Galois field inverse












1












$begingroup$


I'm trying to calculate the inverse of $x^{7}+x^{5}+x^{4}+x^{2}+x+1$ over $mathbb{Z}_{2}[x]/(x^{8}+1)$. I suspect there is something fundemental I'm misunderstanding about this process. I'm doing this using code I'm writing, verified by manual calculation, but it seems the problem is not with my code, but with my understanding of what needs to be done.



First of all, I tried applying the extended Euclidean algorithm, but I get that the GCD is $x+1$ (i.e. - that they are not coprime).



Also, as far as I can tell, $(x^{7}+x^{5}+x^{4}+x^{2}+x+1)*(x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x^{1}+1)=0$, suggesting it divides zero.



What am I misunderstanding?






polynomials finite-fields







share|cite|improve this question











$endgroup$












  • $begingroup$
    Extended Euclidean algorithm gives the answer.
    $endgroup$
    – Wuestenfux
    Dec 16 '18 at 12:50






  • 1




    $begingroup$
    To really get into Galois field inverses you would need to replace $x^8+1$ with an irreducible degree eight polynomial. See for example this thread for a well worked out example about finding the inverse of an element in a degree eight extension field.
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 13:46










  • $begingroup$
    +1 for having a correct example. Your other degree seven polynomial is $$x^7+x^6+x^5+x^4+x^3+x^2+x+1=(x+1)^7.$$ Basically because all the binomial coefficients $binom 7k, k=0,1,ldots,7$ are odd. The six term polynomial is divisible by $x+1$, so the product of those two degree seven polynomials is divisible by $(x+1)^8=x^8+1$, and hence equal to zero in the quotient ring.
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 13:54


















1












$begingroup$


I'm trying to calculate the inverse of $x^{7}+x^{5}+x^{4}+x^{2}+x+1$ over $mathbb{Z}_{2}[x]/(x^{8}+1)$. I suspect there is something fundemental I'm misunderstanding about this process. I'm doing this using code I'm writing, verified by manual calculation, but it seems the problem is not with my code, but with my understanding of what needs to be done.



First of all, I tried applying the extended Euclidean algorithm, but I get that the GCD is $x+1$ (i.e. - that they are not coprime).



Also, as far as I can tell, $(x^{7}+x^{5}+x^{4}+x^{2}+x+1)*(x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x^{1}+1)=0$, suggesting it divides zero.



What am I misunderstanding?






polynomials finite-fields







share|cite|improve this question











$endgroup$












  • $begingroup$
    Extended Euclidean algorithm gives the answer.
    $endgroup$
    – Wuestenfux
    Dec 16 '18 at 12:50






  • 1




    $begingroup$
    To really get into Galois field inverses you would need to replace $x^8+1$ with an irreducible degree eight polynomial. See for example this thread for a well worked out example about finding the inverse of an element in a degree eight extension field.
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 13:46










  • $begingroup$
    +1 for having a correct example. Your other degree seven polynomial is $$x^7+x^6+x^5+x^4+x^3+x^2+x+1=(x+1)^7.$$ Basically because all the binomial coefficients $binom 7k, k=0,1,ldots,7$ are odd. The six term polynomial is divisible by $x+1$, so the product of those two degree seven polynomials is divisible by $(x+1)^8=x^8+1$, and hence equal to zero in the quotient ring.
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 13:54
















1












1








1


1



$begingroup$


I'm trying to calculate the inverse of $x^{7}+x^{5}+x^{4}+x^{2}+x+1$ over $mathbb{Z}_{2}[x]/(x^{8}+1)$. I suspect there is something fundemental I'm misunderstanding about this process. I'm doing this using code I'm writing, verified by manual calculation, but it seems the problem is not with my code, but with my understanding of what needs to be done.



First of all, I tried applying the extended Euclidean algorithm, but I get that the GCD is $x+1$ (i.e. - that they are not coprime).



Also, as far as I can tell, $(x^{7}+x^{5}+x^{4}+x^{2}+x+1)*(x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x^{1}+1)=0$, suggesting it divides zero.



