Present $frac{1}{x^p-x}$ as a sum of simple fractions in $mathbb{Z}_p$
$begingroup$
I've done some (maybe incorrect) algebra, which has led me to a strange equality:
$$frac{1}{x^p-x}$$
$$frac{1}{(1+(x+p-1))^p+(p-1)x}$$
$$frac{1}{1+p(x+p-1)+cdots+p(x+p-1)^{p-1}+(x+p-1)^p+(p-1)x}$$
$p(x+p-1)+cdots+p(x+p-1)$ are $0$ in $mathbb{Z}_p$.
$$frac{1}{(x-1)^p +1 -x}$$
$$frac{1}{(x-1)^p -(x-1)}$$
Which gave me an idea to use induction on $x$ (I don't know if this idea is legal), but I'm struggling to find the base.
I also see that $frac{1}{x^p-x}=frac{1}{x(x^{p-1}-1)}$, but that doesn't seem to lead anywhere.
The answer in the book is $-sum_{a=0}^{p-1}frac{1}{x-a}$.
Hints or full answers, I will be very grateful for any help.
Thank you.
polynomials modular-arithmetic
$endgroup$
add a comment |
$begingroup$
I've done some (maybe incorrect) algebra, which has led me to a strange equality:
$$frac{1}{x^p-x}$$
$$frac{1}{(1+(x+p-1))^p+(p-1)x}$$
$$frac{1}{1+p(x+p-1)+cdots+p(x+p-1)^{p-1}+(x+p-1)^p+(p-1)x}$$
$p(x+p-1)+cdots+p(x+p-1)$ are $0$ in $mathbb{Z}_p$.
$$frac{1}{(x-1)^p +1 -x}$$
$$frac{1}{(x-1)^p -(x-1)}$$
Which gave me an idea to use induction on $x$ (I don't know if this idea is legal), but I'm struggling to find the base.
I also see that $frac{1}{x^p-x}=frac{1}{x(x^{p-1}-1)}$, but that doesn't seem to lead anywhere.
The answer in the book is $-sum_{a=0}^{p-1}frac{1}{x-a}$.
Hints or full answers, I will be very grateful for any help.
Thank you.
polynomials modular-arithmetic
$endgroup$
1
$begingroup$
What is the definition of a simple fraction?
$endgroup$
– Frpzzd
Dec 4 '18 at 14:50
$begingroup$
@Frpzzd $frac{r(x)}{q(x)}$ such that $q=(f(x))^k$, $f(x)$ is an irreducible polynomial (i.e.$f neq a cdot b$, where $a, b$ are polynomials), and degree of $r(x)<$ degree of $f(x)$.
$endgroup$
– fragileradius
Dec 4 '18 at 14:53
add a comment |
$begingroup$
I've done some (maybe incorrect) algebra, which has led me to a strange equality:
$$frac{1}{x^p-x}$$
$$frac{1}{(1+(x+p-1))^p+(p-1)x}$$
$$frac{1}{1+p(x+p-1)+cdots+p(x+p-1)^{p-1}+(x+p-1)^p+(p-1)x}$$
$p(x+p-1)+cdots+p(x+p-1)$ are $0$ in $mathbb{Z}_p$.
$$frac{1}{(x-1)^p +1 -x}$$
$$frac{1}{(x-1)^p -(x-1)}$$
Which gave me an idea to use induction on $x$ (I don't know if this idea is legal), but I'm struggling to find the base.
I also see that $frac{1}{x^p-x}=frac{1}{x(x^{p-1}-1)}$, but that doesn't seem to lead anywhere.
The answer in the book is $-sum_{a=0}^{p-1}frac{1}{x-a}$.
Hints or full answers, I will be very grateful for any help.
Thank you.
polynomials modular-arithmetic
$endgroup$
I've done some (maybe incorrect) algebra, which has led me to a strange equality:
$$frac{1}{x^p-x}$$
$$frac{1}{(1+(x+p-1))^p+(p-1)x}$$
$$frac{1}{1+p(x+p-1)+cdots+p(x+p-1)^{p-1}+(x+p-1)^p+(p-1)x}$$
$p(x+p-1)+cdots+p(x+p-1)$ are $0$ in $mathbb{Z}_p$.
$$frac{1}{(x-1)^p +1 -x}$$
$$frac{1}{(x-1)^p -(x-1)}$$
Which gave me an idea to use induction on $x$ (I don't know if this idea is legal), but I'm struggling to find the base.
