Unramified primes inside $mathbb{F}_q(t)$
3
$begingroup$
I have some questions about the following statement:
Let $P subset mathbb{F}_q(t)$ be a prime, with $q=p^e$ for a prime number $p$ and $f$ a polynomial of degree $n$ with coefficients inside $mathbb{F}_q(t)$.
Let $K$ be the residue field of $P$ and $overline{f}$ the image of $f$ over $K$.
If $overline{f}$ is squarefree over $K$, $P$ is unramified over $mathbb{F}_q(t)[X] / f$.
Until now I have seen ramified/unramified primes only in the case of number fields, so perhaps I am missing something.
I would appreciate it if someone could check my thoughts about this statement.
- $P in mathbb{F}_q[t]$
$K = mathbb{F}_q[t] / (p) = mathbb{F}_{q^k}[t]$ for some $k$.
Also I would appreciate a definition of ramification/unramified in this context. Now I can not see, why this statement should be true.
finite-fields function-fields
asked Dec 4 '18 at 14:30
SqyuliSqyuli
1689
$endgroup$
|
show 10 more comments
3
$begingroup$
I have some questions about the following statement:
Let $P subset mathbb{F}_q(t)$ be a prime, with $q=p^e$ for a prime number $p$ and $f$ a polynomial of degree $n$ with coefficients inside $mathbb{F}_q(t)$.
Let $K$ be the residue field of $P$ and $overline{f}$ the image of $f$ over $K$.
If $overline{f}$ is squarefree over $K$, $P$ is unramified over $mathbb{F}_q(t)[X] / f$.
Until now I have seen ramified/unramified primes only in the case of number fields, so perhaps I am missing something.
I would appreciate it if someone could check my thoughts about this statement.
- $P in mathbb{F}_q[t]$
$K = mathbb{F}_q[t] / (p) = mathbb{F}_{q^k}[t]$ for some $k$.
Also I would appreciate a definition of ramification/unramified in this context. Now I can not see, why this statement should be true.
finite-fields function-fields
asked Dec 4 '18 at 14:30
SqyuliSqyuli
1689
$endgroup$
1
$begingroup$
Anyway, in case you haven't already done so, I recommend that you also familiarize yourself with the geometric point of view here. You can think of $f$ as a polynomial in $Bbb{F}_q[t,X]$ (clear the eventual denominators), and that defines a curve in the $(t,X)$-plane (and in its projective closure to include the points at infinity, or the primes lying above $P=(1/t)$. You need to go to a smooth model to get bijective correspondence between the primes of the function field and the points on the projective curve, but that can wait.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 15:08
1
$begingroup$
With $F(t,X):=X^8+t+1$ you get ramification at the place corresponding to the irreducible polynomial $t+1$. Or, if you prefer, at the point $Q=(t,X)=(-1,0)$. At that point only $X$ is a local parameter, and the valuation $nu_Q(t+1)=8$. So we get $e=8$. Calculus shows the same thing. The curve $t+1=-X^8$ has a vertical tangent at $(t,X)=(-1,0)$. The implicit function theorem says the same thing as $$dfrac{partial F}{partial X}(Q)=0.$$ IIRC the place $t=infty$ also ramifies in $Bbb{F}_q(t,X)$.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 19:20
1
$begingroup$
Yes. WHen taking the quotient ring by moding out the ideal $t-a$ you effectively set $t=a$. When you mod out a higher degree irreducible polynomial you get a higher degree place. Geometrically you then look at points whose coordinates are Galois conjugates of each other - all lumped together to form a single place.
$endgroup$
– Jyrki Lahtonen
Dec 5 '18 at 4:12
1
$begingroup$
On the curve $X^8+t+1=0$ we have the ring $mathcal{O}_Q$ of rational functions defined at $(t,X)=(-1,0)$. It has a maximal ideal $m_Q$ of those functions that vanish at $Q$. That ideal is generated by $t+1$ and $X$. An actually we see that $X$ alone suffices. Basicall because the equation $X^8=-(t+1)$ tells us how to write $t+1$ as a multiple of $X$. So $m_Q=langle Xrangle$. The general theory says that $mathcal{O}_Q$ is a DVR, and as a generator of the maximal ideal we can use $X$ to define the valuation. We have $t+1=(-1)cdot X^8$. So here $u=-1$ and $s=X$.
