Reducible and Irreducible polynomials are confusing me
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The definition claims that a polynomial in a field of positive degree is a reducible polynomial when it can be written as the product of 2 polynomials in the field with positive degrees. Other wise it is irreducible.
So if a polynomial $f(x)$ can be written as the product of say $41(x^2 + x)$, is that considered not reducible because 41 is really $41x^0$, and 0 isn't technically positive, but by the definition of a polynomial in a field it is a polynomial if $a_n$ isn't 0 for the highest degree $n$ where $n geq 0$.
So would the example of the polynomial example I gave be reducible or irreducible?
linear-algebra
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The definition claims that a polynomial in a field of positive degree is a reducible polynomial when it can be written as the product of 2 polynomials in the field with positive degrees. Other wise it is irreducible.
So if a polynomial $f(x)$ can be written as the product of say $41(x^2 + x)$, is that considered not reducible because 41 is really $41x^0$, and 0 isn't technically positive, but by the definition of a polynomial in a field it is a polynomial if $a_n$ isn't 0 for the highest degree $n$ where $n geq 0$.
So would the example of the polynomial example I gave be reducible or irreducible?
linear-algebra
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up vote
3
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favorite
up vote
3
down vote
favorite
The definition claims that a polynomial in a field of positive degree is a reducible polynomial when it can be written as the product of 2 polynomials in the field with positive degrees. Other wise it is irreducible.
So if a polynomial $f(x)$ can be written as the product of say $41(x^2 + x)$, is that considered not reducible because 41 is really $41x^0$, and 0 isn't technically positive, but by the definition of a polynomial in a field it is a polynomial if $a_n$ isn't 0 for the highest degree $n$ where $n geq 0$.
So would the example of the polynomial example I gave be reducible or irreducible?
linear-algebra
The definition claims that a polynomial in a field of positive degree is a reducible polynomial when it can be written as the product of 2 polynomials in the field with positive degrees. Other wise it is irreducible.
So if a polynomial $f(x)$ can be written as the product of say $41(x^2 + x)$, is that considered not reducible because 41 is really $41x^0$, and 0 isn't technically positive, but by the definition of a polynomial in a field it is a polynomial if $a_n$ isn't 0 for the highest degree $n$ where $n geq 0$.
So would the example of the polynomial example I gave be reducible or irreducible?
linear-algebra
linear-algebra
asked 1 hour ago
ming
424
424
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3 Answers
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9
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This particular example would be reducible, but not because of the reason you give -- it would be because you could write it as $41x(x+1)$. Note that $0$ is not positive, so the factorization you give is not a product of two polynomials with positive degree ($41$ is a polynomial of degree $0$).
2
So if it was $41(x + 1)$ it would be irreducible right? Because 41 cannot be written with a degree of anything greater than 0, and 0 is not positive?
– ming
1 hour ago
Correct, $41x + 41$ is irreducible (assuming you are working over $mathbb{Z}[x]$).
– platty
1 hour ago
It is reducible in Z[x] as 41 is not a unit.
– Hans
9 mins ago
I think the definition of irreducible as a polynomial requires that both factors be non-constant; I'm not sure why I added that stipulation. Otherwise you run into uniqueness issues with something like $4x+4$.
– platty
4 mins ago
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up vote
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Definition : Given an integral domain $R$, a non-zero non-unit element $r in R$ is said to be irreducible if it cannot be written as a product of non-units i.e. whenever it is written as a product of two elements, at least one of them is a unit in $R$.
For polynomials, this becomes :
Definition : Given an integral domain $R$, the ring $R[X]$ also is an integral domain, and $f$ is an irreducible polynomial over $R$ if it is an irreducible element of $R[X]$.
So, it is that simple. Let us take some examples to clarify.
The polynomial $f(x) = x$ is irreducible over any ring, since if $a(x)b(x) =x$, then WLOG $a$ must be a constant polynomial with the constant being a unit(use the rules for multiplication of polynomials), so $a$ is a unit in $R[X]$, hence $x$ is irreducible.
