Mixing
![Prove that if $P[Xleq Y] =1$ then $E[X]leq E[Y]$](https://lh4.googleusercontent.com/-YFYIJxwAX_g/AAAAAAAAAAI/AAAAAAAClZw/reL-HBAoiOs/s72-c/photo.jpg?sz=32)
Meromorphic Functions on Riemann Surfaces
$begingroup$
My question refers to a step in the proof of Prop. 3.3.5 Szamuely and Tamásin's "Galois groups and fundamental groups":
Here the statement and Thm 3.3.3 & lemma 3.3.6: The main ingredients for the proof:
AND here the proof of 3.3.5 with red tagged unclear argument:
My question is why $f$ satisfies an irreducible polynomial equation $(phi^*a_n)f^n + ... +(phi^*a_0)=0$? Why irreducible?
Lemma 3.3.6 says that the polynomial equation is not neccessary irreducible.
Intuitively I guess that should have something to do with the assumption that all values $f(y_i)$ are distinct. Why does it suffice? I don't find a clear argument.
irreducible-polynomials riemann-surfaces complex-manifolds function-fields
asked Jan 18 at 17:54
KarlPeterKarlPeter
3541315
$endgroup$
add a comment |
$begingroup$
My question refers to a step in the proof of Prop. 3.3.5 Szamuely and Tamásin's "Galois groups and fundamental groups":
Here the statement and Thm 3.3.3 & lemma 3.3.6: The main ingredients for the proof:
AND here the proof of 3.3.5 with red tagged unclear argument:
My question is why $f$ satisfies an irreducible polynomial equation $(phi^*a_n)f^n + ... +(phi^*a_0)=0$? Why irreducible?
Lemma 3.3.6 says that the polynomial equation is not neccessary irreducible.
Intuitively I guess that should have something to do with the assumption that all values $f(y_i)$ are distinct. Why does it suffice? I don't find a clear argument.
irreducible-polynomials riemann-surfaces complex-manifolds function-fields
asked Jan 18 at 17:54
KarlPeterKarlPeter
3541315
$endgroup$
add a comment |
$begingroup$
My question refers to a step in the proof of Prop. 3.3.5 Szamuely and Tamásin's "Galois groups and fundamental groups":
Here the statement and Thm 3.3.3 & lemma 3.3.6: The main ingredients for the proof:
AND here the proof of 3.3.5 with red tagged unclear argument:
My question is why $f$ satisfies an irreducible polynomial equation $(phi^*a_n)f^n + ... +(phi^*a_0)=0$? Why irreducible?
Lemma 3.3.6 says that the polynomial equation is not neccessary irreducible.
Intuitively I guess that should have something to do with the assumption that all values $f(y_i)$ are distinct. Why does it suffice? I don't find a clear argument.
irreducible-polynomials riemann-surfaces complex-manifolds function-fields
asked Jan 18 at 17:54
KarlPeterKarlPeter
3541315
$endgroup$
My question refers to a step in the proof of Prop. 3.3.5 Szamuely and Tamásin's "Galois groups and fundamental groups":
Here the statement and Thm 3.3.3 & lemma 3.3.6: The main ingredients for the proof:
AND here the proof of 3.3.5 with red tagged unclear argument:
My question is why $f$ satisfies an irreducible polynomial equation $(phi^*a_n)f^n + ... +(phi^*a_0)=0$? Why irreducible?
Lemma 3.3.6 says that the polynomial equation is not neccessary irreducible.
Intuitively I guess that should have something to do with the assumption that all values $f(y_i)$ are distinct. Why does it suffice? I don't find a clear argument.
irreducible-polynomials riemann-surfaces complex-manifolds function-fields
irreducible-polynomials riemann-surfaces complex-manifolds function-fields
asked Jan 18 at 17:54
KarlPeterKarlPeter
3541315
asked Jan 18 at 17:54
KarlPeterKarlPeter
3541315
asked Jan 18 at 17:54
KarlPeterKarlPeter
3541315
asked Jan 18 at 17:54
KarlPeterKarlPeter
3541315
asked Jan 18 at 17:54
KarlPeterKarlPeter
3541315
3541315
add a comment |
add a comment |
1 Answer
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I haven't looked into the whole question in detail, but this is just a general algebraic principle. The Lemma tells you that there is some non-zero polynomial $P$ with coefficients in $M(Y)$ that kills $f$, i.e. satisfies $P(f) =0$. Now take such a non-ozero polynomial $P_0$ of smallest degree $n leq d$. If $P_0$ were not irreducible, you could factor $P_0$ as $P_0 = RS$ for some non-zero polynoimals $R,S$, both of strictly smaller degree than $P_0$. But then $0=P_0(f) = R(f) S(f)$ implies that either $R(f) =0$ or $S(f) = 0$, contradicting minimiality of the degree.
