Stuck on Ex. 2.5 - Prove that the sum and product of algebraic numbers is algebraic

[ArthurAllshire]

Not really sure how to approach this proof. Help would be greatly appreciated :)

[ArthurAllshire]

(Also, maybe I'm missing something here but I don't think that I can tag this issue as requested because only contributors can add labels)

[j2kun]

This one requires some more elbow-grease than it appears at first glance. I think the best vague hint I can give is to pose the question: how do you know that square-root(2) + cube-root(3) is algebraic (or even simpler, sqrt(2) + sqrt(3))? If you can find a polynomial that proves this is algebraic, I think the steps you took would generalize to a neat proof.

Also, be warned, that a lot of the discussion on the internet about this topic is couched in flowery language about field extensions and Galois groups. Those tools provide a tidy analysis, but I think one shouldn't need any special tools to prove this.

[ArthurAllshire]

So obviously to show say sqrt(2) + sqrt(3) is algebraic you have to find a polynomial with that as a root. To do this, I fiddled about by setting a generic quartic to zero, so I had
0 = a0 + a1(sqrt(2) + sqrt(3)) + a2(sqrt(2) + sqrt(3))^2 + a3(sqrt(2) + sqrt(3))^3 + a4(sqrt(2) + sqrt(3))^4
I then realized that by choosing the coefficients a2 and a4 to get the sqrt(6)s that result from them to cancel, a1 and a3 to zero, and a0 to cancel the rational portion to get the whole expression to zero, giving me
p(x) = x^4 - 10x^2 + 1
I've tried generalizing this but I'm not sure that I'm on the right track because this method doesn't directly depend on the numbers being algebraic (ie use the fact that they are algebraic to find the solution)...

[j2kun]

I think you have the right idea, but being a little bit less direct will help keep the reasoning more focused on the general proof. When I think about sqrt(2) + sqrt(3), I notice that each part is algebraic, and they both have polynomials of degree 2. So My first thought is to square that sucker.

The first two terms on the right where I use that sqrt(2) and sqrt(3) are algebraic, and in this simplified example they already turn out to be integers. So I can ignore them and henceforth reduce any power of (sqrt(2) + sqrt(3)) into a power of (sqrt(2)*sqrt(3)) plus some rationals. Moreover, since a power of sqrt(2)*sqrt(3) can be reduced in each factor—(sqrt(2) sqrt(3))^3 is an integer times (sqrt(2) sqrt(3))—I really just have to worry about (sqrt(2)sqrt(3)). As you noticed, the fourth power also helps.

Here k and m are some rational numbers that one can work out, but it doesn't matter what they are.

Using these two equations together, you can rearrange to get all the rationals and powers of (sqrt(2) + sqrt(3)) on one side, and zero on the other side, because the sqrt(2)sqrt(3) terms can be made to cancel by the suitable choice of a rational coefficient, something like -2/m if I haven't screwed up :)

As you may have noticed, we're taking advantage of my note on page 8 that a polynomial is more general than the raw definition; it can be any combination of multiplication and addition and constants (in this case the constants have to be rational numbers). While I still haven't quite revealed the proof, we're definitely taking advantage of this to make our calculations more compact.

In the general case, what can you say about similarly large powers of (a+b)?

[j2kun]

I misspoke above about "reducing any power", but hopefully you still get the idea? I am trying to convey the idea that you can get from large powers of (sqrt(a) + sqrt(b)) to polynomials involving only sqrt(a)sqrt(b)

[ArthurAllshire]

OK, I sort of get how to prove this now I think, not sure if this is sufficiently rigorous, and it may just be wrong... (also apologies for horrific typography - maybe learning latex is the next project)
Let a be an algebraic number which is the root of a polynomial of degree q, and b an algebraic number the root of a polynomial of degree r.
Now, let n = q * r. Then a^ n, b^n are rational.

Now, take all of the powers (a+b)^n…(a+b)^0, ie
(a+b)^n = a^n + k1 * a^n-1 + … + b^n
(a+b)^(n-1) = a^(n-1) + j1 * a^(n-2) + … + b^(n-1)
…
(a+b)^1 = a + b
(a+b)^0 = 1

Now, it is clear that through a “linear combination” using rational numbers c1...c(n-1) of the equations with powers n-1 through 1 we may eliminate all but a^n and b^n (since coefficients of each term on RHS of each equation are rational apart from a and b to their various powers). Since a^n + b^n is rational, set cn = -c0. c0, c1, … cn now form the coefficients of a polynomial p of degree n with p(a+b) = 0.

[j2kun]

You're very close, and I think what matters more than rigor is that you understand the core idea. However, in what you wrote a^n and b^n are not necessarily rational. For example, if a = 1 + cuberoot(3), then a^3 still has terms involving powers of cuberoot(3). But a^3 can be written entirely in terms of smaller powers of a.

What I think you're saying is that, if a^q can be written in terms of smaller powers, and likewise for b^r, then (a+b)^{qr} can be written in terms of {a^i b^j : i=0...q-1, j=0...r-1}.

