Finding units and zero divisors in a polynomial quotient ring
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I am trying to study for an exam and I am not sure of this solution my professor posted to an exercise. I am given the polynomial quotient ring $mathbb{Z}_6/(x^2+2x)$ and have to find all units and zero divisors.
My initial idea was to do a multiplication table, but with a group this big I am sure there must be another way. For units, I was thinking all polynomials that are coprime to $x$ and $x+2$, but this isn’t what the solutions have.
He lists the units as $1, 5, x+1, 2x+5, 3x+1, 3x+5, 4x+1, 5x+5$, but gives no reason why. I understand these are all coprime to the ideal, but why isn’t $x+3$ an ideal for example?
He then says the zero divisors are everything that isn’t a unit or zero. Why is this? I thought it was possible for an element to be neither a unit nor a zero divisor.
Thank you very much.
abstract-algebra polynomial-rings
asked Dec 12 '18 at 1:34
Tyler6Tyler6
735414
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add a comment |
$begingroup$
I am trying to study for an exam and I am not sure of this solution my professor posted to an exercise. I am given the polynomial quotient ring $mathbb{Z}_6/(x^2+2x)$ and have to find all units and zero divisors.
My initial idea was to do a multiplication table, but with a group this big I am sure there must be another way. For units, I was thinking all polynomials that are coprime to $x$ and $x+2$, but this isn’t what the solutions have.
He lists the units as $1, 5, x+1, 2x+5, 3x+1, 3x+5, 4x+1, 5x+5$, but gives no reason why. I understand these are all coprime to the ideal, but why isn’t $x+3$ an ideal for example?
He then says the zero divisors are everything that isn’t a unit or zero. Why is this? I thought it was possible for an element to be neither a unit nor a zero divisor.
Thank you very much.
abstract-algebra polynomial-rings
asked Dec 12 '18 at 1:34
Tyler6Tyler6
735414
$endgroup$
add a comment |
$begingroup$
I am trying to study for an exam and I am not sure of this solution my professor posted to an exercise. I am given the polynomial quotient ring $mathbb{Z}_6/(x^2+2x)$ and have to find all units and zero divisors.
My initial idea was to do a multiplication table, but with a group this big I am sure there must be another way. For units, I was thinking all polynomials that are coprime to $x$ and $x+2$, but this isn’t what the solutions have.
He lists the units as $1, 5, x+1, 2x+5, 3x+1, 3x+5, 4x+1, 5x+5$, but gives no reason why. I understand these are all coprime to the ideal, but why isn’t $x+3$ an ideal for example?
He then says the zero divisors are everything that isn’t a unit or zero. Why is this? I thought it was possible for an element to be neither a unit nor a zero divisor.
Thank you very much.
abstract-algebra polynomial-rings
asked Dec 12 '18 at 1:34
Tyler6Tyler6
735414
$endgroup$
I am trying to study for an exam and I am not sure of this solution my professor posted to an exercise. I am given the polynomial quotient ring $mathbb{Z}_6/(x^2+2x)$ and have to find all units and zero divisors.
My initial idea was to do a multiplication table, but with a group this big I am sure there must be another way. For units, I was thinking all polynomials that are coprime to $x$ and $x+2$, but this isn’t what the solutions have.
He lists the units as $1, 5, x+1, 2x+5, 3x+1, 3x+5, 4x+1, 5x+5$, but gives no reason why. I understand these are all coprime to the ideal, but why isn’t $x+3$ an ideal for example?
He then says the zero divisors are everything that isn’t a unit or zero. Why is this? I thought it was possible for an element to be neither a unit nor a zero divisor.
Thank you very much.
abstract-algebra polynomial-rings
abstract-algebra polynomial-rings
asked Dec 12 '18 at 1:34
Tyler6Tyler6
735414
asked Dec 12 '18 at 1:34
Tyler6Tyler6
735414
asked Dec 12 '18 at 1:34
Tyler6Tyler6
735414
asked Dec 12 '18 at 1:34
Tyler6Tyler6
735414
asked Dec 12 '18 at 1:34
Tyler6Tyler6
735414
735414
add a comment |
add a comment |
2 Answers
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Hint: first we can think of elements of $mathbb{Z}_6[x]/(x^2+2x)$ is of the form $a+bx$ where $a,bin mathbb{Z}_6$.
Now $a+bx$ is a unit if and only if there is $c+dx$ such that $(a+bx)(c+dx)=1$.