answered Jan 18 at 20:20
m.sm.s
1,324313
$endgroup$
$begingroup$
Seems plausible. The point that irritates me is then the formulation of lemma 3.3.6. Then there shouls always exist an irreducible polynomial where $f$ vanishs. Since $X,Y$ connected $M(X), MY)$ are fields so no problem with zero divisor. Do you maybe have an idea why in 3.3.6 there is told that the exiting vanishing polynomial might be not irreducible. Why this remark if we know with the principle ypu described above that there exist indeed an irreducible one with desired properties.
$endgroup$
– KarlPeter
Jan 18 at 20:35
$begingroup$
It's not too weak, all that is needed for the general argument I presented is that there is some non-constant polynomial in $M(Y)[T]$ annihilating $f$.
$endgroup$
– m.s
Jan 18 at 20:46
add a comment |
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$begingroup$
I haven't looked into the whole question in detail, but this is just a general algebraic principle. The Lemma tells you that there is some non-zero polynomial $P$ with coefficients in $M(Y)$ that kills $f$, i.e. satisfies $P(f) =0$. Now take such a non-ozero polynomial $P_0$ of smallest degree $n leq d$. If $P_0$ were not irreducible, you could factor $P_0$ as $P_0 = RS$ for some non-zero polynoimals $R,S$, both of strictly smaller degree than $P_0$. But then $0=P_0(f) = R(f) S(f)$ implies that either $R(f) =0$ or $S(f) = 0$, contradicting minimiality of the degree.
answered Jan 18 at 20:20
m.sm.s
1,324313
$endgroup$
$begingroup$
Seems plausible. The point that irritates me is then the formulation of lemma 3.3.6. Then there shouls always exist an irreducible polynomial where $f$ vanishs. Since $X,Y$ connected $M(X), MY)$ are fields so no problem with zero divisor. Do you maybe have an idea why in 3.3.6 there is told that the exiting vanishing polynomial might be not irreducible. Why this remark if we know with the principle ypu described above that there exist indeed an irreducible one with desired properties.
$endgroup$
– KarlPeter
Jan 18 at 20:35
$begingroup$
It's not too weak, all that is needed for the general argument I presented is that there is some non-constant polynomial in $M(Y)[T]$ annihilating $f$.
$endgroup$
– m.s
Jan 18 at 20:46
add a comment |
$begingroup$
I haven't looked into the whole question in detail, but this is just a general algebraic principle. The Lemma tells you that there is some non-zero polynomial $P$ with coefficients in $M(Y)$ that kills $f$, i.e. satisfies $P(f) =0$. Now take such a non-ozero polynomial $P_0$ of smallest degree $n leq d$. If $P_0$ were not irreducible, you could factor $P_0$ as $P_0 = RS$ for some non-zero polynoimals $R,S$, both of strictly smaller degree than $P_0$. But then $0=P_0(f) = R(f) S(f)$ implies that either $R(f) =0$ or $S(f) = 0$, contradicting minimiality of the degree.
answered Jan 18 at 20:20
m.sm.s
1,324313
$endgroup$
$begingroup$
Seems plausible. The point that irritates me is then the formulation of lemma 3.3.6. Then there shouls always exist an irreducible polynomial where $f$ vanishs. Since $X,Y$ connected $M(X), MY)$ are fields so no problem with zero divisor. Do you maybe have an idea why in 3.3.6 there is told that the exiting vanishing polynomial might be not irreducible. Why this remark if we know with the principle ypu described above that there exist indeed an irreducible one with desired properties.
$endgroup$
– KarlPeter
Jan 18 at 20:35
$begingroup$
It's not too weak, all that is needed for the general argument I presented is that there is some non-constant polynomial in $M(Y)[T]$ annihilating $f$.
