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Show that the ideal is finitely generated.
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Suppose $R$ is a commutative ring with unity $1_R$. Let $X= f_i$ where I is some index set and $f_i in R$. I want to show that if the ideal generated by $X$ is all of $R$, then $X$ is actually finitely generated.
My attempt: If I could show that $1_R$ is represented by finitely many elements of $X$, then we're done. If $I$ is finite, then this is trivial. Otherwise, in the infinite case, when we consider an element $r$ is in a so-called "span" of a set, then it is a finite linear combination of these elements- hence set $r = 1_R$. However, this seems a little gimmicky and not a rigorous proof.
abstract-algebra
asked Jul 28 at 22:39
Sean Nemetz
12418
 |Â
show 2 more comments
up vote
0
down vote
favorite
Suppose $R$ is a commutative ring with unity $1_R$. Let $X= f_i$ where I is some index set and $f_i in R$. I want to show that if the ideal generated by $X$ is all of $R$, then $X$ is actually finitely generated.
My attempt: If I could show that $1_R$ is represented by finitely many elements of $X$, then we're done. If $I$ is finite, then this is trivial. Otherwise, in the infinite case, when we consider an element $r$ is in a so-called "span" of a set, then it is a finite linear combination of these elements- hence set $r = 1_R$. However, this seems a little gimmicky and not a rigorous proof.
abstract-algebra
asked Jul 28 at 22:39
Sean Nemetz
12418
Do you mean to say "if the ideal generated by $X$ is all of $R$, then $X$ is finite", as opposed to finitely generated? $X$ is just a set, right?
– Matt
Jul 28 at 22:42
I'm not sure I understand the question. Not every commutative ring with unity is finitely generated; take $R=k[t]$ with your favourite field $k$. Is this what you're trying to do?
– Dave
Jul 28 at 22:43
I might be interpreting the question incorrectly, it's #2 in the link math.tamu.edu/graduate/phd/quals/nalgebra/j12.pdf
– Sean Nemetz
Jul 28 at 22:46
2
Your idea is the right one. If the ideal $I$ generated by $X$ is $R$, then by definition, $1_R$ belongs to $I$, and so must be a finite $R$-linear combination of elements of $X$. (Remember what the definition of the ideal generated by $X$ is - it is the set of all $R$-linear combinations of elements of $X$, where all but finitely many of the coefficients are $0$!) These finitely many elements generating $1$ will be your generators of $I$.
– Alex Wertheim
Jul 28 at 22:50
Exactly what I was thinking Alex! Thank you
– Sean Nemetz
Jul 28 at 22:56
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose $R$ is a commutative ring with unity $1_R$. Let $X= f_i$ where I is some index set and $f_i in R$. I want to show that if the ideal generated by $X$ is all of $R$, then $X$ is actually finitely generated.
My attempt: If I could show that $1_R$ is represented by finitely many elements of $X$, then we're done. If $I$ is finite, then this is trivial. Otherwise, in the infinite case, when we consider an element $r$ is in a so-called "span" of a set, then it is a finite linear combination of these elements- hence set $r = 1_R$. However, this seems a little gimmicky and not a rigorous proof.
abstract-algebra
asked Jul 28 at 22:39
Sean Nemetz
12418
Suppose $R$ is a commutative ring with unity $1_R$. Let $X= f_i$ where I is some index set and $f_i in R$. I want to show that if the ideal generated by $X$ is all of $R$, then $X$ is actually finitely generated.
My attempt: If I could show that $1_R$ is represented by finitely many elements of $X$, then we're done. If $I$ is finite, then this is trivial. Otherwise, in the infinite case, when we consider an element $r$ is in a so-called "span" of a set, then it is a finite linear combination of these elements- hence set $r = 1_R$. However, this seems a little gimmicky and not a rigorous proof.
abstract-algebra
asked Jul 28 at 22:39
Sean Nemetz
12418
asked Jul 28 at 22:39
Sean Nemetz
12418
asked Jul 28 at 22:39
Sean Nemetz
12418
asked Jul 28 at 22:39
Sean Nemetz
12418
12418
Do you mean to say "if the ideal generated by $X$ is all of $R$, then $X$ is finite", as opposed to finitely generated? $X$ is just a set, right?
– Matt
Jul 28 at 22:42
I'm not sure I understand the question. Not every commutative ring with unity is finitely generated; take $R=k[t]$ with your favourite field $k$. Is this what you're trying to do?
– Dave
Jul 28 at 22:43
I might be interpreting the question incorrectly, it's #2 in the link math.tamu.edu/graduate/phd/quals/nalgebra/j12.pdf
– Sean Nemetz
Jul 28 at 22:46
2
Your idea is the right one. If the ideal $I$ generated by $X$ is $R$, then by definition, $1_R$ belongs to $I$, and so must be a finite $R$-linear combination of elements of $X$. (Remember what the definition of the ideal generated by $X$ is - it is the set of all $R$-linear combinations of elements of $X$, where all but finitely many of the coefficients are $0$!) These finitely many elements generating $1$ will be your generators of $I$.