The core idea of the proof is that these terms are "enough." I.e., as you said, the powers of (a+b) admit a linear combination that's zero.

And since you said "linear combination", I don't feel so bad looking ahead to the linear algebra chapter: for an irrational number a, the set which is the union of rational numbers and "all possible fractions involving a" (i.e., a, 1/a, (a^2 - 1)/(a+1), etc.) is a vector space with rational scalars. Call it Q(a). The dimension of Q(a) is the degree of the smallest-degree polynomial p for which p(a) = 0. Indeed, this exercise is essentially showing that the basis for Q(a+b) can be built from the bases of Q(a) and Q(b) by taking all possible products.

[j2kun]

Thanks for the discussion. I think I will close this, and if any readers have additional questions on this exercise, feel free to reopen.

[sritchie]

This discussion was really helpful in making progress, BUT I'm still stuck, so I'll drop a comment here.

First, the state of where I am after a few hours of noodling over this (love the book so far, by the way):

1. @j2kun, I'm with you on what seems like an assumption that algebraic numbers are nth roots of rational numbers. (Of course these can be roots of polynomials with rational coefficients... but are these the ONLY things that can be algebraic numbers? I get nervous wading into the math waters that, without a compiler, I've forgotten to shore up some critical piece of the raft I'm floating on.)
2. Given that assumption, a cube root, a^(1/3), is an algebraic number that is the root of a cubic polynomial. This is always true, correct?
3. Okay, now the steps I've taken in the proof.

If

• a is an algebraic number that's the root of a polynomial of degree q, and
• b is an algebraic number that's the root of a polynomial of degree r,
  then the goal is to prove that we can construct a polynomial of degree n = (q * r) with a root equal to (a + b).

That polynomial will have the form

f(x) = 
  c0
+ c1 * (x + y)
+ c2 * (x + y)^2
+ ...
+ cn * (x + y)^n

NOW I also follow your argument here:

What I think you're saying is that, if a^q can be written in terms of smaller powers, and likewise for b^r, then (a+b)^{qr} can be written in terms of {a^i b^j : i=0...q-1, j=0...r-1}.

because a^q can be written in terms of smaller powers then we can always expand out any a with an exponent >= q, and same with b and r.

These new terms that we've decomposed from, say, cn * (x + y)^n can always find a place to fit down into the expansion of the polynomial terms with exponent at most max(q, r). So I get that we can write the entire polynomial f(x) as the sum of the terms in

 {a^i b^j : i=0...q-1, j=0...r-1}

each with its own rational coefficient. ( :) I don't know how to write this as math!)

OKAY! Here's where I'm stuck. How can we claim that there exist rational numbers c0, c1,... cn that will cause of all these reduced terms to sum to zero? I don't know how to think about this - I've been looking for a way to show what the constants should be by construction, but can't manage it.

I also don't understand this note from @ArthurAllshire:

Now, take all of the powers (a+b)^n…(a+b)^0, ie
(a+b)^n = a^n + k1 * a^n-1 + … + b^n
(a+b)^(n-1) = a^(n-1) + j1 * a^(n-2) + … + b^(n-1)
…
(a+b)^1 = a + b
(a+b)^0 = 1

It seems like he's lost all powers of b except the final power in each line and added in constants k, j etc instead?

This final comment may be the clue that I'm not following:

Now, it is clear that through a “linear combination” using rational numbers c1...c(n-1) of the equations with powers n-1 through 1 we may eliminate all but a^n and b^n (since coefficients of each term on RHS of each equation are rational apart from a and b to their various powers). Since a^n + b^n is rational, set cn = -c0. c0, c1, … cn now form the coefficients of a polynomial p of degree n with p(a+b) = 0.

It's not clear!

Thanks again for the fantastic book. Hope this is enough information for a clue to help me along where I'm stuck.

[sritchie]

(I can't re-open the issue, but I think the subscription should work anyway for notifications. Thanks!)

[ArthurAllshire]

Addressing your first point - the roots of polynomials with rational coefficients are the only things that can be algebraic numbers. This is not an assumption, it is the definition of an algebraic number. Now, these numbers aren't only nth roots of rational numbers - eg (the golden ratio) is a root of is thus algebraic.

You are right about my losing the powers of b in that snippet - I think it was a typo on my part (otherwise I'm not sure what I was thinking). The numbers k, j etc that I added are integers which represent the binomial coefficients in the expansion of those powers.

The logic in that comment sort of goes like this:
We don't actually need to construct coefficients c0,...,cn to show that they exist such that they make the equations sum to zero. Saying that we can express in terms of smaller powers in is the same as saying that each term in the expansion of appears in more than one equation. Since the coefficients of these terms are all rational (because they are all just terms of a binomial expansion), we know that by choosing rational coefficients c0,...cn to multiply each equation by we can make all the terms in the different equations cancel out when we add the different equations together. We don't need to know what these terms are per se. to know that they actually exist.
I hope this helps!

[j2kun]

I suppose there'e no point in ever closing these issues, and when they're open they're easier to discover.