Integer valued polynomials in several variables
$begingroup$
For simplicity this is about polynomials in just two variables.
Any $finmathbb Q[X,Y]$ can be written as a linear combination of monomials
$X^iY^j$ and therefore as a sum of polynomials $p_{ij}inmathbb Q[X^iY^j]$ over one variable:
$$displaystyle f(X,Y)=p_{00}+sum_{gcd(i,j)=1}p_{ij}(X^iY^j).$$
My question: is the subring of all integer valued polynomials $f$ over
two variables identical with the subring of all sums of integer valued
polynomials $p_{ij}$ over one variable as above?
An integer valued polynomial in one variable is a polynomial $p$ with rational coefficients such that $p(mathbb Z)subseteq mathbb Z$. And corresponding for polynomials over several variables.
It might be some abuse of language to call $p_{ij}$ polynomials over one variable, they may rather be polynomials over one variable applied to monomials $X^iY^j$.
The answer is that there is a counter-example is
$frac{X(X-1)Y(Y-1)}{4}$, as a user of Mathematics Stack Overflow
found. The sums doesn't form a ring, just a group.
number-theory elementary-number-theory polynomials ring-theory
asked Jan 7 at 13:44
LehsLehs
7,07631664
$endgroup$
add a comment |
$begingroup$
For simplicity this is about polynomials in just two variables.
Any $finmathbb Q[X,Y]$ can be written as a linear combination of monomials
$X^iY^j$ and therefore as a sum of polynomials $p_{ij}inmathbb Q[X^iY^j]$ over one variable:
$$displaystyle f(X,Y)=p_{00}+sum_{gcd(i,j)=1}p_{ij}(X^iY^j).$$
My question: is the subring of all integer valued polynomials $f$ over
two variables identical with the subring of all sums of integer valued
polynomials $p_{ij}$ over one variable as above?
An integer valued polynomial in one variable is a polynomial $p$ with rational coefficients such that $p(mathbb Z)subseteq mathbb Z$. And corresponding for polynomials over several variables.
It might be some abuse of language to call $p_{ij}$ polynomials over one variable, they may rather be polynomials over one variable applied to monomials $X^iY^j$.
The answer is that there is a counter-example is
$frac{X(X-1)Y(Y-1)}{4}$, as a user of Mathematics Stack Overflow
found. The sums doesn't form a ring, just a group.
number-theory elementary-number-theory polynomials ring-theory
asked Jan 7 at 13:44
LehsLehs
7,07631664
$endgroup$
1
$begingroup$
How is your displayed formula to be viewed as a sum of polynomials "over one variable"? Also what does the $(i,j))=1$ under the sum mean? That $i,j$ are coprime, or that they are any pairs of positive integers?
$endgroup$
– coffeemath
Jan 7 at 13:52
add a comment |
$begingroup$
For simplicity this is about polynomials in just two variables.
Any $finmathbb Q[X,Y]$ can be written as a linear combination of monomials
$X^iY^j$ and therefore as a sum of polynomials $p_{ij}inmathbb Q[X^iY^j]$ over one variable:
$$displaystyle f(X,Y)=p_{00}+sum_{gcd(i,j)=1}p_{ij}(X^iY^j).$$
My question: is the subring of all integer valued polynomials $f$ over
two variables identical with the subring of all sums of integer valued
polynomials $p_{ij}$ over one variable as above?
An integer valued polynomial in one variable is a polynomial $p$ with rational coefficients such that $p(mathbb Z)subseteq mathbb Z$. And corresponding for polynomials over several variables.
It might be some abuse of language to call $p_{ij}$ polynomials over one variable, they may rather be polynomials over one variable applied to monomials $X^iY^j$.
The answer is that there is a counter-example is
$frac{X(X-1)Y(Y-1)}{4}$, as a user of Mathematics Stack Overflow
found. The sums doesn't form a ring, just a group.
number-theory elementary-number-theory polynomials ring-theory
asked Jan 7 at 13:44
LehsLehs
7,07631664
$endgroup$
For simplicity this is about polynomials in just two variables.
Any $finmathbb Q[X,Y]$ can be written as a linear combination of monomials
$X^iY^j$ and therefore as a sum of polynomials $p_{ij}inmathbb Q[X^iY^j]$ over one variable:
$$displaystyle f(X,Y)=p_{00}+sum_{gcd(i,j)=1}p_{ij}(X^iY^j).$$
My question: is the subring of all integer valued polynomials $f$ over
two variables identical with the subring of all sums of integer valued
polynomials $p_{ij}$ over one variable as above?