The polynomial $f(x) = 2x+2$ is irreducible over $mathbb R[X]$. This is because if $a(x)b(x)$ divides $2(x+1)$ then at least one of $a(x)$ or $b(x)$ is a constant polynomial, but every constant polynomial is a unit in $mathbb R$. However, this polynomial is reducible over $mathbb Z[X]$, since here, $2(x+1)$ counts as a non-unit factorization, because $2$ is not a unit.
Therefore, reducibility depends on "over which ring/field"? For example, $41 = 41x^0$ is a constant polynomial, but it isn't a unit in $mathbb Z[X]$, while it is one in $mathbb R[X]$. So, a polynomial like $41(x+1)$ is irreducible over the latter but not over the former.
While working over a field, it turns out that the set of units of $F[X]$ is equal to the non-zero constant polynomials. Therefore, any polynomial is irreducible in $F[X]$ if and only if it can be written as the product of two non-constant polynomials. Things will change for a ring which is not a field, since some non-constant polynomials may possibly not be units.
Also, note that $41(x^2+x)$ is reducible in every field, since it can be written as $(41 x) times (x+1)$ which is the product of two non-constant polynomials, which are always non-units by the fact that the degree is multiplicative.
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A polynomial $f(x)$ is irreducible if $f(x)=g(x)h(x)$ with $deg(g(x)) ge 1$, and $deg(h(x)) ge 1$
Do you mean reducible? I don't see how your answer helps as OP has clearly mentioned the definition.
– Thomas Shelby
41 mins ago
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
This particular example would be reducible, but not because of the reason you give -- it would be because you could write it as $41x(x+1)$. Note that $0$ is not positive, so the factorization you give is not a product of two polynomials with positive degree ($41$ is a polynomial of degree $0$).
2
So if it was $41(x + 1)$ it would be irreducible right? Because 41 cannot be written with a degree of anything greater than 0, and 0 is not positive?
– ming
1 hour ago
Correct, $41x + 41$ is irreducible (assuming you are working over $mathbb{Z}[x]$).
– platty
1 hour ago
It is reducible in Z[x] as 41 is not a unit.
– Hans
9 mins ago
I think the definition of irreducible as a polynomial requires that both factors be non-constant; I'm not sure why I added that stipulation. Otherwise you run into uniqueness issues with something like $4x+4$.
– platty
4 mins ago
add a comment |
up vote
9
down vote
This particular example would be reducible, but not because of the reason you give -- it would be because you could write it as $41x(x+1)$. Note that $0$ is not positive, so the factorization you give is not a product of two polynomials with positive degree ($41$ is a polynomial of degree $0$).
2
So if it was $41(x + 1)$ it would be irreducible right? Because 41 cannot be written with a degree of anything greater than 0, and 0 is not positive?
– ming
1 hour ago
Correct, $41x + 41$ is irreducible (assuming you are working over $mathbb{Z}[x]$).
– platty
1 hour ago
It is reducible in Z[x] as 41 is not a unit.
– Hans
9 mins ago
I think the definition of irreducible as a polynomial requires that both factors be non-constant; I'm not sure why I added that stipulation. Otherwise you run into uniqueness issues with something like $4x+4$.
– platty
4 mins ago
add a comment |
up vote
9
down vote
up vote
9
down vote
This particular example would be reducible, but not because of the reason you give -- it would be because you could write it as $41x(x+1)$. Note that $0$ is not positive, so the factorization you give is not a product of two polynomials with positive degree ($41$ is a polynomial of degree $0$).
This particular example would be reducible, but not because of the reason you give -- it would be because you could write it as $41x(x+1)$. Note that $0$ is not positive, so the factorization you give is not a product of two polynomials with positive degree ($41$ is a polynomial of degree $0$).
answered 1 hour ago
platty
1,877211
1,877211
2
So if it was $41(x + 1)$ it would be irreducible right? Because 41 cannot be written with a degree of anything greater than 0, and 0 is not positive?
