What is an irreducible element in $Bbb{Z}_6[x]$?
$begingroup$
What is an irreducible element in $Bbb{Z}_6[x]$?
This was a problem on our final and no one knew how to solve it. Does anyone have a method for solving this?
abstract-algebra polynomials ring-theory modular-arithmetic factoring
asked Dec 13 '18 at 22:37
Riley HRiley H
1046
$endgroup$
add a comment |
$begingroup$
What is an irreducible element in $Bbb{Z}_6[x]$?
This was a problem on our final and no one knew how to solve it. Does anyone have a method for solving this?
abstract-algebra polynomials ring-theory modular-arithmetic factoring
asked Dec 13 '18 at 22:37
Riley HRiley H
1046
$endgroup$
$begingroup$
Is the question asking for a definition or an example?
$endgroup$
– Peter Taylor
Dec 13 '18 at 22:47
$begingroup$
An example with proof @PeterTaylor
$endgroup$
– Riley H
Dec 13 '18 at 22:48
2
$begingroup$
@PeterTaylor No, $x$ is reducible here: $x=(2x+3)(3x+2)$. Even $2$ is not reducible: $2=2cdot 4$. The same goes for $3$: $3=3cdot 3$.
$endgroup$
– user614671
Dec 13 '18 at 22:53
$begingroup$
Factorization of polynomials (and consequently also irreducibility) becomes rather delicate when the modulus is not a power of a prime. I have been linking several questions of this type to this answer by Bill Dubuque where more material can be found. For more incarnations of this theme on our site see the several questions linked to that.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 11:17
$begingroup$
Modulo a prime power the theory is quite a bit simpler. Even though we still have zero divisors and intricacies like $x^2equiv(x-2)^2pmod 4$. In the prime power case the subset of polynomials with zeros that are roots of unity of order prime to $p$ behave reasonably well (because Hensel lifting), and give rise to a theory of Galois rings extending some of the facts from the theory of finite fields.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 11:21
add a comment |
$begingroup$
What is an irreducible element in $Bbb{Z}_6[x]$?
This was a problem on our final and no one knew how to solve it. Does anyone have a method for solving this?
abstract-algebra polynomials ring-theory modular-arithmetic factoring
asked Dec 13 '18 at 22:37
Riley HRiley H
1046
$endgroup$
What is an irreducible element in $Bbb{Z}_6[x]$?
This was a problem on our final and no one knew how to solve it. Does anyone have a method for solving this?
abstract-algebra polynomials ring-theory modular-arithmetic factoring
abstract-algebra polynomials ring-theory modular-arithmetic factoring
asked Dec 13 '18 at 22:37
Riley HRiley H
1046
asked Dec 13 '18 at 22:37
Riley HRiley H
1046
edited Dec 13 '18 at 23:44
user614671
asked Dec 13 '18 at 22:37
Riley HRiley H
1046
asked Dec 13 '18 at 22:37
Riley HRiley H
1046
asked Dec 13 '18 at 22:37
Riley HRiley H
1046
1046
$begingroup$
Is the question asking for a definition or an example?
$endgroup$
– Peter Taylor
Dec 13 '18 at 22:47
$begingroup$
An example with proof @PeterTaylor
$endgroup$
– Riley H
Dec 13 '18 at 22:48
2
$begingroup$
@PeterTaylor No, $x$ is reducible here: $x=(2x+3)(3x+2)$. Even $2$ is not reducible: $2=2cdot 4$. The same goes for $3$: $3=3cdot 3$.
$endgroup$
– user614671
Dec 13 '18 at 22:53
$begingroup$
Factorization of polynomials (and consequently also irreducibility) becomes rather delicate when the modulus is not a power of a prime. I have been linking several questions of this type to this answer by Bill Dubuque where more material can be found. For more incarnations of this theme on our site see the several questions linked to that.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 11:17
$begingroup$
Modulo a prime power the theory is quite a bit simpler. Even though we still have zero divisors and intricacies like $x^2equiv(x-2)^2pmod 4$. In the prime power case the subset of polynomials with zeros that are roots of unity of order prime to $p$ behave reasonably well (because Hensel lifting), and give rise to a theory of Galois rings extending some of the facts from the theory of finite fields.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 11:21
add a comment |
$begingroup$
Is the question asking for a definition or an example?
