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edited Sep 17, 2021 at 18:43

Ofir Gorodetsky

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Here is a proof of the experimental fact Joe Silverman discovered.

Let $f$ be a monic polynomial of degree $d$ in $R:=\mathbb{F}_q[x]$. For $I=(f)$, the sum $$1+\sum_{a \in R/I} a^{\# R/I - 1}$$ is equal, modulo $f$, to $$1+\sum_{g \in R,\, \deg g < d} g^{q^d-1}.$$ Let $$F_d(x) := 1+\sum_{g \in R,\, \deg g < d} g^{q^d-1} \in R.$$ We shall show that if $f$ is irreducible than $f \mid F_d$ and if $f$ is reducible than $f \nmid F_d$.

First, suppose that $f$ is reducible. Let $\alpha$ be a root of $f$, with minimal polynomial $m_{\alpha} \mid f$ and $\deg m_{\alpha} = e < d$. We can write any $g \in R$ of degree $<d$ uniquely as $m_{\alpha} g_0 + g_1$ where $\deg g_1 < e$ and $\deg g_0 < d-e$. Hence $$F_d(\alpha) = 1+\sum_{g_0,g_1 \in R,\, \deg g_0 < d-e, \, \deg g_1 < e} (m_{\alpha}(\alpha) g_0(\alpha)+g_1(\alpha))^{q^d-1} = 1+q^{d-e}\sum_{g_1 \in R,\, \deg g_1 < e} g_1(\alpha)^{q^d-1}=1$$ since we are in characteristic $p \mid q$. Hence $F_d$ cannot be divisible by $f$.

Next suppose that $f$ is irreducible and let $\alpha$ be a root of $f$. The map $g \mapsto g(\alpha)$ from $\mathbb{F}_q[x]/I$ to $\mathbb{F}_q[\alpha] = \mathbb{F}_{q^d}$ is an isomorphism, and so $$F_d(\alpha) = 1+\sum_{\beta \in \mathbb{F}_{q^{d}}} \beta^{q^d-1}=q^d =0$$ by Lagrange's Theorem applied to the multiplicative group of the field. Hence $f$ divides $F_d$, as needed.

Here is a proof of the experimental fact Joe Silverman discovered.

Let $f$ be a monic polynomial of degree $d$ in $R:=\mathbb{F}_q[x]$. For $I=(f)$, the sum $$1+\sum_{a \in R/I} a^{\# R/I - 1}$$ is equal, modulo $f$, to $$1+\sum_{g \in R,\, \deg g < d} g^{q^d-1}.$$ Let $$F_d(x) := 1+\sum_{g \in R,\, \deg g < d} g^{q^d-1} \in R.$$ We shall show that if $f$ is irreducible than $f \mid F_d$ and if $f$ is reducible than $f \nmid F_d$.

First, suppose that $f$ is reducible. Let $\alpha$ be a root of $f$, with minimal polynomial $m_{\alpha} \mid f$ and $\deg m_{\alpha} = e < d$. We can write any $g \in R$ of degree $<d$ uniquely as $m_{\alpha} g_0 + g_1$ where $\deg g_1 < e$ and $\deg g_0 < d-e$. Hence $$F_d(\alpha) = 1+\sum_{g_0,g_1 \in R,\, \deg g_0 < d-e, \, \deg g_1 < e} (m_{\alpha}(\alpha) g_0(\alpha)+g_1(\alpha))^{q^d-1} = 1+q^{d-e}\sum_{g_1 \in R,\, \deg g_1 < e} g_1(\alpha)^{q^d-1}=1$$ since we are in characteristic $p \mid q$. Hence $F_d$ cannot be divisible by $f$.

Next suppose that $f$ is irreducible and let $\alpha$ be a root of $f$. The map $g \mapsto g(\alpha)$ from $\mathbb{F}_q[x]/I$ to $\mathbb{F}_q[\alpha] = \mathbb{F}_{q^d}$ is an isomorphism, and so $$F_d(\alpha) = 1+\sum_{\beta \in \mathbb{F}_{q^{d}}} \beta^{q^d-1}=q^d =0$$ by Lagrange's Theorem applied to the multiplicative group of the field. Hence $f$ divides $F_d$, as needed.