– ming
1 hour ago
Correct, $41x + 41$ is irreducible (assuming you are working over $mathbb{Z}[x]$).
– platty
1 hour ago
It is reducible in Z[x] as 41 is not a unit.
– Hans
9 mins ago
I think the definition of irreducible as a polynomial requires that both factors be non-constant; I'm not sure why I added that stipulation. Otherwise you run into uniqueness issues with something like $4x+4$.
– platty
4 mins ago
add a comment |
2
So if it was $41(x + 1)$ it would be irreducible right? Because 41 cannot be written with a degree of anything greater than 0, and 0 is not positive?
– ming
1 hour ago
Correct, $41x + 41$ is irreducible (assuming you are working over $mathbb{Z}[x]$).
– platty
1 hour ago
It is reducible in Z[x] as 41 is not a unit.
– Hans
9 mins ago
I think the definition of irreducible as a polynomial requires that both factors be non-constant; I'm not sure why I added that stipulation. Otherwise you run into uniqueness issues with something like $4x+4$.
– platty
4 mins ago
2
2
So if it was $41(x + 1)$ it would be irreducible right? Because 41 cannot be written with a degree of anything greater than 0, and 0 is not positive?
– ming
1 hour ago
So if it was $41(x + 1)$ it would be irreducible right? Because 41 cannot be written with a degree of anything greater than 0, and 0 is not positive?
– ming
1 hour ago
Correct, $41x + 41$ is irreducible (assuming you are working over $mathbb{Z}[x]$).
– platty
1 hour ago
Correct, $41x + 41$ is irreducible (assuming you are working over $mathbb{Z}[x]$).
– platty
1 hour ago
It is reducible in Z[x] as 41 is not a unit.
– Hans
9 mins ago
It is reducible in Z[x] as 41 is not a unit.
– Hans
9 mins ago
I think the definition of irreducible as a polynomial requires that both factors be non-constant; I'm not sure why I added that stipulation. Otherwise you run into uniqueness issues with something like $4x+4$.
– platty
4 mins ago
I think the definition of irreducible as a polynomial requires that both factors be non-constant; I'm not sure why I added that stipulation. Otherwise you run into uniqueness issues with something like $4x+4$.
– platty
4 mins ago
add a comment |
up vote
3
down vote
Definition : Given an integral domain $R$, a non-zero non-unit element $r in R$ is said to be irreducible if it cannot be written as a product of non-units i.e. whenever it is written as a product of two elements, at least one of them is a unit in $R$.
For polynomials, this becomes :
Definition : Given an integral domain $R$, the ring $R[X]$ also is an integral domain, and $f$ is an irreducible polynomial over $R$ if it is an irreducible element of $R[X]$.
So, it is that simple. Let us take some examples to clarify.
The polynomial $f(x) = x$ is irreducible over any ring, since if $a(x)b(x) =x$, then WLOG $a$ must be a constant polynomial with the constant being a unit(use the rules for multiplication of polynomials), so $a$ is a unit in $R[X]$, hence $x$ is irreducible.
The polynomial $f(x) = 2x+2$ is irreducible over $mathbb R[X]$. This is because if $a(x)b(x)$ divides $2(x+1)$ then at least one of $a(x)$ or $b(x)$ is a constant polynomial, but every constant polynomial is a unit in $mathbb R$. However, this polynomial is reducible over $mathbb Z[X]$, since here, $2(x+1)$ counts as a non-unit factorization, because $2$ is not a unit.
Therefore, reducibility depends on "over which ring/field"? For example, $41 = 41x^0$ is a constant polynomial, but it isn't a unit in $mathbb Z[X]$, while it is one in $mathbb R[X]$. So, a polynomial like $41(x+1)$ is irreducible over the latter but not over the former.
While working over a field, it turns out that the set of units of $F[X]$ is equal to the non-zero constant polynomials. Therefore, any polynomial is irreducible in $F[X]$ if and only if it can be written as the product of two non-constant polynomials. Things will change for a ring which is not a field, since some non-constant polynomials may possibly not be units.