Note that since $x^2=-2x$ in the ring, we have
begin{align}1=(a+bx)(c+dx)&=ac+(ad+bc)x+bdx^2\&=ac+(ad+bc)x+bd(-2x)\
&=ac+(ad+bc-2bd)x
end{align}
Then you need to solve system of equation $ac=1$ and $ad+bc-2bd=0$.
answered Dec 12 '18 at 1:41
user9077user9077
1,249612
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First, note that this quotient ring is finite (since $x^2+2x$ is monic). In fact it must have $36$ elements.
Now in a finite (commutative) ring $R$, every element is either a unit, zero or a zero-divisor.
Why? Well, for any nonzero element $x$ in $R$, consider the sequence $x^n$ for $nin Bbb{N}$. By pigeonhole $x^n=x^m$ for some $n,m$ with $n<m$. Then $x^nx^{m-n}=x^n$, or $x^n(x^{m-n}-1)=0$. There are two possibilities. Either $x^{m-n}-1=0$ and $x^{m-n}=1$, so $x$ is a unit, or $x^{m-n}-1ne 0$. In this case, let $a$ be the least integer such that $x^a(x^{m-n}-1)=0$. Then $c=x^{a-1}(x^{m-n}-1)ne 0$ but $xc=0$.
For the specifics of finding the actual units, user9077 has already given a good answer, so I won't duplicate that.
answered Dec 12 '18 at 1:46
jgonjgon
16.3k32143
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2 Answers
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2 Answers
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$begingroup$
Hint: first we can think of elements of $mathbb{Z}_6[x]/(x^2+2x)$ is of the form $a+bx$ where $a,bin mathbb{Z}_6$.
Now $a+bx$ is a unit if and only if there is $c+dx$ such that $(a+bx)(c+dx)=1$.
Note that since $x^2=-2x$ in the ring, we have
begin{align}1=(a+bx)(c+dx)&=ac+(ad+bc)x+bdx^2\&=ac+(ad+bc)x+bd(-2x)\
&=ac+(ad+bc-2bd)x
end{align}
Then you need to solve system of equation $ac=1$ and $ad+bc-2bd=0$.
answered Dec 12 '18 at 1:41
user9077user9077
1,249612
$endgroup$
add a comment |
$begingroup$
Hint: first we can think of elements of $mathbb{Z}_6[x]/(x^2+2x)$ is of the form $a+bx$ where $a,bin mathbb{Z}_6$.
Now $a+bx$ is a unit if and only if there is $c+dx$ such that $(a+bx)(c+dx)=1$.
Note that since $x^2=-2x$ in the ring, we have
begin{align}1=(a+bx)(c+dx)&=ac+(ad+bc)x+bdx^2\&=ac+(ad+bc)x+bd(-2x)\
&=ac+(ad+bc-2bd)x
end{align}
Then you need to solve system of equation $ac=1$ and $ad+bc-2bd=0$.
answered Dec 12 '18 at 1:41
user9077user9077
1,249612
$endgroup$
add a comment |
$begingroup$
Hint: first we can think of elements of $mathbb{Z}_6[x]/(x^2+2x)$ is of the form $a+bx$ where $a,bin mathbb{Z}_6$.
Now $a+bx$ is a unit if and only if there is $c+dx$ such that $(a+bx)(c+dx)=1$.
Note that since $x^2=-2x$ in the ring, we have
begin{align}1=(a+bx)(c+dx)&=ac+(ad+bc)x+bdx^2\&=ac+(ad+bc)x+bd(-2x)\
&=ac+(ad+bc-2bd)x
end{align}
Then you need to solve system of equation $ac=1$ and $ad+bc-2bd=0$.
answered Dec 12 '18 at 1:41
user9077user9077
1,249612
$endgroup$
Hint: first we can think of elements of $mathbb{Z}_6[x]/(x^2+2x)$ is of the form $a+bx$ where $a,bin mathbb{Z}_6$.
Now $a+bx$ is a unit if and only if there is $c+dx$ such that $(a+bx)(c+dx)=1$.
Note that since $x^2=-2x$ in the ring, we have
begin{align}1=(a+bx)(c+dx)&=ac+(ad+bc)x+bdx^2\&=ac+(ad+bc)x+bd(-2x)\
&=ac+(ad+bc-2bd)x
end{align}
Then you need to solve system of equation $ac=1$ and $ad+bc-2bd=0$.
answered Dec 12 '18 at 1:41
user9077user9077
1,249612
answered Dec 12 '18 at 1:41
user9077user9077
1,249612
answered Dec 12 '18 at 1:41
user9077user9077
1,249612
answered Dec 12 '18 at 1:41
user9077user9077
1,249612
1,249612
add a comment |
add a comment |
$begingroup$
First, note that this quotient ring is finite (since $x^2+2x$ is monic). In fact it must have $36$ elements.
