Mixing
the ring obtained from integer ring adjoining element
$begingroup$
Here is the problem from Algebra ,Artin : Determine the structure of the ring R' obtained from Z by adjoining element α satisfying relation 2α＝6
I've known what its elements is like. But I don't know how to analyze the structure of the ring obtained by adjoining an element which satisfies a nonmonic polynomial relation and is not the inverse of any element of the original ring. I want a sample to show me how to analyze such a problem. Thankyou.
abstract-algebra
asked Jan 22 at 11:05
JR ChanJR Chan
242
$endgroup$
add a comment |
$begingroup$
Here is the problem from Algebra ,Artin : Determine the structure of the ring R' obtained from Z by adjoining element α satisfying relation 2α＝6
I've known what its elements is like. But I don't know how to analyze the structure of the ring obtained by adjoining an element which satisfies a nonmonic polynomial relation and is not the inverse of any element of the original ring. I want a sample to show me how to analyze such a problem. Thankyou.
abstract-algebra
asked Jan 22 at 11:05
JR ChanJR Chan
242
$endgroup$
$begingroup$
Maybe you missed the meaning of the exercise. It asks to determine the structure of the ring "obtained from $mathbb Z$ by adjoining an element $alpha$ satisfying each set of relations". In your case there is another relation, $6alpha=15$. So you have two relations, not one!
$endgroup$
– user26857
Jan 22 at 20:59
add a comment |
$begingroup$
Here is the problem from Algebra ,Artin : Determine the structure of the ring R' obtained from Z by adjoining element α satisfying relation 2α＝6
I've known what its elements is like. But I don't know how to analyze the structure of the ring obtained by adjoining an element which satisfies a nonmonic polynomial relation and is not the inverse of any element of the original ring. I want a sample to show me how to analyze such a problem. Thankyou.
abstract-algebra
asked Jan 22 at 11:05
JR ChanJR Chan
242
$endgroup$
Here is the problem from Algebra ,Artin : Determine the structure of the ring R' obtained from Z by adjoining element α satisfying relation 2α＝6
I've known what its elements is like. But I don't know how to analyze the structure of the ring obtained by adjoining an element which satisfies a nonmonic polynomial relation and is not the inverse of any element of the original ring. I want a sample to show me how to analyze such a problem. Thankyou.
abstract-algebra
abstract-algebra
asked Jan 22 at 11:05
JR ChanJR Chan
242
asked Jan 22 at 11:05
JR ChanJR Chan
242
asked Jan 22 at 11:05
JR ChanJR Chan
242
asked Jan 22 at 11:05
JR ChanJR Chan
242
asked Jan 22 at 11:05
JR ChanJR Chan
242
242
$begingroup$
Maybe you missed the meaning of the exercise. It asks to determine the structure of the ring "obtained from $mathbb Z$ by adjoining an element $alpha$ satisfying each set of relations". In your case there is another relation, $6alpha=15$. So you have two relations, not one!
$endgroup$
– user26857
Jan 22 at 20:59
add a comment |
$begingroup$
Maybe you missed the meaning of the exercise. It asks to determine the structure of the ring "obtained from $mathbb Z$ by adjoining an element $alpha$ satisfying each set of relations". In your case there is another relation, $6alpha=15$. So you have two relations, not one!
$endgroup$
– user26857
Jan 22 at 20:59
$begingroup$
Maybe you missed the meaning of the exercise. It asks to determine the structure of the ring "obtained from $mathbb Z$ by adjoining an element $alpha$ satisfying each set of relations". In your case there is another relation, $6alpha=15$. So you have two relations, not one!
$endgroup$
– user26857
Jan 22 at 20:59
$begingroup$
Maybe you missed the meaning of the exercise. It asks to determine the structure of the ring "obtained from $mathbb Z$ by adjoining an element $alpha$ satisfying each set of relations". In your case there is another relation, $6alpha=15$. So you have two relations, not one!
$endgroup$
– user26857
Jan 22 at 20:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Generally speaking, in an integral domain you can cancel nonzero elements even if they are not invertible: If $ane 0$ then $acdot b = acdot c implies a(b-c)=0 implies b=c$.