$endgroup$
– Jyrki Lahtonen
Dec 5 '18 at 4:20
1
$begingroup$
At tthe point $R$ with coordinates $t=-2,X=1$ we use the equation $X^8+t+1=0$ in the forms $$t+2=-(X^8-1)=-(X^7+X^6+cdots+X+1)(X-1)$$ and $$(X-1)=-frac{t+2}{X^7+X^6+cdots+X+1}$$ to see that either $t+2$ or $X-1$ alone generates the ideal $m_R$. So either $t+2$ or $X-1$ can serve in the role of the local parameter $s$ in defining the valuation. Anyway, because $t+2$ generates the maximal ideal it follows that $nu_R(t+2)=1$ meaning that the place $R$ lying above $P=t+2$ is unramified.
$endgroup$
– Jyrki Lahtonen
Dec 5 '18 at 4:27
|
show 10 more comments
3
3
3
1
$begingroup$
I have some questions about the following statement:
Let $P subset mathbb{F}_q(t)$ be a prime, with $q=p^e$ for a prime number $p$ and $f$ a polynomial of degree $n$ with coefficients inside $mathbb{F}_q(t)$.
Let $K$ be the residue field of $P$ and $overline{f}$ the image of $f$ over $K$.
If $overline{f}$ is squarefree over $K$, $P$ is unramified over $mathbb{F}_q(t)[X] / f$.
Until now I have seen ramified/unramified primes only in the case of number fields, so perhaps I am missing something.
I would appreciate it if someone could check my thoughts about this statement.
- $P in mathbb{F}_q[t]$
$K = mathbb{F}_q[t] / (p) = mathbb{F}_{q^k}[t]$ for some $k$.
Also I would appreciate a definition of ramification/unramified in this context. Now I can not see, why this statement should be true.
finite-fields function-fields
asked Dec 4 '18 at 14:30
SqyuliSqyuli
1689
$endgroup$
I have some questions about the following statement:
Let $P subset mathbb{F}_q(t)$ be a prime, with $q=p^e$ for a prime number $p$ and $f$ a polynomial of degree $n$ with coefficients inside $mathbb{F}_q(t)$.
Let $K$ be the residue field of $P$ and $overline{f}$ the image of $f$ over $K$.
If $overline{f}$ is squarefree over $K$, $P$ is unramified over $mathbb{F}_q(t)[X] / f$.
Until now I have seen ramified/unramified primes only in the case of number fields, so perhaps I am missing something.
I would appreciate it if someone could check my thoughts about this statement.
- $P in mathbb{F}_q[t]$
$K = mathbb{F}_q[t] / (p) = mathbb{F}_{q^k}[t]$ for some $k$.
Also I would appreciate a definition of ramification/unramified in this context. Now I can not see, why this statement should be true.
finite-fields function-fields
finite-fields function-fields
asked Dec 4 '18 at 14:30
SqyuliSqyuli
1689
asked Dec 4 '18 at 14:30
SqyuliSqyuli
1689
asked Dec 4 '18 at 14:30
SqyuliSqyuli
1689
asked Dec 4 '18 at 14:30
SqyuliSqyuli
1689
asked Dec 4 '18 at 14:30
SqyuliSqyuli
1689
1689
1
$begingroup$
Anyway, in case you haven't already done so, I recommend that you also familiarize yourself with the geometric point of view here. You can think of $f$ as a polynomial in $Bbb{F}_q[t,X]$ (clear the eventual denominators), and that defines a curve in the $(t,X)$-plane (and in its projective closure to include the points at infinity, or the primes lying above $P=(1/t)$. You need to go to a smooth model to get bijective correspondence between the primes of the function field and the points on the projective curve, but that can wait.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 15:08
1
$begingroup$
With $F(t,X):=X^8+t+1$ you get ramification at the place corresponding to the irreducible polynomial $t+1$. Or, if you prefer, at the point $Q=(t,X)=(-1,0)$. At that point only $X$ is a local parameter, and the valuation $nu_Q(t+1)=8$. So we get $e=8$. Calculus shows the same thing. The curve $t+1=-X^8$ has a vertical tangent at $(t,X)=(-1,0)$. The implicit function theorem says the same thing as $$dfrac{partial F}{partial X}(Q)=0.$$ IIRC the place $t=infty$ also ramifies in $Bbb{F}_q(t,X)$.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 19:20
1
$begingroup$
Yes. WHen taking the quotient ring by moding out the ideal $t-a$ you effectively set $t=a$. When you mod out a higher degree irreducible polynomial you get a higher degree place. Geometrically you then look at points whose coordinates are Galois conjugates of each other - all lumped together to form a single place.