What am I misunderstanding?






polynomials finite-fields







share|cite|improve this question











$endgroup$




I'm trying to calculate the inverse of $x^{7}+x^{5}+x^{4}+x^{2}+x+1$ over $mathbb{Z}_{2}[x]/(x^{8}+1)$. I suspect there is something fundemental I'm misunderstanding about this process. I'm doing this using code I'm writing, verified by manual calculation, but it seems the problem is not with my code, but with my understanding of what needs to be done.



First of all, I tried applying the extended Euclidean algorithm, but I get that the GCD is $x+1$ (i.e. - that they are not coprime).



Also, as far as I can tell, $(x^{7}+x^{5}+x^{4}+x^{2}+x+1)*(x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x^{1}+1)=0$, suggesting it divides zero.



What am I misunderstanding?






polynomials finite-fields




polynomials finite-fields






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 13:37









Jyrki Lahtonen

108k12166369




108k12166369










asked Dec 16 '18 at 12:01









Shachar ShemeshShachar Shemesh

1114




1114












  • $begingroup$
    Extended Euclidean algorithm gives the answer.
    $endgroup$
    – Wuestenfux
    Dec 16 '18 at 12:50






  • 1




    $begingroup$
    To really get into Galois field inverses you would need to replace $x^8+1$ with an irreducible degree eight polynomial. See for example this thread for a well worked out example about finding the inverse of an element in a degree eight extension field.
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 13:46










  • $begingroup$
    +1 for having a correct example. Your other degree seven polynomial is $$x^7+x^6+x^5+x^4+x^3+x^2+x+1=(x+1)^7.$$ Basically because all the binomial coefficients $binom 7k, k=0,1,ldots,7$ are odd. The six term polynomial is divisible by $x+1$, so the product of those two degree seven polynomials is divisible by $(x+1)^8=x^8+1$, and hence equal to zero in the quotient ring.
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 13:54




















  • $begingroup$
    Extended Euclidean algorithm gives the answer.
    $endgroup$
    – Wuestenfux
    Dec 16 '18 at 12:50






  • 1




    $begingroup$
    To really get into Galois field inverses you would need to replace $x^8+1$ with an irreducible degree eight polynomial. See for example this thread for a well worked out example about finding the inverse of an element in a degree eight extension field.
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 13:46










  • $begingroup$
    +1 for having a correct example. Your other degree seven polynomial is $$x^7+x^6+x^5+x^4+x^3+x^2+x+1=(x+1)^7.$$ Basically because all the binomial coefficients $binom 7k, k=0,1,ldots,7$ are odd. The six term polynomial is divisible by $x+1$, so the product of those two degree seven polynomials is divisible by $(x+1)^8=x^8+1$, and hence equal to zero in the quotient ring.
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 13:54


















$begingroup$
Extended Euclidean algorithm gives the answer.
$endgroup$
– Wuestenfux
Dec 16 '18 at 12:50




$begingroup$
Extended Euclidean algorithm gives the answer.
$endgroup$
– Wuestenfux
Dec 16 '18 at 12:50




1




1




$begingroup$
To really get into Galois field inverses you would need to replace $x^8+1$ with an irreducible degree eight polynomial. See for example this thread for a well worked out example about finding the inverse of an element in a degree eight extension field.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 13:46




$begingroup$
To really get into Galois field inverses you would need to replace $x^8+1$ with an irreducible degree eight polynomial. See for example this thread for a well worked out example about finding the inverse of an element in a degree eight extension field.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 13:46












$begingroup$
+1 for having a correct example. Your other degree seven polynomial is $$x^7+x^6+x^5+x^4+x^3+x^2+x+1=(x+1)^7.$$ Basically because all the binomial coefficients $binom 7k, k=0,1,ldots,7$ are odd. The six term polynomial is divisible by $x+1$, so the product of those two degree seven polynomials is divisible by $(x+1)^8=x^8+1$, and hence equal to zero in the quotient ring.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 13:54






$begingroup$
+1 for having a correct example. Your other degree seven polynomial is $$x^7+x^6+x^5+x^4+x^3+x^2+x+1=(x+1)^7.$$ Basically because all the binomial coefficients $binom 7k, k=0,1,ldots,7$ are odd. The six term polynomial is divisible by $x+1$, so the product of those two degree seven polynomials is divisible by $(x+1)^8=x^8+1$, and hence equal to zero in the quotient ring.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 13:54












1 Answer
1






active

oldest

votes


















2












$begingroup$

I'm afraid the element you described has no inverse in the quotient ring $Bbb{Z}_2[x]/langle x^8+1rangle$.