I also see that $frac{1}{x^p-x}=frac{1}{x(x^{p-1}-1)}$, but that doesn't seem to lead anywhere.
The answer in the book is $-sum_{a=0}^{p-1}frac{1}{x-a}$.
Hints or full answers, I will be very grateful for any help.
Thank you.
polynomials modular-arithmetic
polynomials modular-arithmetic
asked Dec 4 '18 at 14:44
fragileradiusfragileradius
297114
297114
1
$begingroup$
What is the definition of a simple fraction?
$endgroup$
– Frpzzd
Dec 4 '18 at 14:50
$begingroup$
@Frpzzd $frac{r(x)}{q(x)}$ such that $q=(f(x))^k$, $f(x)$ is an irreducible polynomial (i.e.$f neq a cdot b$, where $a, b$ are polynomials), and degree of $r(x)<$ degree of $f(x)$.
$endgroup$
– fragileradius
Dec 4 '18 at 14:53
add a comment |
1
$begingroup$
What is the definition of a simple fraction?
$endgroup$
– Frpzzd
Dec 4 '18 at 14:50
$begingroup$
@Frpzzd $frac{r(x)}{q(x)}$ such that $q=(f(x))^k$, $f(x)$ is an irreducible polynomial (i.e.$f neq a cdot b$, where $a, b$ are polynomials), and degree of $r(x)<$ degree of $f(x)$.
$endgroup$
– fragileradius
Dec 4 '18 at 14:53
1
1
$begingroup$
What is the definition of a simple fraction?
$endgroup$
– Frpzzd
Dec 4 '18 at 14:50
$begingroup$
What is the definition of a simple fraction?
$endgroup$
– Frpzzd
Dec 4 '18 at 14:50
$begingroup$
@Frpzzd $frac{r(x)}{q(x)}$ such that $q=(f(x))^k$, $f(x)$ is an irreducible polynomial (i.e.$f neq a cdot b$, where $a, b$ are polynomials), and degree of $r(x)<$ degree of $f(x)$.
$endgroup$
– fragileradius
Dec 4 '18 at 14:53
$begingroup$
@Frpzzd $frac{r(x)}{q(x)}$ such that $q=(f(x))^k$, $f(x)$ is an irreducible polynomial (i.e.$f neq a cdot b$, where $a, b$ are polynomials), and degree of $r(x)<$ degree of $f(x)$.
$endgroup$
– fragileradius
Dec 4 '18 at 14:53
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You cannot use induction on $x$, since in $mathbb Z_p[x]$, $x$ does not actually represent a number, let along an integer; it is merely a placeholder, or an algebraic object. You can only induct over things like integers or natural numbers, so this is out of the question.
Here is a more beneficial approach. Because $x^p-x$ equals $0$ for all $xinmathbb Z_p$, you may factor it as $x(x-1)...(x-p+1),$ which suggests (if you are familiar with partial fractions) that the decomposition is of the form
$$frac{a_0}{x}+frac{a_1}{x-1}+...+frac{a_{p-1}}{x-p+1}$$
But by symmetry (or, by substituting $xto x+1$), you can see that all of these coefficients must be equal. So instead, you have a decomposition in the form
$$a_0bigg(frac{1}{x}+frac{1}{x-1}+...+frac{1}{x-p+1}bigg)$$
In which case all you have to do is show that this constant $a_0$ is equal to $-1$. Does this help?
$endgroup$
add a comment |
$begingroup$
I shall find the partial fraction decomposition of $$f(x):=frac{1}{x^q-x}$$
over $mathbb{F}_q$, where $q=p^r$ for some prime natural number $p$ and for some positive integer $r$. In particular, when $r=1$, $mathbb{F}_q=mathbb{F}_p=mathbb{Z}_p$, which is what the OP asks for.