An integer valued polynomial in one variable is a polynomial $p$ with rational coefficients such that $p(mathbb Z)subseteq mathbb Z$. And corresponding for polynomials over several variables.
It might be some abuse of language to call $p_{ij}$ polynomials over one variable, they may rather be polynomials over one variable applied to monomials $X^iY^j$.
The answer is that there is a counter-example is
$frac{X(X-1)Y(Y-1)}{4}$, as a user of Mathematics Stack Overflow
found. The sums doesn't form a ring, just a group.
number-theory elementary-number-theory polynomials ring-theory
number-theory elementary-number-theory polynomials ring-theory
asked Jan 7 at 13:44
LehsLehs
7,07631664
asked Jan 7 at 13:44
LehsLehs
7,07631664
edited Jan 8 at 9:29
Lehs
asked Jan 7 at 13:44
LehsLehs
7,07631664
asked Jan 7 at 13:44
LehsLehs
7,07631664
asked Jan 7 at 13:44
LehsLehs
7,07631664
7,07631664
1
$begingroup$
How is your displayed formula to be viewed as a sum of polynomials "over one variable"? Also what does the $(i,j))=1$ under the sum mean? That $i,j$ are coprime, or that they are any pairs of positive integers?
$endgroup$
– coffeemath
Jan 7 at 13:52
add a comment |
1
$begingroup$
How is your displayed formula to be viewed as a sum of polynomials "over one variable"? Also what does the $(i,j))=1$ under the sum mean? That $i,j$ are coprime, or that they are any pairs of positive integers?
$endgroup$
– coffeemath
Jan 7 at 13:52
1
1
$begingroup$
How is your displayed formula to be viewed as a sum of polynomials "over one variable"? Also what does the $(i,j))=1$ under the sum mean? That $i,j$ are coprime, or that they are any pairs of positive integers?
$endgroup$
– coffeemath
Jan 7 at 13:52
$begingroup$
How is your displayed formula to be viewed as a sum of polynomials "over one variable"? Also what does the $(i,j))=1$ under the sum mean? That $i,j$ are coprime, or that they are any pairs of positive integers?
$endgroup$
– coffeemath
Jan 7 at 13:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If I have understood your structure correctly, each $p_{ij}$ is a polynomial in one variable, but the variable is different for each combination of $i$ and $j$. So $p_{10}$ is a polynomial in $X$, $p_{01}$ is a polynomial in $Y$, $p_{11}$ is a polynomial in $XY$ etc. When you add the $p_{ij}$ terms together you have to keep these variables distinct. For example, $p_{10}(X) + p_{01}(Y) + P_{11}(XY)$ is a function of $X$, $Y$ and $XY$, not just of one "generic" variable.
Another complication is that each of the $p_{ij}$ can have a constant term, so the decomposition is not unique. Do you regard $X+Y+1$ as $(X) + (Y+1)$, or as $(X+1)+(Y)$ or even as $(X) + (Y) + (1)$ where $(1)$ is a (constant) polynomial in $XY$ etc.
answered Jan 7 at 15:53
gandalf61gandalf61
9,164825
$endgroup$
$begingroup$
The constant term belongs to $p_{00}$ but I see know $i,j=0$ has to be added as a separate term. Thanks!
$endgroup$
– Lehs
Jan 7 at 15:59
add a comment |
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$begingroup$
If I have understood your structure correctly, each $p_{ij}$ is a polynomial in one variable, but the variable is different for each combination of $i$ and $j$. So $p_{10}$ is a polynomial in $X$, $p_{01}$ is a polynomial in $Y$, $p_{11}$ is a polynomial in $XY$ etc. When you add the $p_{ij}$ terms together you have to keep these variables distinct. For example, $p_{10}(X) + p_{01}(Y) + P_{11}(XY)$ is a function of $X$, $Y$ and $XY$, not just of one "generic" variable.
Another complication is that each of the $p_{ij}$ can have a constant term, so the decomposition is not unique. Do you regard $X+Y+1$ as $(X) + (Y+1)$, or as $(X+1)+(Y)$ or even as $(X) + (Y) + (1)$ where $(1)$ is a (constant) polynomial in $XY$ etc.
answered Jan 7 at 15:53
gandalf61gandalf61
9,164825
$endgroup$
$begingroup$
The constant term belongs to $p_{00}$ but I see know $i,j=0$ has to be added as a separate term. Thanks!
