Why is $p_{3}(x) = 9x^{2}-3=3(xsqrt{3}-1)(xsqrt{3}+1)$ reducible over integers?
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I am learning Irreducibility of polynomials, and I am reading this entry of Wiki. However, I am confused why the polynomial $p_{3}(x) = 9x^{2}-3=3(xsqrt{3}-1)(xsqrt{3}+1)$ is reducible over integers. I though it falls right into the defintion of irreducibility(over integer), because it could only be written into two polynomials of strictly lower degree with coefficients not being integers. The entry explains that it is reducible because the factor 3 is not invertible in integers, but I just can't see how this makes the polynomial reducible. I am very thankful if someone could help explain it.
abstract-algebra polynomials factoring irreducible-polynomials
edited Mar 10 at 10:29
Sil
5,65721745
asked Mar 10 at 7:47
BM YoonBM Yoon
283
$endgroup$
add a comment |
$begingroup$
I am learning Irreducibility of polynomials, and I am reading this entry of Wiki. However, I am confused why the polynomial $p_{3}(x) = 9x^{2}-3=3(xsqrt{3}-1)(xsqrt{3}+1)$ is reducible over integers. I though it falls right into the defintion of irreducibility(over integer), because it could only be written into two polynomials of strictly lower degree with coefficients not being integers. The entry explains that it is reducible because the factor 3 is not invertible in integers, but I just can't see how this makes the polynomial reducible. I am very thankful if someone could help explain it.
abstract-algebra polynomials factoring irreducible-polynomials
edited Mar 10 at 10:29
Sil
5,65721745
asked Mar 10 at 7:47
BM YoonBM Yoon
283
$endgroup$
$begingroup$
Recall when is a polynomial in $mathbb Z[X]$ reducible in $mathbb Z [X]$
$endgroup$
– Soumik Ghosh
Mar 10 at 7:54
add a comment |
$begingroup$
I am learning Irreducibility of polynomials, and I am reading this entry of Wiki. However, I am confused why the polynomial $p_{3}(x) = 9x^{2}-3=3(xsqrt{3}-1)(xsqrt{3}+1)$ is reducible over integers. I though it falls right into the defintion of irreducibility(over integer), because it could only be written into two polynomials of strictly lower degree with coefficients not being integers. The entry explains that it is reducible because the factor 3 is not invertible in integers, but I just can't see how this makes the polynomial reducible. I am very thankful if someone could help explain it.
abstract-algebra polynomials factoring irreducible-polynomials
edited Mar 10 at 10:29
Sil
5,65721745
asked Mar 10 at 7:47
BM YoonBM Yoon
283
$endgroup$
I am learning Irreducibility of polynomials, and I am reading this entry of Wiki. However, I am confused why the polynomial $p_{3}(x) = 9x^{2}-3=3(xsqrt{3}-1)(xsqrt{3}+1)$ is reducible over integers. I though it falls right into the defintion of irreducibility(over integer), because it could only be written into two polynomials of strictly lower degree with coefficients not being integers. The entry explains that it is reducible because the factor 3 is not invertible in integers, but I just can't see how this makes the polynomial reducible. I am very thankful if someone could help explain it.