Note that $x^q-xinmathbb{F}_q[x]$ factors into linear factors $prodlimits_{tinmathbb{F}_q},(x-t)$, with each linear factor occurring with multiplicity $1$. Therefore, the partial fraction decomposition of $f(x)$ is a sum of simple fractions
$$f(x)=sum_{tinmathbb{F}_q},frac{s_t}{x-t}$$
for some $s_tinmathbb{F}_q$ for each $tinmathbb{F}_q$.
Now, $g_tau(x):=dfrac{1}{(x-tau),f(x)}$ is a polynomial in $mathbb{F}_q[x]$ for each $tauinmathbb{F}_q$. It can be easily seen that $g_tau(tau)$ is the (first) derivative of $x^q-x$ evaluated at $x:=tau$, which is $-1$. From
$$frac{1}{g_tau(x)}=s_tau+sum_{tinmathbb{F}_qsetminus{tau}},frac{s_t(x-tau)}{x-t}text{ for all }tauinmathbb{F}_q,,$$
we evaluate this expression at $x:=tau$ to get
$$s_tau=-1text{ for every }tauinmathbb{F}_q,.$$
This shows that
$$frac{1}{x^q-x}=f(x)=-sum_{tinmathbb{F}_q},frac{1}{x-t},.$$
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You cannot use induction on $x$, since in $mathbb Z_p[x]$, $x$ does not actually represent a number, let along an integer; it is merely a placeholder, or an algebraic object. You can only induct over things like integers or natural numbers, so this is out of the question.
Here is a more beneficial approach. Because $x^p-x$ equals $0$ for all $xinmathbb Z_p$, you may factor it as $x(x-1)...(x-p+1),$ which suggests (if you are familiar with partial fractions) that the decomposition is of the form
$$frac{a_0}{x}+frac{a_1}{x-1}+...+frac{a_{p-1}}{x-p+1}$$
But by symmetry (or, by substituting $xto x+1$), you can see that all of these coefficients must be equal. So instead, you have a decomposition in the form
$$a_0bigg(frac{1}{x}+frac{1}{x-1}+...+frac{1}{x-p+1}bigg)$$
In which case all you have to do is show that this constant $a_0$ is equal to $-1$. Does this help?
$endgroup$
add a comment |
$begingroup$
You cannot use induction on $x$, since in $mathbb Z_p[x]$, $x$ does not actually represent a number, let along an integer; it is merely a placeholder, or an algebraic object. You can only induct over things like integers or natural numbers, so this is out of the question.
Here is a more beneficial approach. Because $x^p-x$ equals $0$ for all $xinmathbb Z_p$, you may factor it as $x(x-1)...(x-p+1),$ which suggests (if you are familiar with partial fractions) that the decomposition is of the form
$$frac{a_0}{x}+frac{a_1}{x-1}+...+frac{a_{p-1}}{x-p+1}$$
But by symmetry (or, by substituting $xto x+1$), you can see that all of these coefficients must be equal. So instead, you have a decomposition in the form
$$a_0bigg(frac{1}{x}+frac{1}{x-1}+...+frac{1}{x-p+1}bigg)$$
In which case all you have to do is show that this constant $a_0$ is equal to $-1$. Does this help?
$endgroup$
add a comment |
$begingroup$
You cannot use induction on $x$, since in $mathbb Z_p[x]$, $x$ does not actually represent a number, let along an integer; it is merely a placeholder, or an algebraic object. You can only induct over things like integers or natural numbers, so this is out of the question.
Here is a more beneficial approach. Because $x^p-x$ equals $0$ for all $xinmathbb Z_p$, you may factor it as $x(x-1)...(x-p+1),$ which suggests (if you are familiar with partial fractions) that the decomposition is of the form
$$frac{a_0}{x}+frac{a_1}{x-1}+...+frac{a_{p-1}}{x-p+1}$$
But by symmetry (or, by substituting $xto x+1$), you can see that all of these coefficients must be equal. So instead, you have a decomposition in the form
$$a_0bigg(frac{1}{x}+frac{1}{x-1}+...+frac{1}{x-p+1}bigg)$$
In which case all you have to do is show that this constant $a_0$ is equal to $-1$. Does this help?
