$P(z,w)=exp(w)-z=0$ polynomial in $z$ and $w$ and irreducible?
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Considering the set $$X=(z,w)in mathbbC^2 mid P(z,w)=0,$$ $P:mathbbC²rightarrow mathbbC$ holomorphic in each variable separately, including any polynomial in $z$ and $w$ and $P$ is non-singular. Consequently $X$ is a Riemann surface.
If I take $P(z,w)=w²-z$ it's clear that this function is holomorphic in each variable and $P$ is polynomial. But if I take $P(z,w)=exp(w)-z$, is this also a polynomial in $z$ and $w$ (because $exp(w)$ can be represented as a power series, but is this enough to say $exp(w)$ is a polynomial?)
Furthermore is $P(z,w)=w²-z$ irreducible and why?
Thank you very much! :)
polynomials irreducible-polynomials riemann-surfaces
edited Sep 5 at 11:34
Bernard
112k635104
asked Sep 5 at 11:30
matzzzz
93
add a comment |Â
up vote
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Considering the set $$X=(z,w)in mathbbC^2 mid P(z,w)=0,$$ $P:mathbbC²rightarrow mathbbC$ holomorphic in each variable separately, including any polynomial in $z$ and $w$ and $P$ is non-singular. Consequently $X$ is a Riemann surface.
If I take $P(z,w)=w²-z$ it's clear that this function is holomorphic in each variable and $P$ is polynomial. But if I take $P(z,w)=exp(w)-z$, is this also a polynomial in $z$ and $w$ (because $exp(w)$ can be represented as a power series, but is this enough to say $exp(w)$ is a polynomial?)
Furthermore is $P(z,w)=w²-z$ irreducible and why?
Thank you very much! :)
polynomials irreducible-polynomials riemann-surfaces
edited Sep 5 at 11:34
Bernard
112k635104
asked Sep 5 at 11:30
matzzzz
93
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Considering the set $$X=(z,w)in mathbbC^2 mid P(z,w)=0,$$ $P:mathbbC²rightarrow mathbbC$ holomorphic in each variable separately, including any polynomial in $z$ and $w$ and $P$ is non-singular. Consequently $X$ is a Riemann surface.
If I take $P(z,w)=w²-z$ it's clear that this function is holomorphic in each variable and $P$ is polynomial. But if I take $P(z,w)=exp(w)-z$, is this also a polynomial in $z$ and $w$ (because $exp(w)$ can be represented as a power series, but is this enough to say $exp(w)$ is a polynomial?)
Furthermore is $P(z,w)=w²-z$ irreducible and why?
Thank you very much! :)
polynomials irreducible-polynomials riemann-surfaces
edited Sep 5 at 11:34
Bernard
112k635104
asked Sep 5 at 11:30
matzzzz
93
Considering the set $$X=(z,w)in mathbbC^2 mid P(z,w)=0,$$ $P:mathbbC²rightarrow mathbbC$ holomorphic in each variable separately, including any polynomial in $z$ and $w$ and $P$ is non-singular. Consequently $X$ is a Riemann surface.
If I take $P(z,w)=w²-z$ it's clear that this function is holomorphic in each variable and $P$ is polynomial. But if I take $P(z,w)=exp(w)-z$, is this also a polynomial in $z$ and $w$ (because $exp(w)$ can be represented as a power series, but is this enough to say $exp(w)$ is a polynomial?)
Furthermore is $P(z,w)=w²-z$ irreducible and why?
Thank you very much! :)
polynomials irreducible-polynomials riemann-surfaces
polynomials irreducible-polynomials riemann-surfaces
edited Sep 5 at 11:34
Bernard
112k635104
asked Sep 5 at 11:30
matzzzz
93
edited Sep 5 at 11:34
Bernard
112k635104
asked Sep 5 at 11:30
matzzzz
93
edited Sep 5 at 11:34
Bernard
112k635104
edited Sep 5 at 11:34
Bernard
112k635104
edited Sep 5 at 11:34
Bernard
112k635104
112k635104
asked Sep 5 at 11:30
matzzzz
93
asked Sep 5 at 11:30
matzzzz
93
asked Sep 5 at 11:30
matzzzz
93
93
add a comment |Â
add a comment |Â
1 Answer
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No, $exp w$ is not a polynomial: if it were, its derivatives of sufficiently high order would be $0$ – and $exp^(n)(w)=exp w$ for all $n$.