Also, note that $41(x^2+x)$ is reducible in every field, since it can be written as $(41 x) times (x+1)$ which is the product of two non-constant polynomials, which are always non-units by the fact that the degree is multiplicative.
add a comment |
up vote
3
down vote
Definition : Given an integral domain $R$, a non-zero non-unit element $r in R$ is said to be irreducible if it cannot be written as a product of non-units i.e. whenever it is written as a product of two elements, at least one of them is a unit in $R$.
For polynomials, this becomes :
Definition : Given an integral domain $R$, the ring $R[X]$ also is an integral domain, and $f$ is an irreducible polynomial over $R$ if it is an irreducible element of $R[X]$.
So, it is that simple. Let us take some examples to clarify.
The polynomial $f(x) = x$ is irreducible over any ring, since if $a(x)b(x) =x$, then WLOG $a$ must be a constant polynomial with the constant being a unit(use the rules for multiplication of polynomials), so $a$ is a unit in $R[X]$, hence $x$ is irreducible.
The polynomial $f(x) = 2x+2$ is irreducible over $mathbb R[X]$. This is because if $a(x)b(x)$ divides $2(x+1)$ then at least one of $a(x)$ or $b(x)$ is a constant polynomial, but every constant polynomial is a unit in $mathbb R$. However, this polynomial is reducible over $mathbb Z[X]$, since here, $2(x+1)$ counts as a non-unit factorization, because $2$ is not a unit.
Therefore, reducibility depends on "over which ring/field"? For example, $41 = 41x^0$ is a constant polynomial, but it isn't a unit in $mathbb Z[X]$, while it is one in $mathbb R[X]$. So, a polynomial like $41(x+1)$ is irreducible over the latter but not over the former.
While working over a field, it turns out that the set of units of $F[X]$ is equal to the non-zero constant polynomials. Therefore, any polynomial is irreducible in $F[X]$ if and only if it can be written as the product of two non-constant polynomials. Things will change for a ring which is not a field, since some non-constant polynomials may possibly not be units.
Also, note that $41(x^2+x)$ is reducible in every field, since it can be written as $(41 x) times (x+1)$ which is the product of two non-constant polynomials, which are always non-units by the fact that the degree is multiplicative.
add a comment |
up vote
3
down vote
up vote
3
down vote
Definition : Given an integral domain $R$, a non-zero non-unit element $r in R$ is said to be irreducible if it cannot be written as a product of non-units i.e. whenever it is written as a product of two elements, at least one of them is a unit in $R$.
For polynomials, this becomes :
Definition : Given an integral domain $R$, the ring $R[X]$ also is an integral domain, and $f$ is an irreducible polynomial over $R$ if it is an irreducible element of $R[X]$.
So, it is that simple. Let us take some examples to clarify.
The polynomial $f(x) = x$ is irreducible over any ring, since if $a(x)b(x) =x$, then WLOG $a$ must be a constant polynomial with the constant being a unit(use the rules for multiplication of polynomials), so $a$ is a unit in $R[X]$, hence $x$ is irreducible.
The polynomial $f(x) = 2x+2$ is irreducible over $mathbb R[X]$. This is because if $a(x)b(x)$ divides $2(x+1)$ then at least one of $a(x)$ or $b(x)$ is a constant polynomial, but every constant polynomial is a unit in $mathbb R$. However, this polynomial is reducible over $mathbb Z[X]$, since here, $2(x+1)$ counts as a non-unit factorization, because $2$ is not a unit.
Therefore, reducibility depends on "over which ring/field"? For example, $41 = 41x^0$ is a constant polynomial, but it isn't a unit in $mathbb Z[X]$, while it is one in $mathbb R[X]$. So, a polynomial like $41(x+1)$ is irreducible over the latter but not over the former.
While working over a field, it turns out that the set of units of $F[X]$ is equal to the non-zero constant polynomials. Therefore, any polynomial is irreducible in $F[X]$ if and only if it can be written as the product of two non-constant polynomials. Things will change for a ring which is not a field, since some non-constant polynomials may possibly not be units.