$endgroup$
– Peter Taylor
Dec 13 '18 at 22:47
$begingroup$
An example with proof @PeterTaylor
$endgroup$
– Riley H
Dec 13 '18 at 22:48
2
$begingroup$
@PeterTaylor No, $x$ is reducible here: $x=(2x+3)(3x+2)$. Even $2$ is not reducible: $2=2cdot 4$. The same goes for $3$: $3=3cdot 3$.
$endgroup$
– user614671
Dec 13 '18 at 22:53
$begingroup$
Factorization of polynomials (and consequently also irreducibility) becomes rather delicate when the modulus is not a power of a prime. I have been linking several questions of this type to this answer by Bill Dubuque where more material can be found. For more incarnations of this theme on our site see the several questions linked to that.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 11:17
$begingroup$
Modulo a prime power the theory is quite a bit simpler. Even though we still have zero divisors and intricacies like $x^2equiv(x-2)^2pmod 4$. In the prime power case the subset of polynomials with zeros that are roots of unity of order prime to $p$ behave reasonably well (because Hensel lifting), and give rise to a theory of Galois rings extending some of the facts from the theory of finite fields.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 11:21
$begingroup$
Is the question asking for a definition or an example?
$endgroup$
– Peter Taylor
Dec 13 '18 at 22:47
$begingroup$
Is the question asking for a definition or an example?
$endgroup$
– Peter Taylor
Dec 13 '18 at 22:47
$begingroup$
An example with proof @PeterTaylor
$endgroup$
– Riley H
Dec 13 '18 at 22:48
$begingroup$
An example with proof @PeterTaylor
$endgroup$
– Riley H
Dec 13 '18 at 22:48
2
2
$begingroup$
@PeterTaylor No, $x$ is reducible here: $x=(2x+3)(3x+2)$. Even $2$ is not reducible: $2=2cdot 4$. The same goes for $3$: $3=3cdot 3$.
$endgroup$
– user614671
Dec 13 '18 at 22:53
$begingroup$
@PeterTaylor No, $x$ is reducible here: $x=(2x+3)(3x+2)$. Even $2$ is not reducible: $2=2cdot 4$. The same goes for $3$: $3=3cdot 3$.
$endgroup$
– user614671
Dec 13 '18 at 22:53
$begingroup$
Factorization of polynomials (and consequently also irreducibility) becomes rather delicate when the modulus is not a power of a prime. I have been linking several questions of this type to this answer by Bill Dubuque where more material can be found. For more incarnations of this theme on our site see the several questions linked to that.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 11:17
$begingroup$
Factorization of polynomials (and consequently also irreducibility) becomes rather delicate when the modulus is not a power of a prime. I have been linking several questions of this type to this answer by Bill Dubuque where more material can be found. For more incarnations of this theme on our site see the several questions linked to that.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 11:17
$begingroup$
Modulo a prime power the theory is quite a bit simpler. Even though we still have zero divisors and intricacies like $x^2equiv(x-2)^2pmod 4$. In the prime power case the subset of polynomials with zeros that are roots of unity of order prime to $p$ behave reasonably well (because Hensel lifting), and give rise to a theory of Galois rings extending some of the facts from the theory of finite fields.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 11:21
$begingroup$
Modulo a prime power the theory is quite a bit simpler. Even though we still have zero divisors and intricacies like $x^2equiv(x-2)^2pmod 4$. In the prime power case the subset of polynomials with zeros that are roots of unity of order prime to $p$ behave reasonably well (because Hensel lifting), and give rise to a theory of Galois rings extending some of the facts from the theory of finite fields.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 11:21
add a comment |
1 Answer
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Claim. Assume that $Bbb Z_m=Bbb Z/mBbb Z$. The polynomial $f(x)inBbb Z_6[x]$ is irreducible if and only if exactly one of these is true:
$(a)$ $f(x)$ is irreducible over $Bbb Z_2$ and $f(x)equiv pm1pmod{3}$ or
$(b)$ $f(x)$ is irreducible over $Bbb Z_3$ and $f(x)equiv 1pmod{2}$.