$endgroup$
– Jyrki Lahtonen
Dec 5 '18 at 4:12
1
$begingroup$
On the curve $X^8+t+1=0$ we have the ring $mathcal{O}_Q$ of rational functions defined at $(t,X)=(-1,0)$. It has a maximal ideal $m_Q$ of those functions that vanish at $Q$. That ideal is generated by $t+1$ and $X$. An actually we see that $X$ alone suffices. Basicall because the equation $X^8=-(t+1)$ tells us how to write $t+1$ as a multiple of $X$. So $m_Q=langle Xrangle$. The general theory says that $mathcal{O}_Q$ is a DVR, and as a generator of the maximal ideal we can use $X$ to define the valuation. We have $t+1=(-1)cdot X^8$. So here $u=-1$ and $s=X$.
$endgroup$
– Jyrki Lahtonen
Dec 5 '18 at 4:20
1
$begingroup$
At tthe point $R$ with coordinates $t=-2,X=1$ we use the equation $X^8+t+1=0$ in the forms $$t+2=-(X^8-1)=-(X^7+X^6+cdots+X+1)(X-1)$$ and $$(X-1)=-frac{t+2}{X^7+X^6+cdots+X+1}$$ to see that either $t+2$ or $X-1$ alone generates the ideal $m_R$. So either $t+2$ or $X-1$ can serve in the role of the local parameter $s$ in defining the valuation. Anyway, because $t+2$ generates the maximal ideal it follows that $nu_R(t+2)=1$ meaning that the place $R$ lying above $P=t+2$ is unramified.
$endgroup$
– Jyrki Lahtonen
Dec 5 '18 at 4:27
|
show 10 more comments
1
$begingroup$
Anyway, in case you haven't already done so, I recommend that you also familiarize yourself with the geometric point of view here. You can think of $f$ as a polynomial in $Bbb{F}_q[t,X]$ (clear the eventual denominators), and that defines a curve in the $(t,X)$-plane (and in its projective closure to include the points at infinity, or the primes lying above $P=(1/t)$. You need to go to a smooth model to get bijective correspondence between the primes of the function field and the points on the projective curve, but that can wait.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 15:08
1
$begingroup$
With $F(t,X):=X^8+t+1$ you get ramification at the place corresponding to the irreducible polynomial $t+1$. Or, if you prefer, at the point $Q=(t,X)=(-1,0)$. At that point only $X$ is a local parameter, and the valuation $nu_Q(t+1)=8$. So we get $e=8$. Calculus shows the same thing. The curve $t+1=-X^8$ has a vertical tangent at $(t,X)=(-1,0)$. The implicit function theorem says the same thing as $$dfrac{partial F}{partial X}(Q)=0.$$ IIRC the place $t=infty$ also ramifies in $Bbb{F}_q(t,X)$.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 19:20
1
$begingroup$
Yes. WHen taking the quotient ring by moding out the ideal $t-a$ you effectively set $t=a$. When you mod out a higher degree irreducible polynomial you get a higher degree place. Geometrically you then look at points whose coordinates are Galois conjugates of each other - all lumped together to form a single place.
$endgroup$
– Jyrki Lahtonen
Dec 5 '18 at 4:12
1
$begingroup$
On the curve $X^8+t+1=0$ we have the ring $mathcal{O}_Q$ of rational functions defined at $(t,X)=(-1,0)$. It has a maximal ideal $m_Q$ of those functions that vanish at $Q$. That ideal is generated by $t+1$ and $X$. An actually we see that $X$ alone suffices. Basicall because the equation $X^8=-(t+1)$ tells us how to write $t+1$ as a multiple of $X$. So $m_Q=langle Xrangle$. The general theory says that $mathcal{O}_Q$ is a DVR, and as a generator of the maximal ideal we can use $X$ to define the valuation. We have $t+1=(-1)cdot X^8$. So here $u=-1$ and $s=X$.
$endgroup$
– Jyrki Lahtonen
Dec 5 '18 at 4:20
1
$begingroup$
At tthe point $R$ with coordinates $t=-2,X=1$ we use the equation $X^8+t+1=0$ in the forms $$t+2=-(X^8-1)=-(X^7+X^6+cdots+X+1)(X-1)$$ and $$(X-1)=-frac{t+2}{X^7+X^6+cdots+X+1}$$ to see that either $t+2$ or $X-1$ alone generates the ideal $m_R$. So either $t+2$ or $X-1$ can serve in the role of the local parameter $s$ in defining the valuation. Anyway, because $t+2$ generates the maximal ideal it follows that $nu_R(t+2)=1$ meaning that the place $R$ lying above $P=t+2$ is unramified.