This is because your degree seven polynomial $f(x)=x^7+x^5+x^4+x^2+x+1$ shares a factor $x+1$ with $g(x)=x^8+1$. The easiest way of seeing this is to observe that $f(1)=0=g(1)$ so they are both divisible by $x-1=x+1$.



You can tell it at a glance because both polynomials have an even number of terms.




The quotient ring $Bbb{Z}_2[x]/langle g(x)rangle$ is a field if and only if $g(x)$ is irreducible. When $g(x)$ is irreducible it obviously cannot have common factors with lower degree polynomials, and extended Euclid works like charm and finds the inverse.



A popular choice for a degree $8$ irreducible polynomial is $g(x)=x^8+x^4+x^3+x+1$ specified in Rijndael/AES crypto standard<.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    If $h(x)$ is any polynomial we still have $h(1)f(1)=0$ and this makes it impossible for $h(x)f(x)$ to be congruent to $1$ modulo $g(x)$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 13:36






  • 1




    $begingroup$
    so this is what I misunderstood. Not all polynoms create fields under GF.
    $endgroup$
    – Shachar Shemesh
    Dec 16 '18 at 13:42













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1 Answer
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1 Answer
1






active

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2












$begingroup$

I'm afraid the element you described has no inverse in the quotient ring $Bbb{Z}_2[x]/langle x^8+1rangle$.



This is because your degree seven polynomial $f(x)=x^7+x^5+x^4+x^2+x+1$ shares a factor $x+1$ with $g(x)=x^8+1$. The easiest way of seeing this is to observe that $f(1)=0=g(1)$ so they are both divisible by $x-1=x+1$.



You can tell it at a glance because both polynomials have an even number of terms.




The quotient ring $Bbb{Z}_2[x]/langle g(x)rangle$ is a field if and only if $g(x)$ is irreducible. When $g(x)$ is irreducible it obviously cannot have common factors with lower degree polynomials, and extended Euclid works like charm and finds the inverse.



A popular choice for a degree $8$ irreducible polynomial is $g(x)=x^8+x^4+x^3+x+1$ specified in Rijndael/AES crypto standard<.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    If $h(x)$ is any polynomial we still have $h(1)f(1)=0$ and this makes it impossible for $h(x)f(x)$ to be congruent to $1$ modulo $g(x)$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 13:36






  • 1




    $begingroup$
    so this is what I misunderstood. Not all polynoms create fields under GF.
    $endgroup$
    – Shachar Shemesh
    Dec 16 '18 at 13:42


















2












$begingroup$

I'm afraid the element you described has no inverse in the quotient ring $Bbb{Z}_2[x]/langle x^8+1rangle$.



This is because your degree seven polynomial $f(x)=x^7+x^5+x^4+x^2+x+1$ shares a factor $x+1$ with $g(x)=x^8+1$. The easiest way of seeing this is to observe that $f(1)=0=g(1)$ so they are both divisible by $x-1=x+1$.



You can tell it at a glance because both polynomials have an even number of terms.




The quotient ring $Bbb{Z}_2[x]/langle g(x)rangle$ is a field if and only if $g(x)$ is irreducible. When $g(x)$ is irreducible it obviously cannot have common factors with lower degree polynomials, and extended Euclid works like charm and finds the inverse.



