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Martin Brandenburg
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Martin Brandenburg
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Let $n>1$ be a natural number, letand $p$ be an odd prime number with $p-1 \mid n-1$ such that $p^k - 1 \mid n-1$ does not hold for any $k>1$. In the following, we work in the ring $\mathbb{F}_p[T]$. Notice that, since $p-1 \mid n-1$, we have $T^p - T \mid T^n-T$ in $\mathbb{F}_p[T]$ and hence also $T^p - T = (T+u)^p - (T+u) \mid (T+u)^n - (T+u)$ for all $u \in \mathbb{F}_p$.

Question. Is $T^p - T$ actually the gcd of $\{(T+u)^n - (T+u) : u \in \mathbb{F}_p\}$ in $\mathbb{F}_p[T]$?

I have verified this with computer algebra software for $n \leq 7000$ (code link). For many $n$ actually $u=0,1$ are sufficient. 

I tried to find a proof, but did not succeedmy first idea so fardidn't work. The only thing I know so far is that the gcd is invariant under $T \mapsto T+1$ and therefore contained in $\mathbb{F}_p[T^p-T]$. I expect that there are two proofs (if the statement is true at all), namely one using finite fields $\mathbb{F}_{p^m}$, and one using a direct calculation with polynomials. I am more interested in a direct calculation here. The background is a new proof of Jacobson's theorem I am working on. Notice that the statement is false for $p=2$ (but still true for many $n$ in this case) and that it is absolutelyclearly false without the $p^k-1$-requirement.

Let $n>1$ be a natural number, let $p$ be an odd prime number with $p-1 \mid n-1$ such that $p^k - 1 \mid n-1$ does not hold for any $k>1$. In the following, we work in the ring $\mathbb{F}_p[T]$. Notice that, since $p-1 \mid n-1$, we have $T^p - T \mid T^n-T$ and hence also $T^p - T = (T+u)^p - (T+u) \mid (T+u)^n - (T+u)$ for all $u \in \mathbb{F}_p$.

Question. Is $T^p - T$ actually the gcd of $\{(T+u)^n - (T+u) : u \in \mathbb{F}_p\}$?

I have verified this with computer algebra software for $n \leq 7000$. For many $n$ actually $u=0,1$ are sufficient. I tried to find a proof, but did not succeed so far. The only thing I know is that the gcd is invariant under $T \mapsto T+1$ and therefore contained in $\mathbb{F}_p[T^p-T]$. I expect that there are two proofs (if the statement is true at all), namely one using finite fields $\mathbb{F}_{p^m}$, and one using a direct calculation with polynomials. I am more interested in a direct calculation here. The background is a new proof of Jacobson's theorem I am working on. Notice that the statement is false for $p=2$ (but still true for many $n$ in this case) and that it is absolutely false without the $p^k-1$-requirement.

Let $n>1$ and $p$ be an odd prime with $p-1 \mid n-1$ such that $p^k - 1 \mid n-1$ does not hold for any $k>1$. Notice that, since $p-1 \mid n-1$, we have $T^p - T \mid T^n-T$ in $\mathbb{F}_p[T]$ and hence also $T^p - T = (T+u)^p - (T+u) \mid (T+u)^n - (T+u)$ for all $u \in \mathbb{F}_p$.

Question. Is $T^p - T$ actually the gcd of $\{(T+u)^n - (T+u) : u \in \mathbb{F}_p\}$ in $\mathbb{F}_p[T]$?

I have verified this with computer algebra software for $n \leq 7000$ (code link). For many $n$ actually $u=0,1$ are sufficient. 

I tried to find a proof, but my first idea didn't work. The only thing I know so far is that the gcd is invariant under $T \mapsto T+1$ and therefore contained in $\mathbb{F}_p[T^p-T]$. I expect that there are two proofs (if the statement is true at all), namely one using finite fields $\mathbb{F}_{p^m}$, and one using a direct calculation with polynomials. I am more interested in a direct calculation here. The background is a new proof of Jacobson's theorem I am working on. Notice that the statement is false for $p=2$ (but still true for many $n$ in this case) and that it is clearly false without the $p^k-1$-requirement.