As to $P(z,w)= w^2-z$ being irreducible; you can apply Eisenstein's criterion in the univariate polynomial ring $mathbf C[z][w]$ over the P.I.D. $mathbf C[z]$.
answered Sep 5 at 11:42
Bernard
112k635104
Thank you @Bernard. So P(z,w)=exp(w)-z is also not irreduzible, right? :)
– matzzzz
Sep 5 at 11:52
In which ring? You're not in a polynomial ring with this expression.
– Bernard
Sep 5 at 11:55
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
No, $exp w$ is not a polynomial: if it were, its derivatives of sufficiently high order would be $0$ – and $exp^(n)(w)=exp w$ for all $n$.
As to $P(z,w)= w^2-z$ being irreducible; you can apply Eisenstein's criterion in the univariate polynomial ring $mathbf C[z][w]$ over the P.I.D. $mathbf C[z]$.
answered Sep 5 at 11:42
Bernard
112k635104
Thank you @Bernard. So P(z,w)=exp(w)-z is also not irreduzible, right? :)
– matzzzz
Sep 5 at 11:52
In which ring? You're not in a polynomial ring with this expression.
– Bernard
Sep 5 at 11:55
add a comment |Â
up vote
1
down vote
No, $exp w$ is not a polynomial: if it were, its derivatives of sufficiently high order would be $0$ – and $exp^(n)(w)=exp w$ for all $n$.
As to $P(z,w)= w^2-z$ being irreducible; you can apply Eisenstein's criterion in the univariate polynomial ring $mathbf C[z][w]$ over the P.I.D. $mathbf C[z]$.
answered Sep 5 at 11:42
Bernard
112k635104
Thank you @Bernard. So P(z,w)=exp(w)-z is also not irreduzible, right? :)
– matzzzz
Sep 5 at 11:52
In which ring? You're not in a polynomial ring with this expression.
– Bernard
Sep 5 at 11:55
add a comment |Â
up vote
1
down vote
up vote
1
down vote
No, $exp w$ is not a polynomial: if it were, its derivatives of sufficiently high order would be $0$ – and $exp^(n)(w)=exp w$ for all $n$.
As to $P(z,w)= w^2-z$ being irreducible; you can apply Eisenstein's criterion in the univariate polynomial ring $mathbf C[z][w]$ over the P.I.D. $mathbf C[z]$.
answered Sep 5 at 11:42
Bernard
112k635104
No, $exp w$ is not a polynomial: if it were, its derivatives of sufficiently high order would be $0$ – and $exp^(n)(w)=exp w$ for all $n$.
As to $P(z,w)= w^2-z$ being irreducible; you can apply Eisenstein's criterion in the univariate polynomial ring $mathbf C[z][w]$ over the P.I.D. $mathbf C[z]$.
answered Sep 5 at 11:42
Bernard
112k635104
answered Sep 5 at 11:42
Bernard
112k635104
answered Sep 5 at 11:42
Bernard
112k635104
answered Sep 5 at 11:42
Bernard
112k635104
112k635104
Thank you @Bernard. So P(z,w)=exp(w)-z is also not irreduzible, right? :)
– matzzzz
Sep 5 at 11:52
In which ring? You're not in a polynomial ring with this expression.
– Bernard
Sep 5 at 11:55
add a comment |Â
Thank you @Bernard. So P(z,w)=exp(w)-z is also not irreduzible, right? :)
– matzzzz
Sep 5 at 11:52
In which ring? You're not in a polynomial ring with this expression.
– Bernard
Sep 5 at 11:55
Thank you @Bernard. So P(z,w)=exp(w)-z is also not irreduzible, right? :)
– matzzzz
Sep 5 at 11:52
Thank you @Bernard. So P(z,w)=exp(w)-z is also not irreduzible, right? :)
– matzzzz
Sep 5 at 11:52
In which ring? You're not in a polynomial ring with this expression.
– Bernard
Sep 5 at 11:55
In which ring? You're not in a polynomial ring with this expression.
– Bernard
Sep 5 at 11:55
add a comment |Â
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