Suppose that $f(x)in Bbb Z_6[x]$ is irreducible. Then, $f(x)$ is either irreducible or invertible over $Bbb{Z}_2$, and over $Bbb{Z}_3$. If $f(x)$ is invertible in both $Bbb Z_2$ and $Bbb Z_3$, then $f(x)$ is one of the constant polynomials $pm1$, which are invertible (and not irreducible). This is a contradiction, so $f(x)$ must either be irreducible over $Bbb Z_2$ or over $Bbb Z_3$.
If $f(x)$ is invertible in neither $Bbb Z_2$ nor $Bbb Z_3$, then we can solve for $$u(x)equiv 1pmod{2} wedge u(x)equiv f(x)pmod 3$$
and
$$v(x)equiv f(x)pmod{2} wedge v(x)equiv 1pmod{3}$$
for $u(x),v(x)inBbb{Z}_6[x]$. These $u(x)$ and $v(x)$ are non-invertible over $Bbb Z_6$ (since $u(x)$ is non-invertible modulo $3$, and $v(x)$ is non-invertible modulo $2$). However, $f(x)=u(x)cdot v(x)$ in $Bbb Z_6[x]$, which is a contradiction. Therefore, either $(a)$ or $(b)$ holds, but not both.
Conversely, suppose that $(a)$ holds. If $f(x)=p(x)cdot q(x)$ for some $p(x),q(x)in Bbb Z_6[x]$, then reduce $p(x)$ and $q(x)$ modulo $2$ and $3$ respectively. As $f(x)$ is irreducible mod $2$, $(p,q)=(f,1)$ or $(p,q)=(1,f)$ modulo $2$. Wlog, $p=f$ and $q=1$. As $f(x)equiv pm 1pmod{3}$, $p(x)$ and $q(x)$ modulo $3$ are constants $pm 1$. Thus, $q(x)equiv 1pmod{2}$ and $q(x)equiv -1pmod{3}$. Therefore, $q(x)= pm 1$ in $Bbb Z_6[x]$. That is, $q(x)$ is constant, and so $f(x)$ is irreducible.
Finally, suppose that $(b)$ holds. If $f(x)=p(x)cdot q(x)$ for some $p(x),q(x)in Bbb Z_6[x]$, then reduce $p(x)$ and $q(x)$ modulo $2$ and $3$ respectively. As $f(x)$ is irreducible mod $3$, $(p,q)=(f,1)$ or $(p,q)=(1,f)$ modulo $3$, up to sign swapping. Wlog, $p=f$ and $q=1$. As $f(x)equiv 1pmod{2}$, $p(x)$ and $q(x)$ modulo $2$ equal the constant $1$. Thus, $q(x)equiv 1pmod{2}$. Therefore, $q(x)= 1$ in $Bbb Z_6[x]$. That is, $q(x)$ is constant, and so $f(x)$ is irreducible.
Examples. The polynomial $f(x)=3x+1$ fits $(a)$, so it is irreducible in $Bbb{Z}_6[x]$. The polynomial $f(x)=2x+1$ fits $(b)$, so it is irreducible in $Bbb{Z}_6[x]$.
edited Dec 16 '18 at 10:46
Batominovski
33.1k33293
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$begingroup$
Claim. Assume that $Bbb Z_m=Bbb Z/mBbb Z$. The polynomial $f(x)inBbb Z_6[x]$ is irreducible if and only if exactly one of these is true:
$(a)$ $f(x)$ is irreducible over $Bbb Z_2$ and $f(x)equiv pm1pmod{3}$ or
$(b)$ $f(x)$ is irreducible over $Bbb Z_3$ and $f(x)equiv 1pmod{2}$.