$endgroup$
– Jyrki Lahtonen
Dec 5 '18 at 4:27
1
1
$begingroup$
Anyway, in case you haven't already done so, I recommend that you also familiarize yourself with the geometric point of view here. You can think of $f$ as a polynomial in $Bbb{F}_q[t,X]$ (clear the eventual denominators), and that defines a curve in the $(t,X)$-plane (and in its projective closure to include the points at infinity, or the primes lying above $P=(1/t)$. You need to go to a smooth model to get bijective correspondence between the primes of the function field and the points on the projective curve, but that can wait.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 15:08
$begingroup$
Anyway, in case you haven't already done so, I recommend that you also familiarize yourself with the geometric point of view here. You can think of $f$ as a polynomial in $Bbb{F}_q[t,X]$ (clear the eventual denominators), and that defines a curve in the $(t,X)$-plane (and in its projective closure to include the points at infinity, or the primes lying above $P=(1/t)$. You need to go to a smooth model to get bijective correspondence between the primes of the function field and the points on the projective curve, but that can wait.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 15:08
1
1
$begingroup$
With $F(t,X):=X^8+t+1$ you get ramification at the place corresponding to the irreducible polynomial $t+1$. Or, if you prefer, at the point $Q=(t,X)=(-1,0)$. At that point only $X$ is a local parameter, and the valuation $nu_Q(t+1)=8$. So we get $e=8$. Calculus shows the same thing. The curve $t+1=-X^8$ has a vertical tangent at $(t,X)=(-1,0)$. The implicit function theorem says the same thing as $$dfrac{partial F}{partial X}(Q)=0.$$ IIRC the place $t=infty$ also ramifies in $Bbb{F}_q(t,X)$.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 19:20
$begingroup$
With $F(t,X):=X^8+t+1$ you get ramification at the place corresponding to the irreducible polynomial $t+1$. Or, if you prefer, at the point $Q=(t,X)=(-1,0)$. At that point only $X$ is a local parameter, and the valuation $nu_Q(t+1)=8$. So we get $e=8$. Calculus shows the same thing. The curve $t+1=-X^8$ has a vertical tangent at $(t,X)=(-1,0)$. The implicit function theorem says the same thing as $$dfrac{partial F}{partial X}(Q)=0.$$ IIRC the place $t=infty$ also ramifies in $Bbb{F}_q(t,X)$.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 19:20
1
1
$begingroup$
Yes. WHen taking the quotient ring by moding out the ideal $t-a$ you effectively set $t=a$. When you mod out a higher degree irreducible polynomial you get a higher degree place. Geometrically you then look at points whose coordinates are Galois conjugates of each other - all lumped together to form a single place.
$endgroup$
– Jyrki Lahtonen
Dec 5 '18 at 4:12
$begingroup$
Yes. WHen taking the quotient ring by moding out the ideal $t-a$ you effectively set $t=a$. When you mod out a higher degree irreducible polynomial you get a higher degree place. Geometrically you then look at points whose coordinates are Galois conjugates of each other - all lumped together to form a single place.
$endgroup$
– Jyrki Lahtonen
Dec 5 '18 at 4:12
1
1
$begingroup$
On the curve $X^8+t+1=0$ we have the ring $mathcal{O}_Q$ of rational functions defined at $(t,X)=(-1,0)$. It has a maximal ideal $m_Q$ of those functions that vanish at $Q$. That ideal is generated by $t+1$ and $X$. An actually we see that $X$ alone suffices. Basicall because the equation $X^8=-(t+1)$ tells us how to write $t+1$ as a multiple of $X$. So $m_Q=langle Xrangle$. The general theory says that $mathcal{O}_Q$ is a DVR, and as a generator of the maximal ideal we can use $X$ to define the valuation. We have $t+1=(-1)cdot X^8$. So here $u=-1$ and $s=X$.
$endgroup$
– Jyrki Lahtonen
Dec 5 '18 at 4:20
$begingroup$
On the curve $X^8+t+1=0$ we have the ring $mathcal{O}_Q$ of rational functions defined at $(t,X)=(-1,0)$. It has a maximal ideal $m_Q$ of those functions that vanish at $Q$. That ideal is generated by $t+1$ and $X$. An actually we see that $X$ alone suffices. Basicall because the equation $X^8=-(t+1)$ tells us how to write $t+1$ as a multiple of $X$. So $m_Q=langle Xrangle$. The general theory says that $mathcal{O}_Q$ is a DVR, and as a generator of the maximal ideal we can use $X$ to define the valuation. We have $t+1=(-1)cdot X^8$. So here $u=-1$ and $s=X$.