Also, note that $41(x^2+x)$ is reducible in every field, since it can be written as $(41 x) times (x+1)$ which is the product of two non-constant polynomials, which are always non-units by the fact that the degree is multiplicative.
Definition : Given an integral domain $R$, a non-zero non-unit element $r in R$ is said to be irreducible if it cannot be written as a product of non-units i.e. whenever it is written as a product of two elements, at least one of them is a unit in $R$.
For polynomials, this becomes :
Definition : Given an integral domain $R$, the ring $R[X]$ also is an integral domain, and $f$ is an irreducible polynomial over $R$ if it is an irreducible element of $R[X]$.
So, it is that simple. Let us take some examples to clarify.
The polynomial $f(x) = x$ is irreducible over any ring, since if $a(x)b(x) =x$, then WLOG $a$ must be a constant polynomial with the constant being a unit(use the rules for multiplication of polynomials), so $a$ is a unit in $R[X]$, hence $x$ is irreducible.
The polynomial $f(x) = 2x+2$ is irreducible over $mathbb R[X]$. This is because if $a(x)b(x)$ divides $2(x+1)$ then at least one of $a(x)$ or $b(x)$ is a constant polynomial, but every constant polynomial is a unit in $mathbb R$. However, this polynomial is reducible over $mathbb Z[X]$, since here, $2(x+1)$ counts as a non-unit factorization, because $2$ is not a unit.
Therefore, reducibility depends on "over which ring/field"? For example, $41 = 41x^0$ is a constant polynomial, but it isn't a unit in $mathbb Z[X]$, while it is one in $mathbb R[X]$. So, a polynomial like $41(x+1)$ is irreducible over the latter but not over the former.
While working over a field, it turns out that the set of units of $F[X]$ is equal to the non-zero constant polynomials. Therefore, any polynomial is irreducible in $F[X]$ if and only if it can be written as the product of two non-constant polynomials. Things will change for a ring which is not a field, since some non-constant polynomials may possibly not be units.
Also, note that $41(x^2+x)$ is reducible in every field, since it can be written as $(41 x) times (x+1)$ which is the product of two non-constant polynomials, which are always non-units by the fact that the degree is multiplicative.
answered 1 hour ago
астон вілла олоф мэллбэрг
36.6k33376
36.6k33376
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up vote
-1
down vote
A polynomial $f(x)$ is irreducible if $f(x)=g(x)h(x)$ with $deg(g(x)) ge 1$, and $deg(h(x)) ge 1$
Do you mean reducible? I don't see how your answer helps as OP has clearly mentioned the definition.
– Thomas Shelby
41 mins ago
add a comment |
up vote
-1
down vote
A polynomial $f(x)$ is irreducible if $f(x)=g(x)h(x)$ with $deg(g(x)) ge 1$, and $deg(h(x)) ge 1$
Do you mean reducible? I don't see how your answer helps as OP has clearly mentioned the definition.
– Thomas Shelby
41 mins ago
add a comment |
up vote
-1
down vote
up vote
-1
down vote
A polynomial $f(x)$ is irreducible if $f(x)=g(x)h(x)$ with $deg(g(x)) ge 1$, and $deg(h(x)) ge 1$
A polynomial $f(x)$ is irreducible if $f(x)=g(x)h(x)$ with $deg(g(x)) ge 1$, and $deg(h(x)) ge 1$
answered 50 mins ago
Fareed AF
35411
35411
Do you mean reducible? I don't see how your answer helps as OP has clearly mentioned the definition.
– Thomas Shelby
41 mins ago
add a comment |
Do you mean reducible? I don't see how your answer helps as OP has clearly mentioned the definition.
– Thomas Shelby
41 mins ago
Do you mean reducible? I don't see how your answer helps as OP has clearly mentioned the definition.
– Thomas Shelby
41 mins ago
Do you mean reducible? I don't see how your answer helps as OP has clearly mentioned the definition.
– Thomas Shelby
41 mins ago
add a comment |
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