$endgroup$
– m.s
Jan 18 at 20:46
add a comment |
$begingroup$
I haven't looked into the whole question in detail, but this is just a general algebraic principle. The Lemma tells you that there is some non-zero polynomial $P$ with coefficients in $M(Y)$ that kills $f$, i.e. satisfies $P(f) =0$. Now take such a non-ozero polynomial $P_0$ of smallest degree $n leq d$. If $P_0$ were not irreducible, you could factor $P_0$ as $P_0 = RS$ for some non-zero polynoimals $R,S$, both of strictly smaller degree than $P_0$. But then $0=P_0(f) = R(f) S(f)$ implies that either $R(f) =0$ or $S(f) = 0$, contradicting minimiality of the degree.
answered Jan 18 at 20:20
m.sm.s
1,324313
$endgroup$
I haven't looked into the whole question in detail, but this is just a general algebraic principle. The Lemma tells you that there is some non-zero polynomial $P$ with coefficients in $M(Y)$ that kills $f$, i.e. satisfies $P(f) =0$. Now take such a non-ozero polynomial $P_0$ of smallest degree $n leq d$. If $P_0$ were not irreducible, you could factor $P_0$ as $P_0 = RS$ for some non-zero polynoimals $R,S$, both of strictly smaller degree than $P_0$. But then $0=P_0(f) = R(f) S(f)$ implies that either $R(f) =0$ or $S(f) = 0$, contradicting minimiality of the degree.
answered Jan 18 at 20:20
m.sm.s
1,324313
answered Jan 18 at 20:20
m.sm.s
1,324313
answered Jan 18 at 20:20
m.sm.s
1,324313
answered Jan 18 at 20:20
m.sm.s
1,324313
1,324313
$begingroup$
Seems plausible. The point that irritates me is then the formulation of lemma 3.3.6. Then there shouls always exist an irreducible polynomial where $f$ vanishs. Since $X,Y$ connected $M(X), MY)$ are fields so no problem with zero divisor. Do you maybe have an idea why in 3.3.6 there is told that the exiting vanishing polynomial might be not irreducible. Why this remark if we know with the principle ypu described above that there exist indeed an irreducible one with desired properties.
$endgroup$
– KarlPeter
Jan 18 at 20:35
$begingroup$
It's not too weak, all that is needed for the general argument I presented is that there is some non-constant polynomial in $M(Y)[T]$ annihilating $f$.
$endgroup$
– m.s
Jan 18 at 20:46
add a comment |
$begingroup$
Seems plausible. The point that irritates me is then the formulation of lemma 3.3.6. Then there shouls always exist an irreducible polynomial where $f$ vanishs. Since $X,Y$ connected $M(X), MY)$ are fields so no problem with zero divisor. Do you maybe have an idea why in 3.3.6 there is told that the exiting vanishing polynomial might be not irreducible. Why this remark if we know with the principle ypu described above that there exist indeed an irreducible one with desired properties.
$endgroup$
– KarlPeter
Jan 18 at 20:35
$begingroup$
It's not too weak, all that is needed for the general argument I presented is that there is some non-constant polynomial in $M(Y)[T]$ annihilating $f$.
$endgroup$
– m.s
Jan 18 at 20:46
$begingroup$
Seems plausible. The point that irritates me is then the formulation of lemma 3.3.6. Then there shouls always exist an irreducible polynomial where $f$ vanishs. Since $X,Y$ connected $M(X), MY)$ are fields so no problem with zero divisor. Do you maybe have an idea why in 3.3.6 there is told that the exiting vanishing polynomial might be not irreducible. Why this remark if we know with the principle ypu described above that there exist indeed an irreducible one with desired properties.
$endgroup$
– KarlPeter
Jan 18 at 20:35
$begingroup$
Seems plausible. The point that irritates me is then the formulation of lemma 3.3.6. Then there shouls always exist an irreducible polynomial where $f$ vanishs. Since $X,Y$ connected $M(X), MY)$ are fields so no problem with zero divisor. Do you maybe have an idea why in 3.3.6 there is told that the exiting vanishing polynomial might be not irreducible. Why this remark if we know with the principle ypu described above that there exist indeed an irreducible one with desired properties.
$endgroup$
– KarlPeter
Jan 18 at 20:35
$begingroup$
It's not too weak, all that is needed for the general argument I presented is that there is some non-constant polynomial in $M(Y)[T]$ annihilating $f$.
$endgroup$
– m.s
Jan 18 at 20:46
$begingroup$
It's not too weak, all that is needed for the general argument I presented is that there is some non-constant polynomial in $M(Y)[T]$ annihilating $f$.
$endgroup$
– m.s
Jan 18 at 20:46
add a comment |
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