Concerning 'a change of variables'
$begingroup$
Let $f: (x_1,ldots,x_n) mapsto (F_1,ldots,F_n)$
be a $k$-algebra endomorphism of $k[x_1,ldots,x_n]$, where $F$ is a field and $n geq 1$.
When one says that, after a change of variables, we can assume that $F_1,ldots,F_n$ has a certain property, does one mean the following:
There exists a $k$-algebra automorphism $g$ of $k[x_1,ldots,x_n]$,
such that $g(F_1),ldots,g(F_n)$ has that certain property.
Please, is this correct?
I guess that my question is quite trivial, but I wish to be sure that I am not missing something. More precisely, I wish to be sure that we are talking about $g(F_i)=g(f(x_i))=(gf)(x_i)$ and not about $(fg)(x_i)$.
Thank you very much!
abstract-algebra polynomials ring-theory commutative-algebra
asked Dec 31 '18 at 11:06
user237522user237522
2,1631617
$endgroup$
add a comment |
$begingroup$
Let $f: (x_1,ldots,x_n) mapsto (F_1,ldots,F_n)$
be a $k$-algebra endomorphism of $k[x_1,ldots,x_n]$, where $F$ is a field and $n geq 1$.
When one says that, after a change of variables, we can assume that $F_1,ldots,F_n$ has a certain property, does one mean the following:
There exists a $k$-algebra automorphism $g$ of $k[x_1,ldots,x_n]$,
such that $g(F_1),ldots,g(F_n)$ has that certain property.
Please, is this correct?
I guess that my question is quite trivial, but I wish to be sure that I am not missing something. More precisely, I wish to be sure that we are talking about $g(F_i)=g(f(x_i))=(gf)(x_i)$ and not about $(fg)(x_i)$.
Thank you very much!
abstract-algebra polynomials ring-theory commutative-algebra
asked Dec 31 '18 at 11:06
user237522user237522
2,1631617
$endgroup$
$begingroup$
First $f$, then $g$. So $(gf)(x_i) = g(f(x_i)) =g(F_i)$ would be fine.
$endgroup$
– Wuestenfux
Dec 31 '18 at 12:12
$begingroup$
Thank you. So what I thought is ok.
$endgroup$
– user237522
Dec 31 '18 at 12:23
add a comment |
$begingroup$
Let $f: (x_1,ldots,x_n) mapsto (F_1,ldots,F_n)$
be a $k$-algebra endomorphism of $k[x_1,ldots,x_n]$, where $F$ is a field and $n geq 1$.
When one says that, after a change of variables, we can assume that $F_1,ldots,F_n$ has a certain property, does one mean the following:
There exists a $k$-algebra automorphism $g$ of $k[x_1,ldots,x_n]$,
such that $g(F_1),ldots,g(F_n)$ has that certain property.
Please, is this correct?
I guess that my question is quite trivial, but I wish to be sure that I am not missing something. More precisely, I wish to be sure that we are talking about $g(F_i)=g(f(x_i))=(gf)(x_i)$ and not about $(fg)(x_i)$.
Thank you very much!
abstract-algebra polynomials ring-theory commutative-algebra
asked Dec 31 '18 at 11:06
user237522user237522
2,1631617
$endgroup$
Let $f: (x_1,ldots,x_n) mapsto (F_1,ldots,F_n)$
be a $k$-algebra endomorphism of $k[x_1,ldots,x_n]$, where $F$ is a field and $n geq 1$.
When one says that, after a change of variables, we can assume that $F_1,ldots,F_n$ has a certain property, does one mean the following:
There exists a $k$-algebra automorphism $g$ of $k[x_1,ldots,x_n]$,
such that $g(F_1),ldots,g(F_n)$ has that certain property.
Please, is this correct?
I guess that my question is quite trivial, but I wish to be sure that I am not missing something. More precisely, I wish to be sure that we are talking about $g(F_i)=g(f(x_i))=(gf)(x_i)$ and not about $(fg)(x_i)$.
Thank you very much!
abstract-algebra polynomials ring-theory commutative-algebra
abstract-algebra polynomials ring-theory commutative-algebra
asked Dec 31 '18 at 11:06
user237522user237522
2,1631617
asked Dec 31 '18 at 11:06
user237522user237522
2,1631617
edited Dec 31 '18 at 11:39
user237522
asked Dec 31 '18 at 11:06
user237522user237522
2,1631617
asked Dec 31 '18 at 11:06
user237522user237522
2,1631617
asked Dec 31 '18 at 11:06
user237522user237522
2,1631617
2,1631617
$begingroup$
First $f$, then $g$. So $(gf)(x_i) = g(f(x_i)) =g(F_i)$ would be fine.
