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Sat, 21 Apr 2007
Degrees of algebraic numbers
For example, all rational numbers have degree 1, since the rational number a/b is a zero of the first-degree polynomial bx - a. √2 has degree 2, since it is a zero of x^{2} - 2, but (as the Greeks showed) not of any first-degree polynomial. It's often pretty easy to guess what degree some number has, just by looking at it. For example, the nth root of a prime number p has degree n. !!\sqrt{1 + \sqrt 2}!! has a square root of a square root, so it's fourth-degree number. If you write !!x = \sqrt{1 + \sqrt 2}!! then eliminate the square roots, you get x^{4} - 2x^{2} - 1, which is the 4th-degree polynomial satisfied by this 4th-degree number. But it's not always quite so simple. One day when I was in high school, I bumped into the fact that !!\sqrt{7 + 4 \sqrt 3}!!, which looks just like a 4th-degree number, is actually a 2nd-degree number. It's numerically equal to !!2 + \sqrt 3!!. At the time, I was totally boggled. I couldn't believe it at first, and I had to get out my calculator and calculate both values numerically to be sure I wasn't hallucinating. I was so sure that the nested square roots in would force it to be 4th-degree. If you eliminate the square roots, as in the other example, you get the 4th-degree polynomial x^{4} - 14x^{2} + 1, which is satisfied by . But unlike the previous 4th-degree polynomial, this one is reducible. It factors into (x^{2} + 4x + 1)(x^{2} - 4x + 1). Since is a zero of the polynomial, it must be a zero of one of the two factors, and so it is second-degree. (It is a zero of the second factor.) I don't know exactly why I was so stunned to discover this. Clearly, the square of any number of the form a + b√c is another number of the same form (namely (a^{2} + b^{2}c) + 2ab√c), so it must be the case that lots of a + b√c numbers must be squares of other such, and so that lots of !!\sqrt{a + b \sqrt c}!! numbers must be second-degree. I must have known this, or at least been capable of knowing it. Socrates says that the truth is within us, and we just don't know it yet; in this case that was certainly true. I think I was so attached to the idea that the nested square roots signified fourth-degreeness that I couldn't stop to realize that they don't always. In the years since, I came to realize that recognizing the degree of an algebraic number could be quite difficult. One method, of course, is the one I used above: eliminate the radical signs, and you have a polynomial; then factor the polynomial and find the irreducible factor of which the original number is a root. But in practice this can be very tricky, even before you get to the "factor the polynomial" stage. For example, let x = 2^{1/2} + 2^{1/3}. Now let's try to eliminate the radicals. Proceeding as before, we do x - 2^{1/3} = 2^{1/2} and then square both sides, getting x^{2} - 2·2^{1/3}x + 2^{2/3} = 2, and then it's not clear what to do next. So we try the other way, starting with x - 2^{1/2} = 2^{1/3} and then cube both sides, getting x^{3} - 3·2^{1/2}x^{2} + 6x - 2·2^{1/2} = 2. Then we move all the 2^{1/2} terms to the other side: x^{3} + 6x - 2 = (3x^{2} + 2)·2^{1/2}. Now squaring both sides eliminates the last radical, giving us x^{6} + 12x^{4} - 4x^{3} + 36x^{2} - 24x + 4 = 18x^{4} + 12x^{2} + 8. Collecting the terms, we see that 2^{1/2} + 2^{1/3} is a root of x^{6} - 6x^{4} - 4x^{3} + 12x^{2} - 24x - 4. Now we need to make sure that this polynomial is irreducible. Ouch. In the course of writing this article, though, I found a much better method. I'll work a simpler example first, √2 + √3. The radical-eliminating method would have us put x - √2 = √3, then x^{2} - 2√2x + 2 = 3, then x^{2} - 1 = 2√2x, then x^{4} - 2x^{2} + 1 = 8x^{2}, so √2 + √3 is a root of x^{4} - 10x^{2} + 1. The new improved method goes like this. Let x = √2 + √3. Now calculate powers of x:
That's a lot of calculating, but it's totally mechanical. All of the powers of x have the form a_{6}√6 + a_{2}√2 + a_{3}√3 + a_{1}. This is easy to see if you write p for √2 and q for √3. Then x = p + q and powers of x are polynomials in p and q. But any time you have p^{2} you replace it with 2, and any time you have q^{2} you replace it with 3, so your polynomials never have any terms in them other than 1, p, q, and pq. This means that you can think of the powers of x as being vectors in a 4-dimensional vector space whose canonical basis is {1, √2, √3, √6}. Any four vectors in this space, such as {1, x, x^{2}, x^{3}}, are either linearly independent, and so can be combined to total up to any other vector, such as x^{4}, or else they are linearly dependent and three of them can be combined to make the fourth. In the former case, we have found a fourth-degree polynomial of which x is a root, and proved that there is no simpler such polynomial; in the latter case, we've found a simpler polynomial of which x is a root. To complete the example above, it is evident that {1, x, x^{2}, x^{3}} are linearly independent, but if you don't believe it you can use any of the usual mechanical tests. This proves that √2 + √3 has degree 4, and not less. Because if √2 + √3 were of degree 2 (say) then we would be able to find a, b, c such that ax^{2} + bx + c = 0, and then the x^{2}, x^{1}, and x^{0} vectors would be dependent. But they aren't, so we can't, so it isn't. Instead, there must be a, b, c, and d such that x^{4} = ax^{3} + bx^{2} + cx + d. To find these we need merely solve a system of four simultaneous equations, one for each column in the table:
And we immediately get a=0, b=10, c=0, d=-1, so x^{4} = 10x^{2} - 1, and our polynomial is x^{4} - 10x^{2} + 1, as before. Yesterday's draft of this article said: I think [2^{1/2} + 2^{1/3}] turns out to be degree 6, but if you try to work it out in the straightforward way, by equating it to x and then trying to get rid of the roots, you get a big mess. I think it turns out that if two numbers have degrees a and b, then their sum has degree at most ab, but I wouldn't even want to swear to that without thinking it over real carefully.Happily, I'm now sure about all of this. I can work through the mechanical method on it. Putting x = 2^{1/2} + 2^{1/3}, we get:
Where the vector [a, b, c, d, e, f] is really shorthand for a2^{1/2}·2^{2/3} + b2^{2/3} + c2^{1/2}·2^{1/3} + d2^{1/3} + e2^{1/2} + f. x^{0}...x^{5} turn out to be linearly independent, almost by inspection, so 2^{1/2} + 2^{1/3} has degree 6. To express x^{6} as a linear combination of x^{0}...x^{5}, we set up the following equations:
Solving these gives [a, b, c, d, e, f]= [0, 6, 4, -12, 24, 4], so x^{6} = 6x^{4} + 4x^{3} - 12x^{2} + 24x + 4, and 2^{1/2} + 2^{1/3} is a root of x^{6} - 6x^{4} - 4x^{3} + 12x^{2} - 24x - 4, which is irreducible. And similarly, using this method, one can calculate in a few minutes that 2^{1/2} + 2^{1/4} has degree 4 and is a root of x^{4} - 4x^{2} - 8x + 2. I wish I had figured this out in high school; it would have delighted me.
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