Suppose that $f(x)in Bbb Z_6[x]$ is irreducible. Then, $f(x)$ is either irreducible or invertible over $Bbb{Z}_2$, and over $Bbb{Z}_3$. If $f(x)$ is invertible in both $Bbb Z_2$ and $Bbb Z_3$, then $f(x)$ is one of the constant polynomials $pm1$, which are invertible (and not irreducible). This is a contradiction, so $f(x)$ must either be irreducible over $Bbb Z_2$ or over $Bbb Z_3$.
If $f(x)$ is invertible in neither $Bbb Z_2$ nor $Bbb Z_3$, then we can solve for $$u(x)equiv 1pmod{2} wedge u(x)equiv f(x)pmod 3$$
and
$$v(x)equiv f(x)pmod{2} wedge v(x)equiv 1pmod{3}$$
for $u(x),v(x)inBbb{Z}_6[x]$. These $u(x)$ and $v(x)$ are non-invertible over $Bbb Z_6$ (since $u(x)$ is non-invertible modulo $3$, and $v(x)$ is non-invertible modulo $2$). However, $f(x)=u(x)cdot v(x)$ in $Bbb Z_6[x]$, which is a contradiction. Therefore, either $(a)$ or $(b)$ holds, but not both.
Conversely, suppose that $(a)$ holds. If $f(x)=p(x)cdot q(x)$ for some $p(x),q(x)in Bbb Z_6[x]$, then reduce $p(x)$ and $q(x)$ modulo $2$ and $3$ respectively. As $f(x)$ is irreducible mod $2$, $(p,q)=(f,1)$ or $(p,q)=(1,f)$ modulo $2$. Wlog, $p=f$ and $q=1$. As $f(x)equiv pm 1pmod{3}$, $p(x)$ and $q(x)$ modulo $3$ are constants $pm 1$. Thus, $q(x)equiv 1pmod{2}$ and $q(x)equiv -1pmod{3}$. Therefore, $q(x)= pm 1$ in $Bbb Z_6[x]$. That is, $q(x)$ is constant, and so $f(x)$ is irreducible.
Finally, suppose that $(b)$ holds. If $f(x)=p(x)cdot q(x)$ for some $p(x),q(x)in Bbb Z_6[x]$, then reduce $p(x)$ and $q(x)$ modulo $2$ and $3$ respectively. As $f(x)$ is irreducible mod $3$, $(p,q)=(f,1)$ or $(p,q)=(1,f)$ modulo $3$, up to sign swapping. Wlog, $p=f$ and $q=1$. As $f(x)equiv 1pmod{2}$, $p(x)$ and $q(x)$ modulo $2$ equal the constant $1$. Thus, $q(x)equiv 1pmod{2}$. Therefore, $q(x)= 1$ in $Bbb Z_6[x]$. That is, $q(x)$ is constant, and so $f(x)$ is irreducible.
Examples. The polynomial $f(x)=3x+1$ fits $(a)$, so it is irreducible in $Bbb{Z}_6[x]$. The polynomial $f(x)=2x+1$ fits $(b)$, so it is irreducible in $Bbb{Z}_6[x]$.
edited Dec 16 '18 at 10:46
Batominovski
33.1k33293
$endgroup$
add a comment |
$begingroup$
Claim. Assume that $Bbb Z_m=Bbb Z/mBbb Z$. The polynomial $f(x)inBbb Z_6[x]$ is irreducible if and only if exactly one of these is true:
$(a)$ $f(x)$ is irreducible over $Bbb Z_2$ and $f(x)equiv pm1pmod{3}$ or
$(b)$ $f(x)$ is irreducible over $Bbb Z_3$ and $f(x)equiv 1pmod{2}$.
