha_k be rationals. (Maybe those rationals all happen to be integers.) Then the various a^k can be used to define a vector space – whatever is the smallest space they span. (To wit: the space of all their linear combinations and nothing else.) Call that space A. In A, the n vectors a^k are linearly dependent. The dimension of A is less than n+1 because we defined A to be the smallest space spanned by the a^k, and there are n+1 of those, and they’re dependent. Also, in a different space (call it B) spanned by the m+1 vectors b_j, the b^j are linearly dependent. (Your mom gave me a set of linearly dependent b^j‘s last night.)

Consider totally new space: the smallest space spanned by the first p powers of (a+b). (By this I mean the smallest space spanned by set \{ 1, a+b, (a+b)^2, (a+b)^3,...(a+b)^p \}. Call this space C_p.

If you expand out all those things like (a+b)^7 and whatever, you get a bunch of things like 21a^2b^5 that come from the binomial coefficients. But say that, for example, m=4. (Recall m+1 is larger than the dimension of B.) It’s true that

b^5 = b^4*b,

but b^4 is can be written as a sum of lower powers of b. Replacing it with that linear combination

(explicitly, b^4 = \frac{1}{\beta_4}\sum_{j=0}^3\beta_jb^j),

we see that we don’t need b^5 at all, and a sum of lower powers will suffice.

Indeed,
b^{27} = b^4*b^4*b^4*b^4*b^4*b^4*b^3.

Replace all those b^4‘s by sums of smaller powers. Then group up all like terms, and for any powers of b that remain greater than 4, do the same thing over again and again. Eventually, you’ll get rid of all powers of b greater than 4.

So to write out all the binomial terms in (a+b)^p, we don’t need any powers bigger than a^n or b^m. Therefore, C_p is a subspace of the tensor product A \otimes B. (That just means it is spanned by the set \{1, a, b, a^2, ab, b^2, a^3, a^2b, ab^2, b^3, ... , a^nb^m\}.)

C_p has dimension less than (m+1)*(n+1), but we have a list of p vectors in it. p was arbitrary, so we take p=(m+1)*(n+1) and it follows that the vectors \{1, (a+b), (a+b)^2, ... (a+b)^p\} are linearly dependent because there are more of them than the dimension of the space in which they reside. We can then write

\sum_{i=0}^p\lambda_i(a+b)^i = 0

for some set of p rationals \lambda_i, proving that a+b is an algebraic number.

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