abstract-algebra polynomials factoring irreducible-polynomials
abstract-algebra polynomials factoring irreducible-polynomials
edited Mar 10 at 10:29
Sil
5,65721745
asked Mar 10 at 7:47
BM YoonBM Yoon
283
edited Mar 10 at 10:29
Sil
5,65721745
asked Mar 10 at 7:47
BM YoonBM Yoon
283
edited Mar 10 at 10:29
Sil
5,65721745
edited Mar 10 at 10:29
Sil
5,65721745
edited Mar 10 at 10:29
Sil
5,65721745
5,65721745
asked Mar 10 at 7:47
BM YoonBM Yoon
283
asked Mar 10 at 7:47
BM YoonBM Yoon
283
asked Mar 10 at 7:47
BM YoonBM Yoon
283
283
$begingroup$
Recall when is a polynomial in $mathbb Z[X]$ reducible in $mathbb Z [X]$
$endgroup$
– Soumik Ghosh
Mar 10 at 7:54
add a comment |
$begingroup$
Recall when is a polynomial in $mathbb Z[X]$ reducible in $mathbb Z [X]$
$endgroup$
– Soumik Ghosh
Mar 10 at 7:54
$begingroup$
Recall when is a polynomial in $mathbb Z[X]$ reducible in $mathbb Z [X]$
$endgroup$
– Soumik Ghosh
Mar 10 at 7:54
$begingroup$
Recall when is a polynomial in $mathbb Z[X]$ reducible in $mathbb Z [X]$
$endgroup$
– Soumik Ghosh
Mar 10 at 7:54
add a comment |
2 Answers
2
active
oldest
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Well, the polynomial $f(x)=3(x^2-1)$ is reducible over the integers, since it factors into a product of non-units in ${Bbb Z}[x]$. The units in ${Bbb Z}[x]$ coincide with the units in ${Bbb Z}$ and are $pm 1$.
The definition says that a polynomial $fin R[x]$ is irreducible if it can only be factored as $f=gh$ with $g,hin R[x]$ such that one of $g,h$ is a unit in $R[x]$. The units in $R[X]$ are the units in $R$.
answered Mar 10 at 7:56
WuestenfuxWuestenfux
5,5131513
$endgroup$
2
$begingroup$
It is a confusing entry though, since the eye leaps to the factorisation into linear factors!
$endgroup$
– ancientmathematician
Mar 10 at 7:57
2
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Indeed, the wiki entry is confusing. It says that if F is a field, a non-constant polynomial is irreducible over F if its coefficients belong to F and it cannot be factored into the product of two NON-CONSTANT polynomials with coefficients in F. Over a ring R, not each constant polynomial is a unit. –
$endgroup$
– Wuestenfux
Mar 10 at 8:33
1
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Your argument is correct, but you should have applied it to $p_3(x) = 3(3x^2-1)$.
$endgroup$
– Paul Frost
Mar 10 at 9:03
add a comment |
$begingroup$
The polynomial p3 (x) is irreducible polynomial over integers but not irreducible element in the ring Z [x] as it can be written as 3 (x^2-1) where neither 3 is unit nor (x^2-1) is unit as definition of irreducible element is non- zero, non- unit element which whenever can be represented as product of two elements of the ring one of the element in that product must be unit.
In wiki , the definition seems unclear then.
answered Mar 10 at 10:22
PriyankaPriyanka
543
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add a comment |
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2 Answers
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active
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2 Answers
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votes
$begingroup$
Well, the polynomial $f(x)=3(x^2-1)$ is reducible over the integers, since it factors into a product of non-units in ${Bbb Z}[x]$. The units in ${Bbb Z}[x]$ coincide with the units in ${Bbb Z}$ and are $pm 1$.
The definition says that a polynomial $fin R[x]$ is irreducible if it can only be factored as $f=gh$ with $g,hin R[x]$ such that one of $g,h$ is a unit in $R[x]$. The units in $R[X]$ are the units in $R$.
answered Mar 10 at 7:56
WuestenfuxWuestenfux
5,5131513
$endgroup$
2
$begingroup$
It is a confusing entry though, since the eye leaps to the factorisation into linear factors!
$endgroup$
– ancientmathematician
Mar 10 at 7:57
2
$begingroup$
Indeed, the wiki entry is confusing. It says that if F is a field, a non-constant polynomial is irreducible over F if its coefficients belong to F and it cannot be factored into the product of two NON-CONSTANT polynomials with coefficients in F. Over a ring R, not each constant polynomial is a unit. –
$endgroup$
– Wuestenfux
Mar 10 at 8:33
1
$begingroup$
Your argument is correct, but you should have applied it to $p_3(x) = 3(3x^2-1)$.