Suppose that $f(x)in Bbb Z_6[x]$ is irreducible. Then, $f(x)$ is either irreducible or invertible over $Bbb{Z}_2$, and over $Bbb{Z}_3$. If $f(x)$ is invertible in both $Bbb Z_2$ and $Bbb Z_3$, then $f(x)$ is one of the constant polynomials $pm1$, which are invertible (and not irreducible). This is a contradiction, so $f(x)$ must either be irreducible over $Bbb Z_2$ or over $Bbb Z_3$.
If $f(x)$ is invertible in neither $Bbb Z_2$ nor $Bbb Z_3$, then we can solve for $$u(x)equiv 1pmod{2} wedge u(x)equiv f(x)pmod 3$$
and
$$v(x)equiv f(x)pmod{2} wedge v(x)equiv 1pmod{3}$$
for $u(x),v(x)inBbb{Z}_6[x]$. These $u(x)$ and $v(x)$ are non-invertible over $Bbb Z_6$ (since $u(x)$ is non-invertible modulo $3$, and $v(x)$ is non-invertible modulo $2$). However, $f(x)=u(x)cdot v(x)$ in $Bbb Z_6[x]$, which is a contradiction. Therefore, either $(a)$ or $(b)$ holds, but not both.
Conversely, suppose that $(a)$ holds. If $f(x)=p(x)cdot q(x)$ for some $p(x),q(x)in Bbb Z_6[x]$, then reduce $p(x)$ and $q(x)$ modulo $2$ and $3$ respectively. As $f(x)$ is irreducible mod $2$, $(p,q)=(f,1)$ or $(p,q)=(1,f)$ modulo $2$. Wlog, $p=f$ and $q=1$. As $f(x)equiv pm 1pmod{3}$, $p(x)$ and $q(x)$ modulo $3$ are constants $pm 1$. Thus, $q(x)equiv 1pmod{2}$ and $q(x)equiv -1pmod{3}$. Therefore, $q(x)= pm 1$ in $Bbb Z_6[x]$. That is, $q(x)$ is constant, and so $f(x)$ is irreducible.
Finally, suppose that $(b)$ holds. If $f(x)=p(x)cdot q(x)$ for some $p(x),q(x)in Bbb Z_6[x]$, then reduce $p(x)$ and $q(x)$ modulo $2$ and $3$ respectively. As $f(x)$ is irreducible mod $3$, $(p,q)=(f,1)$ or $(p,q)=(1,f)$ modulo $3$, up to sign swapping. Wlog, $p=f$ and $q=1$. As $f(x)equiv 1pmod{2}$, $p(x)$ and $q(x)$ modulo $2$ equal the constant $1$. Thus, $q(x)equiv 1pmod{2}$. Therefore, $q(x)= 1$ in $Bbb Z_6[x]$. That is, $q(x)$ is constant, and so $f(x)$ is irreducible.
Examples. The polynomial $f(x)=3x+1$ fits $(a)$, so it is irreducible in $Bbb{Z}_6[x]$. The polynomial $f(x)=2x+1$ fits $(b)$, so it is irreducible in $Bbb{Z}_6[x]$.
edited Dec 16 '18 at 10:46
Batominovski
33.1k33293
$endgroup$
add a comment |
$begingroup$
Claim. Assume that $Bbb Z_m=Bbb Z/mBbb Z$. The polynomial $f(x)inBbb Z_6[x]$ is irreducible if and only if exactly one of these is true:
$(a)$ $f(x)$ is irreducible over $Bbb Z_2$ and $f(x)equiv pm1pmod{3}$ or
$(b)$ $f(x)$ is irreducible over $Bbb Z_3$ and $f(x)equiv 1pmod{2}$.
Suppose that $f(x)in Bbb Z_6[x]$ is irreducible. Then, $f(x)$ is either irreducible or invertible over $Bbb{Z}_2$, and over $Bbb{Z}_3$. If $f(x)$ is invertible in both $Bbb Z_2$ and $Bbb Z_3$, then $f(x)$ is one of the constant polynomials $pm1$, which are invertible (and not irreducible). This is a contradiction, so $f(x)$ must either be irreducible over $Bbb Z_2$ or over $Bbb Z_3$.