Adding algebraic elements to an integral domain gives you a subring of the algebraic closure of the field of fractions of the domain, and a subring of a field is always an integral domain, so $mathbb{Z}[alpha]$ is an integral domain, and you can apply this to get that $alpha = 3$, i.e. $mathbb{Z}[alpha]=mathbb{Z}$ (and the same can be said whenever you adjoint a root of a monic linear polynomial, because you again find out that you've "adjoined" an element already in the ring).
answered Jan 22 at 11:22
Shai DesheShai Deshe
921513
$endgroup$
$begingroup$
$Bbb Z[x]/(2x-6),$ is not a domain since $, 2(x-3) = 0,$ but both factors are nonzero.
$endgroup$
– Bill Dubuque
Jan 22 at 17:54
add a comment |
$begingroup$
The first answer is valid under a natural but not explicitly stated assumption that both $mathbb Z$ and $a$ are contained in a field or an integral domain.
However, without this assumption $mathbb Z$ can be considered as a subring of $mathbb Zoplusmathbb Z_2$ (identifying $n$ with $(n,0)$) and for $a=(0,1)$ we have $mathbb Z[a]=mathbb Zoplusmathbb Z_2$.
So the problem as stated does not have a unique solution.
answered Jan 22 at 12:38
Oleg SmirnovOleg Smirnov
658
$endgroup$
add a comment |
Your Answer
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2 Answers
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
Generally speaking, in an integral domain you can cancel nonzero elements even if they are not invertible: If $ane 0$ then $acdot b = acdot c implies a(b-c)=0 implies b=c$.
Adding algebraic elements to an integral domain gives you a subring of the algebraic closure of the field of fractions of the domain, and a subring of a field is always an integral domain, so $mathbb{Z}[alpha]$ is an integral domain, and you can apply this to get that $alpha = 3$, i.e. $mathbb{Z}[alpha]=mathbb{Z}$ (and the same can be said whenever you adjoint a root of a monic linear polynomial, because you again find out that you've "adjoined" an element already in the ring).
answered Jan 22 at 11:22
Shai DesheShai Deshe
921513
$endgroup$
$begingroup$
$Bbb Z[x]/(2x-6),$ is not a domain since $, 2(x-3) = 0,$ but both factors are nonzero.
$endgroup$
– Bill Dubuque
Jan 22 at 17:54
add a comment |
$begingroup$
Generally speaking, in an integral domain you can cancel nonzero elements even if they are not invertible: If $ane 0$ then $acdot b = acdot c implies a(b-c)=0 implies b=c$.
Adding algebraic elements to an integral domain gives you a subring of the algebraic closure of the field of fractions of the domain, and a subring of a field is always an integral domain, so $mathbb{Z}[alpha]$ is an integral domain, and you can apply this to get that $alpha = 3$, i.e. $mathbb{Z}[alpha]=mathbb{Z}$ (and the same can be said whenever you adjoint a root of a monic linear polynomial, because you again find out that you've "adjoined" an element already in the ring).
answered Jan 22 at 11:22
Shai DesheShai Deshe
921513
$endgroup$
$begingroup$
$Bbb Z[x]/(2x-6),$ is not a domain since $, 2(x-3) = 0,$ but both factors are nonzero.
$endgroup$
– Bill Dubuque
Jan 22 at 17:54
add a comment |
$begingroup$
Generally speaking, in an integral domain you can cancel nonzero elements even if they are not invertible: If $ane 0$ then $acdot b = acdot c implies a(b-c)=0 implies b=c$.
Adding algebraic elements to an integral domain gives you a subring of the algebraic closure of the field of fractions of the domain, and a subring of a field is always an integral domain, so $mathbb{Z}[alpha]$ is an integral domain, and you can apply this to get that $alpha = 3$, i.e. $mathbb{Z}[alpha]=mathbb{Z}$ (and the same can be said whenever you adjoint a root of a monic linear polynomial, because you again find out that you've "adjoined" an element already in the ring).
answered Jan 22 at 11:22
Shai DesheShai Deshe
921513
$endgroup$
Generally speaking, in an integral domain you can cancel nonzero elements even if they are not invertible: If $ane 0$ then $acdot b = acdot c implies a(b-c)=0 implies b=c$.
