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An interesting application of Lagrange's four-square theorem.
The four-square theorem states that every positive integer is the sum of four perfect squares (allowing repetition and allowing 0).
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The four-square theorem states that every positive integer is the sum of four perfect squares (allowing repetition and allowing 0).
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For example,
3 = 1^2 + 1^2 + 1^2 + 0^2
7 = 2^2 + 1^2 + 1^2 + 1^2
I've mainly known this result as a number-theoretic curiosity before, but there's an interesting application I've recently been made aware of.
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3 = 1^2 + 1^2 + 1^2 + 0^2
7 = 2^2 + 1^2 + 1^2 + 1^2
I've mainly known this result as a number-theoretic curiosity before, but there's an interesting application I've recently been made aware of.
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The story begins with a conversation with @weakconvergence. We were talking about division rings, and in particular, the real Hamiltonian quaternions.
The quaternions are the most well-known example of a division ring which is not a field.
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The quaternions are the most well-known example of a division ring which is not a field.
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In other words, the quaternions have all of the properties of a field, except that multiplication (and hence "division") is not commutative.
Quaternions are of the form a+bi+cj+dk where a, b, c, and d are real numbers, and where
i^2 = j^2 = k^2 = ijk = -1.
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Quaternions are of the form a+bi+cj+dk where a, b, c, and d are real numbers, and where
i^2 = j^2 = k^2 = ijk = -1.
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From this relation, ij = k and ji = -k, which shows that the multiplication is not commutative. Exciting!
To demonstrate how "rare" this sort of structure is, I brought up Wedderburn's Little Theorem, which implies that all finite division rings are fields.
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To demonstrate how "rare" this sort of structure is, I brought up Wedderburn's Little Theorem, which implies that all finite division rings are fields.
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So you cannot have a division ring with finitely many elements which isn't commutative!
@weakconvergence then asked me a really wonderful question - what goes wrong with being a division ring if you take the coefficients to be from a finite field, rather than R.
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@weakconvergence then asked me a really wonderful question - what goes wrong with being a division ring if you take the coefficients to be from a finite field, rather than R.
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After looking into this a bit, I was reminded of the concept of the quaternion conjugate. Much like the complex conjugate, we get a nice formula:
(a+bi+cj+dk)(a-bi-cj-dk) = a^2 + b^2 + c^2 + d^2
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(a+bi+cj+dk)(a-bi-cj-dk) = a^2 + b^2 + c^2 + d^2
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Aha! Using this, we can get a nice explanation of what goes wrong!
If the coefficients come from a field of characteristic p, then by Lagrange's four-square theorem, there are nonnegative integers a, b, c, and d with
p = a^2 + b^2 + c^2 + d^2.
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If the coefficients come from a field of characteristic p, then by Lagrange's four-square theorem, there are nonnegative integers a, b, c, and d with
p = a^2 + b^2 + c^2 + d^2.
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Then, we see
(a+bi+cj+dk)(a-bi-cj-dk)
= a^2 + b^2 + c^2 + d^2
= p
= 0.
So, if the coefficients come from a field of positive characteristic (including all finite fields), we actually get zero-divisors! It cannot be a division ring!
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(a+bi+cj+dk)(a-bi-cj-dk)
= a^2 + b^2 + c^2 + d^2
= p
= 0.
So, if the coefficients come from a field of positive characteristic (including all finite fields), we actually get zero-divisors! It cannot be a division ring!
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Lagrange's four-square theorem gives a very compact proof that you cannot get a division ring from quaternions over a field of positive characteristic.
It's interesting to see this result having a use outside of being a number-theoretic curiosity!
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It's interesting to see this result having a use outside of being a number-theoretic curiosity!
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