Mixing
How to define a ring $R$ and two non-zero polynomials $a(x), b(x) in R[x]$
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such that the degree of their product is NOT equal to the sum of their degrees?
Could someone please help with this question?
Thank you.
polynomials ring-theory
asked Jul 19 at 7:45
pramort
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up vote
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such that the degree of their product is NOT equal to the sum of their degrees?
Could someone please help with this question?
Thank you.
polynomials ring-theory
asked Jul 19 at 7:45
pramort
303
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
such that the degree of their product is NOT equal to the sum of their degrees?
Could someone please help with this question?
Thank you.
polynomials ring-theory
asked Jul 19 at 7:45
pramort
303
such that the degree of their product is NOT equal to the sum of their degrees?
Could someone please help with this question?
Thank you.
polynomials ring-theory
asked Jul 19 at 7:45
pramort
303
asked Jul 19 at 7:45
pramort
303
asked Jul 19 at 7:45
pramort
303
asked Jul 19 at 7:45
pramort
303
303
add a comment |Â
add a comment |Â
2 Answers
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In $mathbbZ/4mathbbZ[X]$, consider the polynomials $a=b=2X$. Each of them is of degree $1$ but their product is $4X^2=0X^2=0$.
answered Jul 19 at 7:47
Suzet
2,226527
Thank you Suzet for your help! I understand now the idea, but do I understand correctly that the ring is $Z_4left[xright]$? I'm not sure about the notation..
– pramort
Jul 19 at 7:59
I guess it is $mathbbZ_4$, actually I am not used to this latter notation. In France at least, it is very common to denote explicitly $mathbbZ/4mathbbZ$, which really is the quotient of $mathbbZ$ by the subgroup $mathbb4Z$. It is a cyclic abelian group with four elements $0,1,2,3$.
– Suzet
Jul 19 at 8:02
add a comment |Â
up vote
1
down vote
For two polynomials, the product is given by: $$(a_nx^n + ,dots,+a_0)(b_mx^n + , dots,+ b_0) = a_nb_mx^m+n + Big( textlower degree terms Big) $$
So the only way for the product of these polynomials to have a degree different than their sum is for $a_nb_m$ to be $0$.
One way to do this is to choose a ring that has zero-divisors. Examples have been given in other answers. Another way to do this is to simply let one of the polynomials be the $0$-polynomial
answered Jul 19 at 7:59
Praneet Srivastava
690515
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
In $mathbbZ/4mathbbZ[X]$, consider the polynomials $a=b=2X$. Each of them is of degree $1$ but their product is $4X^2=0X^2=0$.
answered Jul 19 at 7:47
Suzet
2,226527
Thank you Suzet for your help! I understand now the idea, but do I understand correctly that the ring is $Z_4left[xright]$? I'm not sure about the notation..
– pramort
Jul 19 at 7:59
I guess it is $mathbbZ_4$, actually I am not used to this latter notation. In France at least, it is very common to denote explicitly $mathbbZ/4mathbbZ$, which really is the quotient of $mathbbZ$ by the subgroup $mathbb4Z$. It is a cyclic abelian group with four elements $0,1,2,3$.
– Suzet
Jul 19 at 8:02
add a comment |Â
up vote
0
down vote
accepted
In $mathbbZ/4mathbbZ[X]$, consider the polynomials $a=b=2X$. Each of them is of degree $1$ but their product is $4X^2=0X^2=0$.
answered Jul 19 at 7:47
Suzet
2,226527
Thank you Suzet for your help! I understand now the idea, but do I understand correctly that the ring is $Z_4left[xright]$? I'm not sure about the notation..
– pramort
Jul 19 at 7:59
I guess it is $mathbbZ_4$, actually I am not used to this latter notation. In France at least, it is very common to denote explicitly $mathbbZ/4mathbbZ$, which really is the quotient of $mathbbZ$ by the subgroup $mathbb4Z$. It is a cyclic abelian group with four elements $0,1,2,3$.
– Suzet
Jul 19 at 8:02
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
In $mathbbZ/4mathbbZ[X]$, consider the polynomials $a=b=2X$. Each of them is of degree $1$ but their product is $4X^2=0X^2=0$.
answered Jul 19 at 7:47
Suzet
2,226527
In $mathbbZ/4mathbbZ[X]$, consider the polynomials $a=b=2X$. Each of them is of degree $1$ but their product is $4X^2=0X^2=0$.
answered Jul 19 at 7:47
Suzet
2,226527
answered Jul 19 at 7:47
Suzet
2,226527
answered Jul 19 at 7:47
Suzet
2,226527
answered Jul 19 at 7:47
Suzet
2,226527
2,226527
Thank you Suzet for your help! I understand now the idea, but do I understand correctly that the ring is $Z_4left[xright]$? I'm not sure about the notation..