– Alex Wertheim
Jul 28 at 22:50
Exactly what I was thinking Alex! Thank you
– Sean Nemetz
Jul 28 at 22:56
 |Â
show 2 more comments
Do you mean to say "if the ideal generated by $X$ is all of $R$, then $X$ is finite", as opposed to finitely generated? $X$ is just a set, right?
– Matt
Jul 28 at 22:42
I'm not sure I understand the question. Not every commutative ring with unity is finitely generated; take $R=k[t]$ with your favourite field $k$. Is this what you're trying to do?
– Dave
Jul 28 at 22:43
I might be interpreting the question incorrectly, it's #2 in the link math.tamu.edu/graduate/phd/quals/nalgebra/j12.pdf
– Sean Nemetz
Jul 28 at 22:46
2
Your idea is the right one. If the ideal $I$ generated by $X$ is $R$, then by definition, $1_R$ belongs to $I$, and so must be a finite $R$-linear combination of elements of $X$. (Remember what the definition of the ideal generated by $X$ is - it is the set of all $R$-linear combinations of elements of $X$, where all but finitely many of the coefficients are $0$!) These finitely many elements generating $1$ will be your generators of $I$.
– Alex Wertheim
Jul 28 at 22:50
Exactly what I was thinking Alex! Thank you
– Sean Nemetz
Jul 28 at 22:56
Do you mean to say "if the ideal generated by $X$ is all of $R$, then $X$ is finite", as opposed to finitely generated? $X$ is just a set, right?
– Matt
Jul 28 at 22:42
Do you mean to say "if the ideal generated by $X$ is all of $R$, then $X$ is finite", as opposed to finitely generated? $X$ is just a set, right?
– Matt
Jul 28 at 22:42
I'm not sure I understand the question. Not every commutative ring with unity is finitely generated; take $R=k[t]$ with your favourite field $k$. Is this what you're trying to do?
– Dave
Jul 28 at 22:43
I'm not sure I understand the question. Not every commutative ring with unity is finitely generated; take $R=k[t]$ with your favourite field $k$. Is this what you're trying to do?
– Dave
Jul 28 at 22:43
I might be interpreting the question incorrectly, it's #2 in the link math.tamu.edu/graduate/phd/quals/nalgebra/j12.pdf
– Sean Nemetz
Jul 28 at 22:46
I might be interpreting the question incorrectly, it's #2 in the link math.tamu.edu/graduate/phd/quals/nalgebra/j12.pdf
– Sean Nemetz
Jul 28 at 22:46
2
2
Your idea is the right one. If the ideal $I$ generated by $X$ is $R$, then by definition, $1_R$ belongs to $I$, and so must be a finite $R$-linear combination of elements of $X$. (Remember what the definition of the ideal generated by $X$ is - it is the set of all $R$-linear combinations of elements of $X$, where all but finitely many of the coefficients are $0$!) These finitely many elements generating $1$ will be your generators of $I$.
– Alex Wertheim
Jul 28 at 22:50
Your idea is the right one. If the ideal $I$ generated by $X$ is $R$, then by definition, $1_R$ belongs to $I$, and so must be a finite $R$-linear combination of elements of $X$. (Remember what the definition of the ideal generated by $X$ is - it is the set of all $R$-linear combinations of elements of $X$, where all but finitely many of the coefficients are $0$!) These finitely many elements generating $1$ will be your generators of $I$.
– Alex Wertheim
Jul 28 at 22:50
Exactly what I was thinking Alex! Thank you
– Sean Nemetz
Jul 28 at 22:56
Exactly what I was thinking Alex! Thank you
– Sean Nemetz
Jul 28 at 22:56
 |Â
show 2 more comments
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Do you mean to say "if the ideal generated by $X$ is all of $R$, then $X$ is finite", as opposed to finitely generated? $X$ is just a set, right?
– Matt
Jul 28 at 22:42
I'm not sure I understand the question. Not every commutative ring with unity is finitely generated; take $R=k[t]$ with your favourite field $k$. Is this what you're trying to do?
– Dave
Jul 28 at 22:43
I might be interpreting the question incorrectly, it's #2 in the link math.tamu.edu/graduate/phd/quals/nalgebra/j12.pdf
– Sean Nemetz
Jul 28 at 22:46
2
Your idea is the right one. If the ideal $I$ generated by $X$ is $R$, then by definition, $1_R$ belongs to $I$, and so must be a finite $R$-linear combination of elements of $X$. (Remember what the definition of the ideal generated by $X$ is - it is the set of all $R$-linear combinations of elements of $X$, where all but finitely many of the coefficients are $0$!) These finitely many elements generating $1$ will be your generators of $I$.
– Alex Wertheim
Jul 28 at 22:50
Exactly what I was thinking Alex! Thank you
– Sean Nemetz
Jul 28 at 22:56