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Finitely generated R-module is a field iff R is a field? [duplicate]
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Proof that an integral domain that is a finite-dimensional $F$-vector space is in fact a field
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Not sure how to do this one. If $S$ is a field, then I was considering that $exists r_1,ldots, r_nin R$ s.t. $1 = r_1s_1+cdots+r_ns_n$ so for $r = rr_1s_1+cdots+rr_ns_n$ Maybe that is somehow useful for taking inverses of elements.
The assumption that $S$ is an integral domain is necessary because otherwise we could have $S = mathbbZ_p[x]/f(x)$ where $f(x)$ is not irreducible. This is still a finitely generated $mathbbZ_p$ module, but its not a field.
Any hints or solutions would be much appreciated. I feel like this isn't that hard and I'm missing something simple
Source: https://dornsife.usc.edu/assets/sites/363/docs/F17_510ab.pdf
abstract-algebra ring-theory commutative-algebra
asked Aug 12 at 23:44
iYOA
60549
marked as duplicate by rschwieb abstract-algebra
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Aug 13 at 20:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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up vote
6
down vote
favorite
This question already has an answer here:
Proof that an integral domain that is a finite-dimensional $F$-vector space is in fact a field
3 answers
Not sure how to do this one. If $S$ is a field, then I was considering that $exists r_1,ldots, r_nin R$ s.t. $1 = r_1s_1+cdots+r_ns_n$ so for $r = rr_1s_1+cdots+rr_ns_n$ Maybe that is somehow useful for taking inverses of elements.
The assumption that $S$ is an integral domain is necessary because otherwise we could have $S = mathbbZ_p[x]/f(x)$ where $f(x)$ is not irreducible. This is still a finitely generated $mathbbZ_p$ module, but its not a field.
Any hints or solutions would be much appreciated. I feel like this isn't that hard and I'm missing something simple
Source: https://dornsife.usc.edu/assets/sites/363/docs/F17_510ab.pdf
abstract-algebra ring-theory commutative-algebra
asked Aug 12 at 23:44
iYOA
60549
marked as duplicate by rschwieb abstract-algebra
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Aug 13 at 20:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
See Atiyah–Macdonald, Prop 5.1 and 5.7.
– lhf
Aug 13 at 0:11
This is one of the most duplicated ring-theory questions on the site that I know of.
– rschwieb
Aug 13 at 20:12
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up vote
6
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up vote
6
down vote
favorite
This question already has an answer here:
Proof that an integral domain that is a finite-dimensional $F$-vector space is in fact a field
3 answers
Not sure how to do this one. If $S$ is a field, then I was considering that $exists r_1,ldots, r_nin R$ s.t. $1 = r_1s_1+cdots+r_ns_n$ so for $r = rr_1s_1+cdots+rr_ns_n$ Maybe that is somehow useful for taking inverses of elements.
The assumption that $S$ is an integral domain is necessary because otherwise we could have $S = mathbbZ_p[x]/f(x)$ where $f(x)$ is not irreducible. This is still a finitely generated $mathbbZ_p$ module, but its not a field.
Any hints or solutions would be much appreciated. I feel like this isn't that hard and I'm missing something simple
Source: https://dornsife.usc.edu/assets/sites/363/docs/F17_510ab.pdf
abstract-algebra ring-theory commutative-algebra
asked Aug 12 at 23:44
iYOA
60549
This question already has an answer here:
Proof that an integral domain that is a finite-dimensional $F$-vector space is in fact a field
3 answers
Not sure how to do this one. If $S$ is a field, then I was considering that $exists r_1,ldots, r_nin R$ s.t. $1 = r_1s_1+cdots+r_ns_n$ so for $r = rr_1s_1+cdots+rr_ns_n$ Maybe that is somehow useful for taking inverses of elements.
The assumption that $S$ is an integral domain is necessary because otherwise we could have $S = mathbbZ_p[x]/f(x)$ where $f(x)$ is not irreducible. This is still a finitely generated $mathbbZ_p$ module, but its not a field.
Any hints or solutions would be much appreciated. I feel like this isn't that hard and I'm missing something simple
Source: https://dornsife.usc.edu/assets/sites/363/docs/F17_510ab.pdf
This question already has an answer here:
Proof that an integral domain that is a finite-dimensional $F$-vector space is in fact a field
3 answers
abstract-algebra ring-theory commutative-algebra
asked Aug 12 at 23:44
iYOA
60549
asked Aug 12 at 23:44
iYOA
60549
asked Aug 12 at 23:44
iYOA
60549
asked Aug 12 at 23:44
iYOA
60549
60549
marked as duplicate by rschwieb abstract-algebra
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Aug 13 at 20:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
See Atiyah–Macdonald, Prop 5.1 and 5.7.