$endgroup$
You cannot use induction on $x$, since in $mathbb Z_p[x]$, $x$ does not actually represent a number, let along an integer; it is merely a placeholder, or an algebraic object. You can only induct over things like integers or natural numbers, so this is out of the question.
Here is a more beneficial approach. Because $x^p-x$ equals $0$ for all $xinmathbb Z_p$, you may factor it as $x(x-1)...(x-p+1),$ which suggests (if you are familiar with partial fractions) that the decomposition is of the form
$$frac{a_0}{x}+frac{a_1}{x-1}+...+frac{a_{p-1}}{x-p+1}$$
But by symmetry (or, by substituting $xto x+1$), you can see that all of these coefficients must be equal. So instead, you have a decomposition in the form
$$a_0bigg(frac{1}{x}+frac{1}{x-1}+...+frac{1}{x-p+1}bigg)$$
In which case all you have to do is show that this constant $a_0$ is equal to $-1$. Does this help?
answered Dec 4 '18 at 14:58
FrpzzdFrpzzd
22.5k840108
22.5k840108
add a comment |
add a comment |
$begingroup$
I shall find the partial fraction decomposition of $$f(x):=frac{1}{x^q-x}$$
over $mathbb{F}_q$, where $q=p^r$ for some prime natural number $p$ and for some positive integer $r$. In particular, when $r=1$, $mathbb{F}_q=mathbb{F}_p=mathbb{Z}_p$, which is what the OP asks for.
Note that $x^q-xinmathbb{F}_q[x]$ factors into linear factors $prodlimits_{tinmathbb{F}_q},(x-t)$, with each linear factor occurring with multiplicity $1$. Therefore, the partial fraction decomposition of $f(x)$ is a sum of simple fractions
$$f(x)=sum_{tinmathbb{F}_q},frac{s_t}{x-t}$$
for some $s_tinmathbb{F}_q$ for each $tinmathbb{F}_q$.
Now, $g_tau(x):=dfrac{1}{(x-tau),f(x)}$ is a polynomial in $mathbb{F}_q[x]$ for each $tauinmathbb{F}_q$. It can be easily seen that $g_tau(tau)$ is the (first) derivative of $x^q-x$ evaluated at $x:=tau$, which is $-1$. From
$$frac{1}{g_tau(x)}=s_tau+sum_{tinmathbb{F}_qsetminus{tau}},frac{s_t(x-tau)}{x-t}text{ for all }tauinmathbb{F}_q,,$$
we evaluate this expression at $x:=tau$ to get
$$s_tau=-1text{ for every }tauinmathbb{F}_q,.$$
This shows that
$$frac{1}{x^q-x}=f(x)=-sum_{tinmathbb{F}_q},frac{1}{x-t},.$$
$endgroup$
add a comment |
$begingroup$
I shall find the partial fraction decomposition of $$f(x):=frac{1}{x^q-x}$$
over $mathbb{F}_q$, where $q=p^r$ for some prime natural number $p$ and for some positive integer $r$. In particular, when $r=1$, $mathbb{F}_q=mathbb{F}_p=mathbb{Z}_p$, which is what the OP asks for.
Note that $x^q-xinmathbb{F}_q[x]$ factors into linear factors $prodlimits_{tinmathbb{F}_q},(x-t)$, with each linear factor occurring with multiplicity $1$. Therefore, the partial fraction decomposition of $f(x)$ is a sum of simple fractions
$$f(x)=sum_{tinmathbb{F}_q},frac{s_t}{x-t}$$
for some $s_tinmathbb{F}_q$ for each $tinmathbb{F}_q$.
Now, $g_tau(x):=dfrac{1}{(x-tau),f(x)}$ is a polynomial in $mathbb{F}_q[x]$ for each $tauinmathbb{F}_q$. It can be easily seen that $g_tau(tau)$ is the (first) derivative of $x^q-x$ evaluated at $x:=tau$, which is $-1$. From
$$frac{1}{g_tau(x)}=s_tau+sum_{tinmathbb{F}_qsetminus{tau}},frac{s_t(x-tau)}{x-t}text{ for all }tauinmathbb{F}_q,,$$
we evaluate this expression at $x:=tau$ to get
$$s_tau=-1text{ for every }tauinmathbb{F}_q,.$$
This shows that
$$frac{1}{x^q-x}=f(x)=-sum_{tinmathbb{F}_q},frac{1}{x-t},.$$
$endgroup$
add a comment |
$begingroup$
I shall find the partial fraction decomposition of $$f(x):=frac{1}{x^q-x}$$
over $mathbb{F}_q$, where $q=p^r$ for some prime natural number $p$ and for some positive integer $r$. In particular, when $r=1$, $mathbb{F}_q=mathbb{F}_p=mathbb{Z}_p$, which is what the OP asks for.