$endgroup$
– Paul Frost
Mar 10 at 9:03
add a comment |
$begingroup$
Well, the polynomial $f(x)=3(x^2-1)$ is reducible over the integers, since it factors into a product of non-units in ${Bbb Z}[x]$. The units in ${Bbb Z}[x]$ coincide with the units in ${Bbb Z}$ and are $pm 1$.
The definition says that a polynomial $fin R[x]$ is irreducible if it can only be factored as $f=gh$ with $g,hin R[x]$ such that one of $g,h$ is a unit in $R[x]$. The units in $R[X]$ are the units in $R$.
answered Mar 10 at 7:56
WuestenfuxWuestenfux
5,5131513
$endgroup$
2
$begingroup$
It is a confusing entry though, since the eye leaps to the factorisation into linear factors!
$endgroup$
– ancientmathematician
Mar 10 at 7:57
2
$begingroup$
Indeed, the wiki entry is confusing. It says that if F is a field, a non-constant polynomial is irreducible over F if its coefficients belong to F and it cannot be factored into the product of two NON-CONSTANT polynomials with coefficients in F. Over a ring R, not each constant polynomial is a unit. –
$endgroup$
– Wuestenfux
Mar 10 at 8:33
1
$begingroup$
Your argument is correct, but you should have applied it to $p_3(x) = 3(3x^2-1)$.
$endgroup$
– Paul Frost
Mar 10 at 9:03
add a comment |
$begingroup$
Well, the polynomial $f(x)=3(x^2-1)$ is reducible over the integers, since it factors into a product of non-units in ${Bbb Z}[x]$. The units in ${Bbb Z}[x]$ coincide with the units in ${Bbb Z}$ and are $pm 1$.
The definition says that a polynomial $fin R[x]$ is irreducible if it can only be factored as $f=gh$ with $g,hin R[x]$ such that one of $g,h$ is a unit in $R[x]$. The units in $R[X]$ are the units in $R$.
answered Mar 10 at 7:56
WuestenfuxWuestenfux
5,5131513
$endgroup$
Well, the polynomial $f(x)=3(x^2-1)$ is reducible over the integers, since it factors into a product of non-units in ${Bbb Z}[x]$. The units in ${Bbb Z}[x]$ coincide with the units in ${Bbb Z}$ and are $pm 1$.
The definition says that a polynomial $fin R[x]$ is irreducible if it can only be factored as $f=gh$ with $g,hin R[x]$ such that one of $g,h$ is a unit in $R[x]$. The units in $R[X]$ are the units in $R$.
answered Mar 10 at 7:56
WuestenfuxWuestenfux
5,5131513
answered Mar 10 at 7:56
WuestenfuxWuestenfux
5,5131513
answered Mar 10 at 7:56
WuestenfuxWuestenfux
5,5131513
answered Mar 10 at 7:56
WuestenfuxWuestenfux
5,5131513
5,5131513
2
$begingroup$
It is a confusing entry though, since the eye leaps to the factorisation into linear factors!
$endgroup$
– ancientmathematician
Mar 10 at 7:57
2
$begingroup$
Indeed, the wiki entry is confusing. It says that if F is a field, a non-constant polynomial is irreducible over F if its coefficients belong to F and it cannot be factored into the product of two NON-CONSTANT polynomials with coefficients in F. Over a ring R, not each constant polynomial is a unit. –
$endgroup$
– Wuestenfux
Mar 10 at 8:33
1
$begingroup$
Your argument is correct, but you should have applied it to $p_3(x) = 3(3x^2-1)$.
$endgroup$
– Paul Frost
Mar 10 at 9:03
add a comment |
2
$begingroup$
It is a confusing entry though, since the eye leaps to the factorisation into linear factors!
$endgroup$
– ancientmathematician
Mar 10 at 7:57
2
$begingroup$
Indeed, the wiki entry is confusing. It says that if F is a field, a non-constant polynomial is irreducible over F if its coefficients belong to F and it cannot be factored into the product of two NON-CONSTANT polynomials with coefficients in F. Over a ring R, not each constant polynomial is a unit. –
$endgroup$
– Wuestenfux
Mar 10 at 8:33
1
$begingroup$
Your argument is correct, but you should have applied it to $p_3(x) = 3(3x^2-1)$.