– lhf
Aug 13 at 0:11
This is one of the most duplicated ring-theory questions on the site that I know of.
– rschwieb
Aug 13 at 20:12
add a comment |Â
See Atiyah–Macdonald, Prop 5.1 and 5.7.
– lhf
Aug 13 at 0:11
This is one of the most duplicated ring-theory questions on the site that I know of.
– rschwieb
Aug 13 at 20:12
See Atiyah–Macdonald, Prop 5.1 and 5.7.
– lhf
Aug 13 at 0:11
See Atiyah–Macdonald, Prop 5.1 and 5.7.
– lhf
Aug 13 at 0:11
This is one of the most duplicated ring-theory questions on the site that I know of.
– rschwieb
Aug 13 at 20:12
This is one of the most duplicated ring-theory questions on the site that I know of.
– rschwieb
Aug 13 at 20:12
add a comment |Â
2 Answers
2
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up vote
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accepted
Uncover the spoilers for solutions completing the hints below:
Suppose $R$ is a field, and let $s in S$ be a nonzero element. Then multiplication by $s$ is an $R$-linear endomorphism of $S$, which is injective since $s$ is nonzero and $S$ is a domain.
Since $S$ is a finite-dimensional $R$-vector space, it follows that multiplication by $s$ is also surjective, and so $1$ is in the image of this map.
Suppose $S$ is a field. This solution I have in mind for this direction is a bit trickier. Let $r in R$ be nonzero, and let $s$ be the inverse to $r$ in $S$. As before, consider the $R$-linear map $varphi_s colon S to S$ corresponding to multiplication by $S$. Since $S$ is a finitely generated $R$-module, $varphi_s$ satisfies a monic polynomial relation with coefficients in $R$ by Cayley-Hamilton. That is, there exist $r_1, ldots, r_n in R$ such that multiplication by $s^n+r_1s^n-1 + cdots +r_n$ is the zero element of $mathrmEnd_R(S)$.
Since $S$ is a faithful $R$-module ($R$ is a subring of $S$,and so contains $1$), this implies that $s^n+r_1s^n-1 + cdots +r_n = 0$. Now multiply both sides by $r^n-1$ to conclude that $s in R$.
answered Aug 13 at 0:12
Alex Wertheim
15.7k22848
add a comment |Â
up vote
3
down vote
Hints:
$Rightarrow$: If $R$ is a field, let $sin S$, and consider multiplication by $s$ in $S$. Check this is an injective $R$-linear map. What can you conclude, knowing $S$ is a finite dimensional $R$-vector space?
$Leftarrow$: If $S$ is a field, consider $rin R$; you know $r^-1in S$, hence it is a root of a monic polynomial in $R[X]$. Deduce from this polynomial equation that $r^-1$ is a polynomial in $r$, hence it belongs to $R$.
answered Aug 13 at 0:16
Bernard
111k635103
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Uncover the spoilers for solutions completing the hints below:
Suppose $R$ is a field, and let $s in S$ be a nonzero element. Then multiplication by $s$ is an $R$-linear endomorphism of $S$, which is injective since $s$ is nonzero and $S$ is a domain.
Since $S$ is a finite-dimensional $R$-vector space, it follows that multiplication by $s$ is also surjective, and so $1$ is in the image of this map.
Suppose $S$ is a field. This solution I have in mind for this direction is a bit trickier. Let $r in R$ be nonzero, and let $s$ be the inverse to $r$ in $S$. As before, consider the $R$-linear map $varphi_s colon S to S$ corresponding to multiplication by $S$. Since $S$ is a finitely generated $R$-module, $varphi_s$ satisfies a monic polynomial relation with coefficients in $R$ by Cayley-Hamilton. That is, there exist $r_1, ldots, r_n in R$ such that multiplication by $s^n+r_1s^n-1 + cdots +r_n$ is the zero element of $mathrmEnd_R(S)$.
