## 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.