$endgroup$
– Paul Frost
Mar 10 at 9:03
2
2
$begingroup$
It is a confusing entry though, since the eye leaps to the factorisation into linear factors!
$endgroup$
– ancientmathematician
Mar 10 at 7:57
$begingroup$
It is a confusing entry though, since the eye leaps to the factorisation into linear factors!
$endgroup$
– ancientmathematician
Mar 10 at 7:57
2
2
$begingroup$
Indeed, the wiki entry is confusing. It says that if F is a field, a non-constant polynomial is irreducible over F if its coefficients belong to F and it cannot be factored into the product of two NON-CONSTANT polynomials with coefficients in F. Over a ring R, not each constant polynomial is a unit. –
$endgroup$
– Wuestenfux
Mar 10 at 8:33
$begingroup$
Indeed, the wiki entry is confusing. It says that if F is a field, a non-constant polynomial is irreducible over F if its coefficients belong to F and it cannot be factored into the product of two NON-CONSTANT polynomials with coefficients in F. Over a ring R, not each constant polynomial is a unit. –
$endgroup$
– Wuestenfux
Mar 10 at 8:33
1
1
$begingroup$
Your argument is correct, but you should have applied it to $p_3(x) = 3(3x^2-1)$.
$endgroup$
– Paul Frost
Mar 10 at 9:03
$begingroup$
Your argument is correct, but you should have applied it to $p_3(x) = 3(3x^2-1)$.
$endgroup$
– Paul Frost
Mar 10 at 9:03
add a comment |
$begingroup$
The polynomial p3 (x) is irreducible polynomial over integers but not irreducible element in the ring Z [x] as it can be written as 3 (x^2-1) where neither 3 is unit nor (x^2-1) is unit as definition of irreducible element is non- zero, non- unit element which whenever can be represented as product of two elements of the ring one of the element in that product must be unit.
In wiki , the definition seems unclear then.
answered Mar 10 at 10:22
PriyankaPriyanka
543
$endgroup$
add a comment |
$begingroup$
The polynomial p3 (x) is irreducible polynomial over integers but not irreducible element in the ring Z [x] as it can be written as 3 (x^2-1) where neither 3 is unit nor (x^2-1) is unit as definition of irreducible element is non- zero, non- unit element which whenever can be represented as product of two elements of the ring one of the element in that product must be unit.
In wiki , the definition seems unclear then.
answered Mar 10 at 10:22
PriyankaPriyanka
543
$endgroup$
add a comment |
$begingroup$
The polynomial p3 (x) is irreducible polynomial over integers but not irreducible element in the ring Z [x] as it can be written as 3 (x^2-1) where neither 3 is unit nor (x^2-1) is unit as definition of irreducible element is non- zero, non- unit element which whenever can be represented as product of two elements of the ring one of the element in that product must be unit.
In wiki , the definition seems unclear then.
answered Mar 10 at 10:22
PriyankaPriyanka
543
$endgroup$
The polynomial p3 (x) is irreducible polynomial over integers but not irreducible element in the ring Z [x] as it can be written as 3 (x^2-1) where neither 3 is unit nor (x^2-1) is unit as definition of irreducible element is non- zero, non- unit element which whenever can be represented as product of two elements of the ring one of the element in that product must be unit.
In wiki , the definition seems unclear then.
answered Mar 10 at 10:22
PriyankaPriyanka
543
edited Mar 10 at 10:40
answered Mar 10 at 10:22
PriyankaPriyanka
543
answered Mar 10 at 10:22
PriyankaPriyanka
543
answered Mar 10 at 10:22
PriyankaPriyanka
543
543
add a comment |
add a comment |
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$begingroup$
Recall when is a polynomial in $mathbb Z[X]$ reducible in $mathbb Z [X]$
$endgroup$
– Soumik Ghosh
Mar 10 at 7:54