If $f(x)$ is invertible in neither $Bbb Z_2$ nor $Bbb Z_3$, then we can solve for $$u(x)equiv 1pmod{2} wedge u(x)equiv f(x)pmod 3$$
and
$$v(x)equiv f(x)pmod{2} wedge v(x)equiv 1pmod{3}$$
for $u(x),v(x)inBbb{Z}_6[x]$. These $u(x)$ and $v(x)$ are non-invertible over $Bbb Z_6$ (since $u(x)$ is non-invertible modulo $3$, and $v(x)$ is non-invertible modulo $2$). However, $f(x)=u(x)cdot v(x)$ in $Bbb Z_6[x]$, which is a contradiction. Therefore, either $(a)$ or $(b)$ holds, but not both.
Conversely, suppose that $(a)$ holds. If $f(x)=p(x)cdot q(x)$ for some $p(x),q(x)in Bbb Z_6[x]$, then reduce $p(x)$ and $q(x)$ modulo $2$ and $3$ respectively. As $f(x)$ is irreducible mod $2$, $(p,q)=(f,1)$ or $(p,q)=(1,f)$ modulo $2$. Wlog, $p=f$ and $q=1$. As $f(x)equiv pm 1pmod{3}$, $p(x)$ and $q(x)$ modulo $3$ are constants $pm 1$. Thus, $q(x)equiv 1pmod{2}$ and $q(x)equiv -1pmod{3}$. Therefore, $q(x)= pm 1$ in $Bbb Z_6[x]$. That is, $q(x)$ is constant, and so $f(x)$ is irreducible.
Finally, suppose that $(b)$ holds. If $f(x)=p(x)cdot q(x)$ for some $p(x),q(x)in Bbb Z_6[x]$, then reduce $p(x)$ and $q(x)$ modulo $2$ and $3$ respectively. As $f(x)$ is irreducible mod $3$, $(p,q)=(f,1)$ or $(p,q)=(1,f)$ modulo $3$, up to sign swapping. Wlog, $p=f$ and $q=1$. As $f(x)equiv 1pmod{2}$, $p(x)$ and $q(x)$ modulo $2$ equal the constant $1$. Thus, $q(x)equiv 1pmod{2}$. Therefore, $q(x)= 1$ in $Bbb Z_6[x]$. That is, $q(x)$ is constant, and so $f(x)$ is irreducible.
Examples. The polynomial $f(x)=3x+1$ fits $(a)$, so it is irreducible in $Bbb{Z}_6[x]$. The polynomial $f(x)=2x+1$ fits $(b)$, so it is irreducible in $Bbb{Z}_6[x]$.
edited Dec 16 '18 at 10:46
Batominovski
33.1k33293
$endgroup$
Claim. Assume that $Bbb Z_m=Bbb Z/mBbb Z$. The polynomial $f(x)inBbb Z_6[x]$ is irreducible if and only if exactly one of these is true:
$(a)$ $f(x)$ is irreducible over $Bbb Z_2$ and $f(x)equiv pm1pmod{3}$ or
$(b)$ $f(x)$ is irreducible over $Bbb Z_3$ and $f(x)equiv 1pmod{2}$.
Suppose that $f(x)in Bbb Z_6[x]$ is irreducible. Then, $f(x)$ is either irreducible or invertible over $Bbb{Z}_2$, and over $Bbb{Z}_3$. If $f(x)$ is invertible in both $Bbb Z_2$ and $Bbb Z_3$, then $f(x)$ is one of the constant polynomials $pm1$, which are invertible (and not irreducible). This is a contradiction, so $f(x)$ must either be irreducible over $Bbb Z_2$ or over $Bbb Z_3$.