$endgroup$
– Wuestenfux
Dec 31 '18 at 12:12
$begingroup$
Thank you. So what I thought is ok.
$endgroup$
– user237522
Dec 31 '18 at 12:23
add a comment |
$begingroup$
First $f$, then $g$. So $(gf)(x_i) = g(f(x_i)) =g(F_i)$ would be fine.
$endgroup$
– Wuestenfux
Dec 31 '18 at 12:12
$begingroup$
Thank you. So what I thought is ok.
$endgroup$
– user237522
Dec 31 '18 at 12:23
$begingroup$
First $f$, then $g$. So $(gf)(x_i) = g(f(x_i)) =g(F_i)$ would be fine.
$endgroup$
– Wuestenfux
Dec 31 '18 at 12:12
$begingroup$
First $f$, then $g$. So $(gf)(x_i) = g(f(x_i)) =g(F_i)$ would be fine.
$endgroup$
– Wuestenfux
Dec 31 '18 at 12:12
$begingroup$
Thank you. So what I thought is ok.
$endgroup$
– user237522
Dec 31 '18 at 12:23
$begingroup$
Thank you. So what I thought is ok.
$endgroup$
– user237522
Dec 31 '18 at 12:23
add a comment |
1 Answer
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$begingroup$
This phrase is ambiguous and could have several meanings: it could refer to a composition of the form $gf$, or it could refer instead to $fg$, or it could refer to the conjugation $gfg^{-1}$, or it might even refer to something like $gfh$ where $g$ and $h$ are two automorphisms. You'll have to figure out from context what meaning is intended (it should usually be clear in context). Lacking any context, I would guess that the conjugation $gfg^{-1}$ is the most likely meaning.
answered Jan 9 at 5:00
Eric WofseyEric Wofsey
187k14216344
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
This phrase is ambiguous and could have several meanings: it could refer to a composition of the form $gf$, or it could refer instead to $fg$, or it could refer to the conjugation $gfg^{-1}$, or it might even refer to something like $gfh$ where $g$ and $h$ are two automorphisms. You'll have to figure out from context what meaning is intended (it should usually be clear in context). Lacking any context, I would guess that the conjugation $gfg^{-1}$ is the most likely meaning.
answered Jan 9 at 5:00
Eric WofseyEric Wofsey
187k14216344
$endgroup$
add a comment |
$begingroup$
This phrase is ambiguous and could have several meanings: it could refer to a composition of the form $gf$, or it could refer instead to $fg$, or it could refer to the conjugation $gfg^{-1}$, or it might even refer to something like $gfh$ where $g$ and $h$ are two automorphisms. You'll have to figure out from context what meaning is intended (it should usually be clear in context). Lacking any context, I would guess that the conjugation $gfg^{-1}$ is the most likely meaning.
answered Jan 9 at 5:00
Eric WofseyEric Wofsey
187k14216344
$endgroup$
add a comment |
$begingroup$
This phrase is ambiguous and could have several meanings: it could refer to a composition of the form $gf$, or it could refer instead to $fg$, or it could refer to the conjugation $gfg^{-1}$, or it might even refer to something like $gfh$ where $g$ and $h$ are two automorphisms. You'll have to figure out from context what meaning is intended (it should usually be clear in context). Lacking any context, I would guess that the conjugation $gfg^{-1}$ is the most likely meaning.
answered Jan 9 at 5:00
Eric WofseyEric Wofsey
187k14216344
$endgroup$
This phrase is ambiguous and could have several meanings: it could refer to a composition of the form $gf$, or it could refer instead to $fg$, or it could refer to the conjugation $gfg^{-1}$, or it might even refer to something like $gfh$ where $g$ and $h$ are two automorphisms. You'll have to figure out from context what meaning is intended (it should usually be clear in context). Lacking any context, I would guess that the conjugation $gfg^{-1}$ is the most likely meaning.
answered Jan 9 at 5:00
Eric WofseyEric Wofsey
187k14216344
answered Jan 9 at 5:00
Eric WofseyEric Wofsey
187k14216344
answered Jan 9 at 5:00
Eric WofseyEric Wofsey
187k14216344
answered Jan 9 at 5:00
Eric WofseyEric Wofsey
187k14216344
187k14216344
add a comment |
add a comment |
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$begingroup$
First $f$, then $g$. So $(gf)(x_i) = g(f(x_i)) =g(F_i)$ would be fine.
$endgroup$
– Wuestenfux
Dec 31 '18 at 12:12
$begingroup$
Thank you. So what I thought is ok.
$endgroup$
– user237522
Dec 31 '18 at 12:23