Adding algebraic elements to an integral domain gives you a subring of the algebraic closure of the field of fractions of the domain, and a subring of a field is always an integral domain, so $mathbb{Z}[alpha]$ is an integral domain, and you can apply this to get that $alpha = 3$, i.e. $mathbb{Z}[alpha]=mathbb{Z}$ (and the same can be said whenever you adjoint a root of a monic linear polynomial, because you again find out that you've "adjoined" an element already in the ring).
answered Jan 22 at 11:22
Shai DesheShai Deshe
921513
answered Jan 22 at 11:22
Shai DesheShai Deshe
921513
answered Jan 22 at 11:22
Shai DesheShai Deshe
921513
answered Jan 22 at 11:22
Shai DesheShai Deshe
921513
921513
$begingroup$
$Bbb Z[x]/(2x-6),$ is not a domain since $, 2(x-3) = 0,$ but both factors are nonzero.
$endgroup$
– Bill Dubuque
Jan 22 at 17:54
add a comment |
$begingroup$
$Bbb Z[x]/(2x-6),$ is not a domain since $, 2(x-3) = 0,$ but both factors are nonzero.
$endgroup$
– Bill Dubuque
Jan 22 at 17:54
$begingroup$
$Bbb Z[x]/(2x-6),$ is not a domain since $, 2(x-3) = 0,$ but both factors are nonzero.
$endgroup$
– Bill Dubuque
Jan 22 at 17:54
$begingroup$
$Bbb Z[x]/(2x-6),$ is not a domain since $, 2(x-3) = 0,$ but both factors are nonzero.
$endgroup$
– Bill Dubuque
Jan 22 at 17:54
add a comment |
$begingroup$
The first answer is valid under a natural but not explicitly stated assumption that both $mathbb Z$ and $a$ are contained in a field or an integral domain.
However, without this assumption $mathbb Z$ can be considered as a subring of $mathbb Zoplusmathbb Z_2$ (identifying $n$ with $(n,0)$) and for $a=(0,1)$ we have $mathbb Z[a]=mathbb Zoplusmathbb Z_2$.
So the problem as stated does not have a unique solution.
answered Jan 22 at 12:38
Oleg SmirnovOleg Smirnov
658
$endgroup$
add a comment |
$begingroup$
The first answer is valid under a natural but not explicitly stated assumption that both $mathbb Z$ and $a$ are contained in a field or an integral domain.
However, without this assumption $mathbb Z$ can be considered as a subring of $mathbb Zoplusmathbb Z_2$ (identifying $n$ with $(n,0)$) and for $a=(0,1)$ we have $mathbb Z[a]=mathbb Zoplusmathbb Z_2$.
So the problem as stated does not have a unique solution.
answered Jan 22 at 12:38
Oleg SmirnovOleg Smirnov
658
$endgroup$
add a comment |
$begingroup$
The first answer is valid under a natural but not explicitly stated assumption that both $mathbb Z$ and $a$ are contained in a field or an integral domain.
However, without this assumption $mathbb Z$ can be considered as a subring of $mathbb Zoplusmathbb Z_2$ (identifying $n$ with $(n,0)$) and for $a=(0,1)$ we have $mathbb Z[a]=mathbb Zoplusmathbb Z_2$.
So the problem as stated does not have a unique solution.
answered Jan 22 at 12:38
Oleg SmirnovOleg Smirnov
658
$endgroup$
The first answer is valid under a natural but not explicitly stated assumption that both $mathbb Z$ and $a$ are contained in a field or an integral domain.
However, without this assumption $mathbb Z$ can be considered as a subring of $mathbb Zoplusmathbb Z_2$ (identifying $n$ with $(n,0)$) and for $a=(0,1)$ we have $mathbb Z[a]=mathbb Zoplusmathbb Z_2$.
So the problem as stated does not have a unique solution.
answered Jan 22 at 12:38
Oleg SmirnovOleg Smirnov
658
edited Jan 22 at 13:13
answered Jan 22 at 12:38
Oleg SmirnovOleg Smirnov
658
answered Jan 22 at 12:38
Oleg SmirnovOleg Smirnov
658
answered Jan 22 at 12:38
Oleg SmirnovOleg Smirnov
658
658
add a comment |
add a comment |
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$begingroup$
Maybe you missed the meaning of the exercise. It asks to determine the structure of the ring "obtained from $mathbb Z$ by adjoining an element $alpha$ satisfying each set of relations". In your case there is another relation, $6alpha=15$. So you have two relations, not one!
$endgroup$
– user26857
Jan 22 at 20:59