Note that $x^q-xinmathbb{F}_q[x]$ factors into linear factors $prodlimits_{tinmathbb{F}_q},(x-t)$, with each linear factor occurring with multiplicity $1$. Therefore, the partial fraction decomposition of $f(x)$ is a sum of simple fractions
$$f(x)=sum_{tinmathbb{F}_q},frac{s_t}{x-t}$$
for some $s_tinmathbb{F}_q$ for each $tinmathbb{F}_q$.
Now, $g_tau(x):=dfrac{1}{(x-tau),f(x)}$ is a polynomial in $mathbb{F}_q[x]$ for each $tauinmathbb{F}_q$. It can be easily seen that $g_tau(tau)$ is the (first) derivative of $x^q-x$ evaluated at $x:=tau$, which is $-1$. From
$$frac{1}{g_tau(x)}=s_tau+sum_{tinmathbb{F}_qsetminus{tau}},frac{s_t(x-tau)}{x-t}text{ for all }tauinmathbb{F}_q,,$$
we evaluate this expression at $x:=tau$ to get
$$s_tau=-1text{ for every }tauinmathbb{F}_q,.$$
This shows that
$$frac{1}{x^q-x}=f(x)=-sum_{tinmathbb{F}_q},frac{1}{x-t},.$$
$endgroup$
I shall find the partial fraction decomposition of $$f(x):=frac{1}{x^q-x}$$
over $mathbb{F}_q$, where $q=p^r$ for some prime natural number $p$ and for some positive integer $r$. In particular, when $r=1$, $mathbb{F}_q=mathbb{F}_p=mathbb{Z}_p$, which is what the OP asks for.
Note that $x^q-xinmathbb{F}_q[x]$ factors into linear factors $prodlimits_{tinmathbb{F}_q},(x-t)$, with each linear factor occurring with multiplicity $1$. Therefore, the partial fraction decomposition of $f(x)$ is a sum of simple fractions
$$f(x)=sum_{tinmathbb{F}_q},frac{s_t}{x-t}$$
for some $s_tinmathbb{F}_q$ for each $tinmathbb{F}_q$.
Now, $g_tau(x):=dfrac{1}{(x-tau),f(x)}$ is a polynomial in $mathbb{F}_q[x]$ for each $tauinmathbb{F}_q$. It can be easily seen that $g_tau(tau)$ is the (first) derivative of $x^q-x$ evaluated at $x:=tau$, which is $-1$. From
$$frac{1}{g_tau(x)}=s_tau+sum_{tinmathbb{F}_qsetminus{tau}},frac{s_t(x-tau)}{x-t}text{ for all }tauinmathbb{F}_q,,$$
we evaluate this expression at $x:=tau$ to get
$$s_tau=-1text{ for every }tauinmathbb{F}_q,.$$
This shows that
$$frac{1}{x^q-x}=f(x)=-sum_{tinmathbb{F}_q},frac{1}{x-t},.$$
edited Dec 4 '18 at 15:29
answered Dec 4 '18 at 14:59
BatominovskiBatominovski
1
1
add a comment |
add a comment |
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$begingroup$
What is the definition of a simple fraction?
$endgroup$
– Frpzzd
Dec 4 '18 at 14:50
$begingroup$
@Frpzzd $frac{r(x)}{q(x)}$ such that $q=(f(x))^k$, $f(x)$ is an irreducible polynomial (i.e.$f neq a cdot b$, where $a, b$ are polynomials), and degree of $r(x)<$ degree of $f(x)$.
$endgroup$
– fragileradius
Dec 4 '18 at 14:53