$endgroup$
– Jyrki Lahtonen
Dec 5 '18 at 4:20
1
1
$begingroup$
At tthe point $R$ with coordinates $t=-2,X=1$ we use the equation $X^8+t+1=0$ in the forms $$t+2=-(X^8-1)=-(X^7+X^6+cdots+X+1)(X-1)$$ and $$(X-1)=-frac{t+2}{X^7+X^6+cdots+X+1}$$ to see that either $t+2$ or $X-1$ alone generates the ideal $m_R$. So either $t+2$ or $X-1$ can serve in the role of the local parameter $s$ in defining the valuation. Anyway, because $t+2$ generates the maximal ideal it follows that $nu_R(t+2)=1$ meaning that the place $R$ lying above $P=t+2$ is unramified.
$endgroup$
– Jyrki Lahtonen
Dec 5 '18 at 4:27
$begingroup$
At tthe point $R$ with coordinates $t=-2,X=1$ we use the equation $X^8+t+1=0$ in the forms $$t+2=-(X^8-1)=-(X^7+X^6+cdots+X+1)(X-1)$$ and $$(X-1)=-frac{t+2}{X^7+X^6+cdots+X+1}$$ to see that either $t+2$ or $X-1$ alone generates the ideal $m_R$. So either $t+2$ or $X-1$ can serve in the role of the local parameter $s$ in defining the valuation. Anyway, because $t+2$ generates the maximal ideal it follows that $nu_R(t+2)=1$ meaning that the place $R$ lying above $P=t+2$ is unramified.
$endgroup$
– Jyrki Lahtonen
Dec 5 '18 at 4:27
|
show 10 more comments
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1
$begingroup$
Anyway, in case you haven't already done so, I recommend that you also familiarize yourself with the geometric point of view here. You can think of $f$ as a polynomial in $Bbb{F}_q[t,X]$ (clear the eventual denominators), and that defines a curve in the $(t,X)$-plane (and in its projective closure to include the points at infinity, or the primes lying above $P=(1/t)$. You need to go to a smooth model to get bijective correspondence between the primes of the function field and the points on the projective curve, but that can wait.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 15:08
1
$begingroup$
With $F(t,X):=X^8+t+1$ you get ramification at the place corresponding to the irreducible polynomial $t+1$. Or, if you prefer, at the point $Q=(t,X)=(-1,0)$. At that point only $X$ is a local parameter, and the valuation $nu_Q(t+1)=8$. So we get $e=8$. Calculus shows the same thing. The curve $t+1=-X^8$ has a vertical tangent at $(t,X)=(-1,0)$. The implicit function theorem says the same thing as $$dfrac{partial F}{partial X}(Q)=0.$$ IIRC the place $t=infty$ also ramifies in $Bbb{F}_q(t,X)$.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 19:20
1
$begingroup$
Yes. WHen taking the quotient ring by moding out the ideal $t-a$ you effectively set $t=a$. When you mod out a higher degree irreducible polynomial you get a higher degree place. Geometrically you then look at points whose coordinates are Galois conjugates of each other - all lumped together to form a single place.
$endgroup$
– Jyrki Lahtonen
Dec 5 '18 at 4:12
1
$begingroup$
On the curve $X^8+t+1=0$ we have the ring $mathcal{O}_Q$ of rational functions defined at $(t,X)=(-1,0)$. It has a maximal ideal $m_Q$ of those functions that vanish at $Q$. That ideal is generated by $t+1$ and $X$. An actually we see that $X$ alone suffices. Basicall because the equation $X^8=-(t+1)$ tells us how to write $t+1$ as a multiple of $X$. So $m_Q=langle Xrangle$. The general theory says that $mathcal{O}_Q$ is a DVR, and as a generator of the maximal ideal we can use $X$ to define the valuation. We have $t+1=(-1)cdot X^8$. So here $u=-1$ and $s=X$.
$endgroup$
– Jyrki Lahtonen
Dec 5 '18 at 4:20
1
$begingroup$
At tthe point $R$ with coordinates $t=-2,X=1$ we use the equation $X^8+t+1=0$ in the forms $$t+2=-(X^8-1)=-(X^7+X^6+cdots+X+1)(X-1)$$ and $$(X-1)=-frac{t+2}{X^7+X^6+cdots+X+1}$$ to see that either $t+2$ or $X-1$ alone generates the ideal $m_R$. So either $t+2$ or $X-1$ can serve in the role of the local parameter $s$ in defining the valuation. Anyway, because $t+2$ generates the maximal ideal it follows that $nu_R(t+2)=1$ meaning that the place $R$ lying above $P=t+2$ is unramified.
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– Jyrki Lahtonen
Dec 5 '18 at 4:27