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Idly playing around with finite fields one day, I found a curiosity that I can't explain. Given a prime modulus $p$ and one of its primitive roots $r$, define a mapping $f$ from $\mathbb{F}_{p}$ to $\mathbb{C}$ by:

$$f(x) = \zeta^{log_r(x)}$$

and taking $f(0) = 0$, where $\zeta = e^{2\pi i / (p-1)}$, and $log_r$ is the discrete log mod $p$.

Then, empirically it appears that this is some kind of homomorphism-esque thing--that is, it appears to be that when you do some operation to the elements of the finite field, it corresponds to the same operation done in $\mathbb{C}$, except the answer might be off by a distance of a multiple of $\sqrt{p}$. More formally,

$$|f(a+b) - (f(a) + f(b))| = k \sqrt{p}$$

It is easy to prove that the same holds for multiplication exactly:

$$|f(a b) - f(a) f(b)| = 0$$

The same seems to be true for prime power fields $\mathbb{F}_{p^n}$, with the distance being $\sqrt{p}$ again.

It's like finite fields are representable as complex numbers, with a suitable "complex mod". I guess my big question is why is it always off by a multiple of $\sqrt{p}$? Is this a known result? Does the "complex mod" hint I see go any deeper?

I'm no advanced mathematician, and I've used more or less the extent of my algebra vocabulary in posing this question, so keeping the answer as plebian as possible would be appreciated.

Jyrki Lahtonen
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luqui
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    Here $|f(a+b)-f(a)-f(b)|\le3$ by triangle inequality, so I don't quite see how it can be a multiple of $\sqrt p$ for $p\ge11$. What exactly do you mean here? Yes, the discrete logarithm is a homomorphism from the multiplicative group of the finite field to a group of roots of unity, but addition becomes somewhat random. Mind you $\sqrt p$ does appear often in character sums. – Jyrki Lahtonen Dec 22 '22 at 04:59
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    Mind you [tag:representation-theory] is, in my opinion, out of place here as a tag. In abstract algebra a representation has a technical meaning, but you seem to think of the word with a fuzzy meaning from natural language. – Jyrki Lahtonen Dec 22 '22 at 05:02
  • @JyrkiLahtonen, well. Before today I had only checked it by hand on p=5, 7 and the prime power field 9. Getting out the old CAS shows that it indeed fails at p=11. Guess I generalized too soon. – luqui Dec 22 '22 at 05:08
  • Guessed that much :-) It's ok. Have fun with these! – Jyrki Lahtonen Dec 22 '22 at 05:09
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    The instance of $\sqrt p$ that you have seen can possibly be interpreted as coming from Gauss sums. After all, in a sense the Gauss sums are the change of bases -coefficients between the additive and multiplicative characters of finite fields. Much like the $\Gamma$ function for real numbers. – Jyrki Lahtonen Dec 22 '22 at 05:26

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This is not a known result. The function $f$ is not a homomorphism because it does not preserve the operation of the finite field. For example, if we take $p=5$ and $r=2$, then $f(2+3) = f(1) + f(3) + k\sqrt{5}$, but $f(2)+f(3) \neq f(5)$.

However, the function $f$ does have some interesting properties. For example, it is a group homomorphism from the multiplicative group of the finite field to the unit circle in the complex plane. This means that if we take two non-zero elements $x,y$ of the finite field and compute $f(xy)$, it will be equal to $f(x)f(y)$. This can be shown using the property of primitive roots that $r^{a+b}\equiv r^ar^b \pmod{p}$.

Additionally, the function $f$ is periodic with period $p-1$, meaning that $f(x) = f(x+k(p-1))$ for all $x\in \mathbb{F}_p$ and all integers $k$. This follows from the fact that $log_r(x)$ is defined only modulo $p-1$.

Overall, the function $f$ is a kind of "complexification" of the finite field, but it does not preserve the field structure

Pip Lark
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  • That's interesting, thanks. But I was really interested in why it was always off by an exact multiple of sqrt(p), seems like something deeper was going on. Btw my question still had many errors when you answered it, hopefully it's closer to correct now. – luqui Dec 22 '22 at 04:54
  • Hi @luqui I think that $f(i+j) = f(i) + f(j) + k\sqrt{p}$ does not hold in general.

    For example, if we take $p=5$ and $r=2$ as before, then we have $f(0)=0$, $f(1)=1$, $f(2)=\zeta$, $f(3)=\zeta^2$, and $f(4)=\zeta^3$.

    If we take $i=1$, $j=3$, and $k=1$, then we have $f(i+j)=f(4)=\zeta^3$, $f(i)=f(1)=1$, and $f(j)=f(3)=\zeta^2$, so $f(i+j) \neq f(i)+f(j)+k\sqrt{p}$.

    On the other hand, if we take $i=3$, $j=3$, and $k=1$, then we have $f(i+j)=f(6)=\zeta^4$, $f(i)=f(3)=\zeta^2$, and $f(j)=f(3)=\zeta^2$, so $f(i+j) = f(i)+f(j)+k\sqrt{p}$

     – Pip Lark Dec 22 '22 at 05:15
  • you're right, it doesn't hold in general (see comment thread on the question). But it does hold at p=5 and 7. Looks like you got f(2) and f(3) switched; $f(3) = \zeta^3$ and $f(4) = \zeta^2$. But it does break at p=11 and above. – luqui Dec 22 '22 at 05:25