If $f(x)$ is invertible in neither $Bbb Z_2$ nor $Bbb Z_3$, then we can solve for $$u(x)equiv 1pmod{2} wedge u(x)equiv f(x)pmod 3$$
and
$$v(x)equiv f(x)pmod{2} wedge v(x)equiv 1pmod{3}$$
for $u(x),v(x)inBbb{Z}_6[x]$. These $u(x)$ and $v(x)$ are non-invertible over $Bbb Z_6$ (since $u(x)$ is non-invertible modulo $3$, and $v(x)$ is non-invertible modulo $2$). However, $f(x)=u(x)cdot v(x)$ in $Bbb Z_6[x]$, which is a contradiction. Therefore, either $(a)$ or $(b)$ holds, but not both.
Conversely, suppose that $(a)$ holds. If $f(x)=p(x)cdot q(x)$ for some $p(x),q(x)in Bbb Z_6[x]$, then reduce $p(x)$ and $q(x)$ modulo $2$ and $3$ respectively. As $f(x)$ is irreducible mod $2$, $(p,q)=(f,1)$ or $(p,q)=(1,f)$ modulo $2$. Wlog, $p=f$ and $q=1$. As $f(x)equiv pm 1pmod{3}$, $p(x)$ and $q(x)$ modulo $3$ are constants $pm 1$. Thus, $q(x)equiv 1pmod{2}$ and $q(x)equiv -1pmod{3}$. Therefore, $q(x)= pm 1$ in $Bbb Z_6[x]$. That is, $q(x)$ is constant, and so $f(x)$ is irreducible.
Finally, suppose that $(b)$ holds. If $f(x)=p(x)cdot q(x)$ for some $p(x),q(x)in Bbb Z_6[x]$, then reduce $p(x)$ and $q(x)$ modulo $2$ and $3$ respectively. As $f(x)$ is irreducible mod $3$, $(p,q)=(f,1)$ or $(p,q)=(1,f)$ modulo $3$, up to sign swapping. Wlog, $p=f$ and $q=1$. As $f(x)equiv 1pmod{2}$, $p(x)$ and $q(x)$ modulo $2$ equal the constant $1$. Thus, $q(x)equiv 1pmod{2}$. Therefore, $q(x)= 1$ in $Bbb Z_6[x]$. That is, $q(x)$ is constant, and so $f(x)$ is irreducible.
Examples. The polynomial $f(x)=3x+1$ fits $(a)$, so it is irreducible in $Bbb{Z}_6[x]$. The polynomial $f(x)=2x+1$ fits $(b)$, so it is irreducible in $Bbb{Z}_6[x]$.
edited Dec 16 '18 at 10:46
Batominovski
33.1k33293
edited Dec 16 '18 at 10:46
Batominovski
33.1k33293
edited Dec 16 '18 at 10:46
Batominovski
33.1k33293
edited Dec 16 '18 at 10:46
Batominovski
33.1k33293
33.1k33293
answered Dec 13 '18 at 23:20
user614671
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$begingroup$
Is the question asking for a definition or an example?
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– Peter Taylor
Dec 13 '18 at 22:47
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An example with proof @PeterTaylor
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– Riley H
Dec 13 '18 at 22:48
2
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@PeterTaylor No, $x$ is reducible here: $x=(2x+3)(3x+2)$. Even $2$ is not reducible: $2=2cdot 4$. The same goes for $3$: $3=3cdot 3$.
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– user614671
Dec 13 '18 at 22:53
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Factorization of polynomials (and consequently also irreducibility) becomes rather delicate when the modulus is not a power of a prime. I have been linking several questions of this type to this answer by Bill Dubuque where more material can be found. For more incarnations of this theme on our site see the several questions linked to that.
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– Jyrki Lahtonen
Dec 16 '18 at 11:17
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Modulo a prime power the theory is quite a bit simpler. Even though we still have zero divisors and intricacies like $x^2equiv(x-2)^2pmod 4$. In the prime power case the subset of polynomials with zeros that are roots of unity of order prime to $p$ behave reasonably well (because Hensel lifting), and give rise to a theory of Galois rings extending some of the facts from the theory of finite fields.
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– Jyrki Lahtonen
Dec 16 '18 at 11:21