Inverse of multiary composited functions
$begingroup$
The following is given:
$K_0$ a field,
$A$ and $B$ binary functions, algebraic over $K_0$,
$f_1$ and $f_2$ unary bijective functions, algebraically independent over $K_0$ of each other,
$g_1$ and $g_2$ unary bijective functions, algebraically independent over $K_0$ of each other,
$F$ with $Fcolon zmapsto A(f_1(z),f_2(z))$ and $Phi$ with $Phicolon zmapsto B(g_1(z),g_2(z))$ bijective functions.
When exactly is $Phi$ the inverse of $F$?
How can one prove that?
$ $
I already found the following.
If $Phi$ is the inverse of $F$, $F(Phi(z))=z$ and $Phi(F(z))=z$ must apply, that means
$$A(f_1(B(g_1(z),g_2(z))),f_2(B(g_1(z),g_2(z))))=z$$
and
$$B(g_1(A(f_1(z),f_2(z))),g_2(A(f_1(z),f_2(z))))=z.$$
If $f_1$ and $f_2$ would be algebraically dependent over $K_0$ of each other, a $Phi$ would exist which is the inverse of $F$.
But is this the only case?
abstract-algebra analysis closed-form
asked Dec 16 '18 at 14:01
IV_IV_
1,335525
$endgroup$
add a comment |
$begingroup$
The following is given:
$K_0$ a field,
$A$ and $B$ binary functions, algebraic over $K_0$,
$f_1$ and $f_2$ unary bijective functions, algebraically independent over $K_0$ of each other,
$g_1$ and $g_2$ unary bijective functions, algebraically independent over $K_0$ of each other,
$F$ with $Fcolon zmapsto A(f_1(z),f_2(z))$ and $Phi$ with $Phicolon zmapsto B(g_1(z),g_2(z))$ bijective functions.
When exactly is $Phi$ the inverse of $F$?
How can one prove that?
$ $
I already found the following.
If $Phi$ is the inverse of $F$, $F(Phi(z))=z$ and $Phi(F(z))=z$ must apply, that means
$$A(f_1(B(g_1(z),g_2(z))),f_2(B(g_1(z),g_2(z))))=z$$
and
$$B(g_1(A(f_1(z),f_2(z))),g_2(A(f_1(z),f_2(z))))=z.$$
If $f_1$ and $f_2$ would be algebraically dependent over $K_0$ of each other, a $Phi$ would exist which is the inverse of $F$.
But is this the only case?
abstract-algebra analysis closed-form
asked Dec 16 '18 at 14:01
IV_IV_
1,335525
$endgroup$
add a comment |
$begingroup$
The following is given:
$K_0$ a field,
$A$ and $B$ binary functions, algebraic over $K_0$,
$f_1$ and $f_2$ unary bijective functions, algebraically independent over $K_0$ of each other,
$g_1$ and $g_2$ unary bijective functions, algebraically independent over $K_0$ of each other,
$F$ with $Fcolon zmapsto A(f_1(z),f_2(z))$ and $Phi$ with $Phicolon zmapsto B(g_1(z),g_2(z))$ bijective functions.
When exactly is $Phi$ the inverse of $F$?
How can one prove that?
$ $
I already found the following.
If $Phi$ is the inverse of $F$, $F(Phi(z))=z$ and $Phi(F(z))=z$ must apply, that means
$$A(f_1(B(g_1(z),g_2(z))),f_2(B(g_1(z),g_2(z))))=z$$
and
$$B(g_1(A(f_1(z),f_2(z))),g_2(A(f_1(z),f_2(z))))=z.$$
If $f_1$ and $f_2$ would be algebraically dependent over $K_0$ of each other, a $Phi$ would exist which is the inverse of $F$.
But is this the only case?
abstract-algebra analysis closed-form
asked Dec 16 '18 at 14:01
IV_IV_
1,335525
$endgroup$
The following is given:
$K_0$ a field,
$A$ and $B$ binary functions, algebraic over $K_0$,
$f_1$ and $f_2$ unary bijective functions, algebraically independent over $K_0$ of each other,
$g_1$ and $g_2$ unary bijective functions, algebraically independent over $K_0$ of each other,
$F$ with $Fcolon zmapsto A(f_1(z),f_2(z))$ and $Phi$ with $Phicolon zmapsto B(g_1(z),g_2(z))$ bijective functions.
When exactly is $Phi$ the inverse of $F$?
How can one prove that?
$ $
I already found the following.
If $Phi$ is the inverse of $F$, $F(Phi(z))=z$ and $Phi(F(z))=z$ must apply, that means
$$A(f_1(B(g_1(z),g_2(z))),f_2(B(g_1(z),g_2(z))))=z$$
and
$$B(g_1(A(f_1(z),f_2(z))),g_2(A(f_1(z),f_2(z))))=z.$$
If $f_1$ and $f_2$ would be algebraically dependent over $K_0$ of each other, a $Phi$ would exist which is the inverse of $F$.
But is this the only case?
abstract-algebra analysis closed-form
abstract-algebra analysis closed-form
asked Dec 16 '18 at 14:01
IV_IV_
1,335525
asked Dec 16 '18 at 14:01
IV_IV_
1,335525
asked Dec 16 '18 at 14:01
IV_IV_
1,335525
asked Dec 16 '18 at 14:01
IV_IV_
1,335525
asked Dec 16 '18 at 14:01
IV_IV_
1,335525
1,335525
add a comment |
add a comment |
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