$endgroup$
– Lehs
Jan 7 at 15:59
add a comment |
$begingroup$
If I have understood your structure correctly, each $p_{ij}$ is a polynomial in one variable, but the variable is different for each combination of $i$ and $j$. So $p_{10}$ is a polynomial in $X$, $p_{01}$ is a polynomial in $Y$, $p_{11}$ is a polynomial in $XY$ etc. When you add the $p_{ij}$ terms together you have to keep these variables distinct. For example, $p_{10}(X) + p_{01}(Y) + P_{11}(XY)$ is a function of $X$, $Y$ and $XY$, not just of one "generic" variable.
Another complication is that each of the $p_{ij}$ can have a constant term, so the decomposition is not unique. Do you regard $X+Y+1$ as $(X) + (Y+1)$, or as $(X+1)+(Y)$ or even as $(X) + (Y) + (1)$ where $(1)$ is a (constant) polynomial in $XY$ etc.
answered Jan 7 at 15:53
gandalf61gandalf61
9,164825
$endgroup$
$begingroup$
The constant term belongs to $p_{00}$ but I see know $i,j=0$ has to be added as a separate term. Thanks!
$endgroup$
– Lehs
Jan 7 at 15:59
add a comment |
$begingroup$
If I have understood your structure correctly, each $p_{ij}$ is a polynomial in one variable, but the variable is different for each combination of $i$ and $j$. So $p_{10}$ is a polynomial in $X$, $p_{01}$ is a polynomial in $Y$, $p_{11}$ is a polynomial in $XY$ etc. When you add the $p_{ij}$ terms together you have to keep these variables distinct. For example, $p_{10}(X) + p_{01}(Y) + P_{11}(XY)$ is a function of $X$, $Y$ and $XY$, not just of one "generic" variable.
Another complication is that each of the $p_{ij}$ can have a constant term, so the decomposition is not unique. Do you regard $X+Y+1$ as $(X) + (Y+1)$, or as $(X+1)+(Y)$ or even as $(X) + (Y) + (1)$ where $(1)$ is a (constant) polynomial in $XY$ etc.
answered Jan 7 at 15:53
gandalf61gandalf61
9,164825
$endgroup$
If I have understood your structure correctly, each $p_{ij}$ is a polynomial in one variable, but the variable is different for each combination of $i$ and $j$. So $p_{10}$ is a polynomial in $X$, $p_{01}$ is a polynomial in $Y$, $p_{11}$ is a polynomial in $XY$ etc. When you add the $p_{ij}$ terms together you have to keep these variables distinct. For example, $p_{10}(X) + p_{01}(Y) + P_{11}(XY)$ is a function of $X$, $Y$ and $XY$, not just of one "generic" variable.
Another complication is that each of the $p_{ij}$ can have a constant term, so the decomposition is not unique. Do you regard $X+Y+1$ as $(X) + (Y+1)$, or as $(X+1)+(Y)$ or even as $(X) + (Y) + (1)$ where $(1)$ is a (constant) polynomial in $XY$ etc.
answered Jan 7 at 15:53
gandalf61gandalf61
9,164825
answered Jan 7 at 15:53
gandalf61gandalf61
9,164825
answered Jan 7 at 15:53
gandalf61gandalf61
9,164825
answered Jan 7 at 15:53
gandalf61gandalf61
9,164825
9,164825
$begingroup$
The constant term belongs to $p_{00}$ but I see know $i,j=0$ has to be added as a separate term. Thanks!
$endgroup$
– Lehs
Jan 7 at 15:59
add a comment |
$begingroup$
The constant term belongs to $p_{00}$ but I see know $i,j=0$ has to be added as a separate term. Thanks!
$endgroup$
– Lehs
Jan 7 at 15:59
$begingroup$
The constant term belongs to $p_{00}$ but I see know $i,j=0$ has to be added as a separate term. Thanks!
$endgroup$
– Lehs
Jan 7 at 15:59
$begingroup$
The constant term belongs to $p_{00}$ but I see know $i,j=0$ has to be added as a separate term. Thanks!
$endgroup$
– Lehs
Jan 7 at 15:59
add a comment |
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$begingroup$
How is your displayed formula to be viewed as a sum of polynomials "over one variable"? Also what does the $(i,j))=1$ under the sum mean? That $i,j$ are coprime, or that they are any pairs of positive integers?
$endgroup$
– coffeemath
Jan 7 at 13:52