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Martin Brandenburg
  • 63.1k
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  • 424

Let $n>1$ be a natural number, let $p$ be an odd prime number with $p-1 \mid n-1$ such that $p^k - 1 \mid n-1$ does not hold for any $k>1$. In the following, we work in the ring $\mathbb{F}_p[T]$. Notice that, since $p-1 \mid n-1$, we have $T^p - T \mid T^n-T$ and hence also $T^p - T = (T+u)^p - (T+u) \mid (T+u)^n - (T+u)$ for all $u \in \mathbb{F}_p$.

ConjectureQuestion. Is $T^p - T$ is actually the gcd of $\{(T+u)^n - (T+u) : u \in \mathbb{F}_p\}$.?

I have verified this with computer algebra software for $n \leq 7000$. For many $n$ actually $u=0,1$ are sufficient. I tried to find a proof, but did not succeednot succeed so far. The only thing I know so far isis that the gcd is invariant under $T \mapsto T+1$ and therefore contained in $\mathbb{F}_p[T^p-T]$. I expect that there are two proofs (if the statement is true at all), namely one using finite fields $\mathbb{F}_{p^m}$, and one using a direct calculation with polynomials. I am more interested in a direct calculation here. The background is a new proof of Jacobson's theorem I am working on. Notice that the statement is false for $p=2$ (but still true for many $n$ in this case) and that it is absolutely false without the $p^k-1$-requirement.

Let $n>1$ be a natural number, let $p$ be an odd prime number with $p-1 \mid n-1$ such that $p^k - 1 \mid n-1$ does not hold for any $k>1$. In the following, we work in the ring $\mathbb{F}_p[T]$. Notice that, since $p-1 \mid n-1$, we have $T^p - T \mid T^n-T$ and hence also $T^p - T = (T+u)^p - (T+u) \mid (T+u)^n - (T+u)$ for all $u \in \mathbb{F}_p$.

Conjecture. $T^p - T$ is actually the gcd of $\{(T+u)^n - (T+u) : u \in \mathbb{F}_p\}$.

I have verified this with computer algebra software for $n \leq 7000$. For many $n$ actually $u=0,1$ are sufficient. I tried to find a proof, but did not succeed so far. The only thing I know so far is that the gcd is invariant under $T \mapsto T+1$ and therefore contained in $\mathbb{F}_p[T^p-T]$. I expect that there are two proofs (if the statement is true at all), namely one using finite fields $\mathbb{F}_{p^m}$, and one using a direct calculation with polynomials. I am more interested in a direct calculation here. The background is a new proof of Jacobson's theorem I am working on. Notice that the statement is false for $p=2$ (but still true for many $n$ in this case) and that it is absolutely false without the $p^k-1$-requirement.

Let $n>1$ be a natural number, let $p$ be an odd prime number with $p-1 \mid n-1$ such that $p^k - 1 \mid n-1$ does not hold for any $k>1$. In the following, we work in the ring $\mathbb{F}_p[T]$. Notice that, since $p-1 \mid n-1$, we have $T^p - T \mid T^n-T$ and hence also $T^p - T = (T+u)^p - (T+u) \mid (T+u)^n - (T+u)$ for all $u \in \mathbb{F}_p$.

Question. Is $T^p - T$ actually the gcd of $\{(T+u)^n - (T+u) : u \in \mathbb{F}_p\}$?

I have verified this with computer algebra software for $n \leq 7000$. For many $n$ actually $u=0,1$ are sufficient. I tried to find a proof, but did not succeed so far. The only thing I know is that the gcd is invariant under $T \mapsto T+1$ and therefore contained in $\mathbb{F}_p[T^p-T]$. I expect that there are two proofs (if the statement is true at all), namely one using finite fields $\mathbb{F}_{p^m}$, and one using a direct calculation with polynomials. I am more interested in a direct calculation here. The background is a new proof of Jacobson's theorem I am working on. Notice that the statement is false for $p=2$ (but still true for many $n$ in this case) and that it is absolutely false without the $p^k-1$-requirement.

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Martin Brandenburg
  • 63.1k
  • 11
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  • 424
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Martin Brandenburg
  • 63.1k
  • 11
  • 207
  • 424
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