Here is a proof of the experimental fact Joe Silverman discovered.

Let $f$ be a polynomial of degree $d$ in $R:=\mathbb{F}_q[x]$. For $I=(f)$, the sum $$1+\sum_{a \in R/I} a^{\# R/I - 1}$$ is equal, modulo $f$, to $$1+\sum_{g \in R,\, \deg g < d} g^{q^d-1}.$$ Let $$F_d(x) := 1+\sum_{g \in R,\, \deg g < d} g^{q^d-1} \in R.$$ We shall show that if $f$ is irreducible than $f \mid F_d$ and if $f$ is reducible than $f \nmid F_d$.

First, suppose that $f$ is reducible. Let $\alpha$ be a root of $f$, with minimal polynomial $m_{\alpha} \mid f$ and $\deg m_{\alpha} = e < d$. We can write any $g \in R$ of degree $<d$ uniquely as $m_{\alpha} g_0 + g_1$ where $\deg g_1 < e$ and $\deg g_0 < d-e$. Hence $$F_d(\alpha) = 1+\sum_{g_0,g_1 \in R,\, \deg g_0 < d-e, \, \deg g_1 < e} (m_{\alpha}(\alpha) g_0(\alpha)+g_1(\alpha))^{q^d-1} = 1+q^{d-e}\sum_{g_1 \in R,\, \deg g_1 < e} g_1(\alpha)^{q^d-1}=1$$ since we are in characteristic $p \mid q$. Hence $F_d$ cannot be divisible by $f$.

Next suppose that $f$ is irreducible and let $\alpha$ be a root of $f$. The map $g \mapsto g(\alpha)$ from $\mathbb{F}_q[x]/I$ to $\mathbb{F}_q[\alpha] = \mathbb{F}_{q^d}$ is an isomorphism, and so $$F_d(\alpha) = 1+\sum_{\beta \in \mathbb{F}_{q^{d}}} \beta^{q^d-1}=q^d =0$$ by Lagrange's Theorem applied to the multiplicative group of the field. Hence $f$ divides $F_d$, as needed.

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created Sep 17, 2021 at 18:36

Ofir Gorodetsky

14.6k
1
66
79

Here is a proof of the experimental fact Joe Silverman discovered.

Let $f$ be a monic polynomial of degree $d$ in $R:=\mathbb{F}_q[x]$. For $I=(f)$, the sum $$1+\sum_{a \in R/I} a^{\# R/I - 1}$$ is equal, modulo $f$, to $$1+\sum_{g \in R,\, \deg g < d} g^{q^d-1}.$$ Let $$F_d(x) := 1+\sum_{g \in R,\, \deg g < d} g^{q^d-1} \in R.$$ We shall show that if $f$ is irreducible than $f \mid F_d$ and if $f$ is reducible than $f \nmid F_d$.

First, suppose that $f$ is reducible. Let $\alpha$ be a root of $f$, with minimal polynomial $m_{\alpha} \mid f$ and $\deg m_{\alpha} = e < d$. We can write any $g \in R$ of degree $<d$ uniquely as $m_{\alpha} g_0 + g_1$ where $\deg g_1 < e$ and $\deg g_0 < d-e$. Hence $$F_d(\alpha) = 1+\sum_{g_0,g_1 \in R,\, \deg g_0 < d-e, \, \deg g_1 < e} (m_{\alpha}(\alpha) g_0(\alpha)+g_1(\alpha))^{q^d-1} = 1+q^{d-e}\sum_{g_1 \in R,\, \deg g_1 < e} g_1(\alpha)^{q^d-1}=1$$ since we are in characteristic $p \mid q$. Hence $F_d$ cannot be divisible by $f$.

Next suppose that $f$ is irreducible and let $\alpha$ be a root of $f$. The map $g \mapsto g(\alpha)$ from $\mathbb{F}_q[x]/I$ to $\mathbb{F}_q[\alpha] = \mathbb{F}_{q^d}$ is an isomorphism, and so $$F_d(\alpha) = 1+\sum_{\beta \in \mathbb{F}_{q^{d}}} \beta^{q^d-1}=q^d =0$$ by Lagrange's Theorem applied to the multiplicative group of the field. Hence $f$ divides $F_d$, as needed.