– pramort
Jul 19 at 7:59
I guess it is $mathbbZ_4$, actually I am not used to this latter notation. In France at least, it is very common to denote explicitly $mathbbZ/4mathbbZ$, which really is the quotient of $mathbbZ$ by the subgroup $mathbb4Z$. It is a cyclic abelian group with four elements $0,1,2,3$.
– Suzet
Jul 19 at 8:02
add a comment |Â
Thank you Suzet for your help! I understand now the idea, but do I understand correctly that the ring is $Z_4left[xright]$? I'm not sure about the notation..
– pramort
Jul 19 at 7:59
I guess it is $mathbbZ_4$, actually I am not used to this latter notation. In France at least, it is very common to denote explicitly $mathbbZ/4mathbbZ$, which really is the quotient of $mathbbZ$ by the subgroup $mathbb4Z$. It is a cyclic abelian group with four elements $0,1,2,3$.
– Suzet
Jul 19 at 8:02
Thank you Suzet for your help! I understand now the idea, but do I understand correctly that the ring is $Z_4left[xright]$? I'm not sure about the notation..
– pramort
Jul 19 at 7:59
Thank you Suzet for your help! I understand now the idea, but do I understand correctly that the ring is $Z_4left[xright]$? I'm not sure about the notation..
– pramort
Jul 19 at 7:59
I guess it is $mathbbZ_4$, actually I am not used to this latter notation. In France at least, it is very common to denote explicitly $mathbbZ/4mathbbZ$, which really is the quotient of $mathbbZ$ by the subgroup $mathbb4Z$. It is a cyclic abelian group with four elements $0,1,2,3$.
– Suzet
Jul 19 at 8:02
I guess it is $mathbbZ_4$, actually I am not used to this latter notation. In France at least, it is very common to denote explicitly $mathbbZ/4mathbbZ$, which really is the quotient of $mathbbZ$ by the subgroup $mathbb4Z$. It is a cyclic abelian group with four elements $0,1,2,3$.
– Suzet
Jul 19 at 8:02
add a comment |Â
up vote
1
down vote
For two polynomials, the product is given by: $$(a_nx^n + ,dots,+a_0)(b_mx^n + , dots,+ b_0) = a_nb_mx^m+n + Big( textlower degree terms Big) $$
So the only way for the product of these polynomials to have a degree different than their sum is for $a_nb_m$ to be $0$.
One way to do this is to choose a ring that has zero-divisors. Examples have been given in other answers. Another way to do this is to simply let one of the polynomials be the $0$-polynomial
answered Jul 19 at 7:59
Praneet Srivastava
690515
add a comment |Â
up vote
1
down vote
For two polynomials, the product is given by: $$(a_nx^n + ,dots,+a_0)(b_mx^n + , dots,+ b_0) = a_nb_mx^m+n + Big( textlower degree terms Big) $$
So the only way for the product of these polynomials to have a degree different than their sum is for $a_nb_m$ to be $0$.
One way to do this is to choose a ring that has zero-divisors. Examples have been given in other answers. Another way to do this is to simply let one of the polynomials be the $0$-polynomial
answered Jul 19 at 7:59
Praneet Srivastava
690515
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For two polynomials, the product is given by: $$(a_nx^n + ,dots,+a_0)(b_mx^n + , dots,+ b_0) = a_nb_mx^m+n + Big( textlower degree terms Big) $$
So the only way for the product of these polynomials to have a degree different than their sum is for $a_nb_m$ to be $0$.
One way to do this is to choose a ring that has zero-divisors. Examples have been given in other answers. Another way to do this is to simply let one of the polynomials be the $0$-polynomial
answered Jul 19 at 7:59
Praneet Srivastava
690515
For two polynomials, the product is given by: $$(a_nx^n + ,dots,+a_0)(b_mx^n + , dots,+ b_0) = a_nb_mx^m+n + Big( textlower degree terms Big) $$
So the only way for the product of these polynomials to have a degree different than their sum is for $a_nb_m$ to be $0$.
One way to do this is to choose a ring that has zero-divisors. Examples have been given in other answers. Another way to do this is to simply let one of the polynomials be the $0$-polynomial
answered Jul 19 at 7:59
Praneet Srivastava
690515
answered Jul 19 at 7:59
Praneet Srivastava
690515
answered Jul 19 at 7:59
Praneet Srivastava
690515
answered Jul 19 at 7:59
Praneet Srivastava
690515
690515
add a comment |Â
add a comment |Â
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