Now in a finite (commutative) ring $R$, every element is either a unit, zero or a zero-divisor.
Why? Well, for any nonzero element $x$ in $R$, consider the sequence $x^n$ for $nin Bbb{N}$. By pigeonhole $x^n=x^m$ for some $n,m$ with $n<m$. Then $x^nx^{m-n}=x^n$, or $x^n(x^{m-n}-1)=0$. There are two possibilities. Either $x^{m-n}-1=0$ and $x^{m-n}=1$, so $x$ is a unit, or $x^{m-n}-1ne 0$. In this case, let $a$ be the least integer such that $x^a(x^{m-n}-1)=0$. Then $c=x^{a-1}(x^{m-n}-1)ne 0$ but $xc=0$.
For the specifics of finding the actual units, user9077 has already given a good answer, so I won't duplicate that.
answered Dec 12 '18 at 1:46
jgonjgon
16.3k32143
$endgroup$
add a comment |
$begingroup$
First, note that this quotient ring is finite (since $x^2+2x$ is monic). In fact it must have $36$ elements.
Now in a finite (commutative) ring $R$, every element is either a unit, zero or a zero-divisor.
Why? Well, for any nonzero element $x$ in $R$, consider the sequence $x^n$ for $nin Bbb{N}$. By pigeonhole $x^n=x^m$ for some $n,m$ with $n<m$. Then $x^nx^{m-n}=x^n$, or $x^n(x^{m-n}-1)=0$. There are two possibilities. Either $x^{m-n}-1=0$ and $x^{m-n}=1$, so $x$ is a unit, or $x^{m-n}-1ne 0$. In this case, let $a$ be the least integer such that $x^a(x^{m-n}-1)=0$. Then $c=x^{a-1}(x^{m-n}-1)ne 0$ but $xc=0$.
For the specifics of finding the actual units, user9077 has already given a good answer, so I won't duplicate that.
answered Dec 12 '18 at 1:46
jgonjgon
16.3k32143
$endgroup$
add a comment |
$begingroup$
First, note that this quotient ring is finite (since $x^2+2x$ is monic). In fact it must have $36$ elements.
Now in a finite (commutative) ring $R$, every element is either a unit, zero or a zero-divisor.
Why? Well, for any nonzero element $x$ in $R$, consider the sequence $x^n$ for $nin Bbb{N}$. By pigeonhole $x^n=x^m$ for some $n,m$ with $n<m$. Then $x^nx^{m-n}=x^n$, or $x^n(x^{m-n}-1)=0$. There are two possibilities. Either $x^{m-n}-1=0$ and $x^{m-n}=1$, so $x$ is a unit, or $x^{m-n}-1ne 0$. In this case, let $a$ be the least integer such that $x^a(x^{m-n}-1)=0$. Then $c=x^{a-1}(x^{m-n}-1)ne 0$ but $xc=0$.
For the specifics of finding the actual units, user9077 has already given a good answer, so I won't duplicate that.
answered Dec 12 '18 at 1:46
jgonjgon
16.3k32143
$endgroup$
First, note that this quotient ring is finite (since $x^2+2x$ is monic). In fact it must have $36$ elements.
Now in a finite (commutative) ring $R$, every element is either a unit, zero or a zero-divisor.
Why? Well, for any nonzero element $x$ in $R$, consider the sequence $x^n$ for $nin Bbb{N}$. By pigeonhole $x^n=x^m$ for some $n,m$ with $n<m$. Then $x^nx^{m-n}=x^n$, or $x^n(x^{m-n}-1)=0$. There are two possibilities. Either $x^{m-n}-1=0$ and $x^{m-n}=1$, so $x$ is a unit, or $x^{m-n}-1ne 0$. In this case, let $a$ be the least integer such that $x^a(x^{m-n}-1)=0$. Then $c=x^{a-1}(x^{m-n}-1)ne 0$ but $xc=0$.
For the specifics of finding the actual units, user9077 has already given a good answer, so I won't duplicate that.
answered Dec 12 '18 at 1:46
jgonjgon
16.3k32143
answered Dec 12 '18 at 1:46
jgonjgon
16.3k32143
answered Dec 12 '18 at 1:46
jgonjgon
16.3k32143
answered Dec 12 '18 at 1:46
jgonjgon
16.3k32143
16.3k32143
add a comment |
add a comment |
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