A popular choice for a degree $8$ irreducible polynomial is $g(x)=x^8+x^4+x^3+x+1$ specified in Rijndael/AES crypto standard<.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    If $h(x)$ is any polynomial we still have $h(1)f(1)=0$ and this makes it impossible for $h(x)f(x)$ to be congruent to $1$ modulo $g(x)$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 13:36






  • 1




    $begingroup$
    so this is what I misunderstood. Not all polynoms create fields under GF.
    $endgroup$
    – Shachar Shemesh
    Dec 16 '18 at 13:42
















2












2








2





$begingroup$

I'm afraid the element you described has no inverse in the quotient ring $Bbb{Z}_2[x]/langle x^8+1rangle$.



This is because your degree seven polynomial $f(x)=x^7+x^5+x^4+x^2+x+1$ shares a factor $x+1$ with $g(x)=x^8+1$. The easiest way of seeing this is to observe that $f(1)=0=g(1)$ so they are both divisible by $x-1=x+1$.



You can tell it at a glance because both polynomials have an even number of terms.




The quotient ring $Bbb{Z}_2[x]/langle g(x)rangle$ is a field if and only if $g(x)$ is irreducible. When $g(x)$ is irreducible it obviously cannot have common factors with lower degree polynomials, and extended Euclid works like charm and finds the inverse.



A popular choice for a degree $8$ irreducible polynomial is $g(x)=x^8+x^4+x^3+x+1$ specified in Rijndael/AES crypto standard<.






share|cite|improve this answer











$endgroup$



I'm afraid the element you described has no inverse in the quotient ring $Bbb{Z}_2[x]/langle x^8+1rangle$.



This is because your degree seven polynomial $f(x)=x^7+x^5+x^4+x^2+x+1$ shares a factor $x+1$ with $g(x)=x^8+1$. The easiest way of seeing this is to observe that $f(1)=0=g(1)$ so they are both divisible by $x-1=x+1$.



You can tell it at a glance because both polynomials have an even number of terms.




The quotient ring $Bbb{Z}_2[x]/langle g(x)rangle$ is a field if and only if $g(x)$ is irreducible. When $g(x)$ is irreducible it obviously cannot have common factors with lower degree polynomials, and extended Euclid works like charm and finds the inverse.



A popular choice for a degree $8$ irreducible polynomial is $g(x)=x^8+x^4+x^3+x+1$ specified in Rijndael/AES crypto standard<.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 23 '18 at 4:16

























answered Dec 16 '18 at 13:35









Jyrki LahtonenJyrki Lahtonen

108k12166369




108k12166369








  • 1




    $begingroup$
    If $h(x)$ is any polynomial we still have $h(1)f(1)=0$ and this makes it impossible for $h(x)f(x)$ to be congruent to $1$ modulo $g(x)$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 13:36






  • 1




    $begingroup$
    so this is what I misunderstood. Not all polynoms create fields under GF.
    $endgroup$
    – Shachar Shemesh
    Dec 16 '18 at 13:42
















  • 1




    $begingroup$
    If $h(x)$ is any polynomial we still have $h(1)f(1)=0$ and this makes it impossible for $h(x)f(x)$ to be congruent to $1$ modulo $g(x)$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 16 '18 at 13:36






  • 1




    $begingroup$
    so this is what I misunderstood. Not all polynoms create fields under GF.
    $endgroup$
    – Shachar Shemesh
    Dec 16 '18 at 13:42










1




1




$begingroup$
If $h(x)$ is any polynomial we still have $h(1)f(1)=0$ and this makes it impossible for $h(x)f(x)$ to be congruent to $1$ modulo $g(x)$.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 13:36




$begingroup$
If $h(x)$ is any polynomial we still have $h(1)f(1)=0$ and this makes it impossible for $h(x)f(x)$ to be congruent to $1$ modulo $g(x)$.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 13:36




1




1




$begingroup$
so this is what I misunderstood. Not all polynoms create fields under GF.
$endgroup$
– Shachar Shemesh
Dec 16 '18 at 13:42






$begingroup$
so this is what I misunderstood. Not all polynoms create fields under GF.
$endgroup$
– Shachar Shemesh
Dec 16 '18 at 13:42




















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