Since $S$ is a faithful $R$-module ($R$ is a subring of $S$,and so contains $1$), this implies that $s^n+r_1s^n-1 + cdots +r_n = 0$. Now multiply both sides by $r^n-1$ to conclude that $s in R$.
answered Aug 13 at 0:12
Alex Wertheim
15.7k22848
add a comment |Â
up vote
5
down vote
accepted
Uncover the spoilers for solutions completing the hints below:
Suppose $R$ is a field, and let $s in S$ be a nonzero element. Then multiplication by $s$ is an $R$-linear endomorphism of $S$, which is injective since $s$ is nonzero and $S$ is a domain.
Since $S$ is a finite-dimensional $R$-vector space, it follows that multiplication by $s$ is also surjective, and so $1$ is in the image of this map.
Suppose $S$ is a field. This solution I have in mind for this direction is a bit trickier. Let $r in R$ be nonzero, and let $s$ be the inverse to $r$ in $S$. As before, consider the $R$-linear map $varphi_s colon S to S$ corresponding to multiplication by $S$. Since $S$ is a finitely generated $R$-module, $varphi_s$ satisfies a monic polynomial relation with coefficients in $R$ by Cayley-Hamilton. That is, there exist $r_1, ldots, r_n in R$ such that multiplication by $s^n+r_1s^n-1 + cdots +r_n$ is the zero element of $mathrmEnd_R(S)$.
Since $S$ is a faithful $R$-module ($R$ is a subring of $S$,and so contains $1$), this implies that $s^n+r_1s^n-1 + cdots +r_n = 0$. Now multiply both sides by $r^n-1$ to conclude that $s in R$.
answered Aug 13 at 0:12
Alex Wertheim
15.7k22848
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Uncover the spoilers for solutions completing the hints below:
Suppose $R$ is a field, and let $s in S$ be a nonzero element. Then multiplication by $s$ is an $R$-linear endomorphism of $S$, which is injective since $s$ is nonzero and $S$ is a domain.
Since $S$ is a finite-dimensional $R$-vector space, it follows that multiplication by $s$ is also surjective, and so $1$ is in the image of this map.
Suppose $S$ is a field. This solution I have in mind for this direction is a bit trickier. Let $r in R$ be nonzero, and let $s$ be the inverse to $r$ in $S$. As before, consider the $R$-linear map $varphi_s colon S to S$ corresponding to multiplication by $S$. Since $S$ is a finitely generated $R$-module, $varphi_s$ satisfies a monic polynomial relation with coefficients in $R$ by Cayley-Hamilton. That is, there exist $r_1, ldots, r_n in R$ such that multiplication by $s^n+r_1s^n-1 + cdots +r_n$ is the zero element of $mathrmEnd_R(S)$.
Since $S$ is a faithful $R$-module ($R$ is a subring of $S$,and so contains $1$), this implies that $s^n+r_1s^n-1 + cdots +r_n = 0$. Now multiply both sides by $r^n-1$ to conclude that $s in R$.
answered Aug 13 at 0:12
Alex Wertheim
15.7k22848
Uncover the spoilers for solutions completing the hints below:
Suppose $R$ is a field, and let $s in S$ be a nonzero element. Then multiplication by $s$ is an $R$-linear endomorphism of $S$, which is injective since $s$ is nonzero and $S$ is a domain.
Since $S$ is a finite-dimensional $R$-vector space, it follows that multiplication by $s$ is also surjective, and so $1$ is in the image of this map.
Suppose $S$ is a field. This solution I have in mind for this direction is a bit trickier. Let $r in R$ be nonzero, and let $s$ be the inverse to $r$ in $S$. As before, consider the $R$-linear map $varphi_s colon S to S$ corresponding to multiplication by $S$. Since $S$ is a finitely generated $R$-module, $varphi_s$ satisfies a monic polynomial relation with coefficients in $R$ by Cayley-Hamilton. That is, there exist $r_1, ldots, r_n in R$ such that multiplication by $s^n+r_1s^n-1 + cdots +r_n$ is the zero element of $mathrmEnd_R(S)$.
Since $S$ is a faithful $R$-module ($R$ is a subring of $S$,and so contains $1$), this implies that $s^n+r_1s^n-1 + cdots +r_n = 0$. Now multiply both sides by $r^n-1$ to conclude that $s in R$.
answered Aug 13 at 0:12
Alex Wertheim
15.7k22848
edited Aug 13 at 0:27
answered Aug 13 at 0:12
Alex Wertheim
15.7k22848
answered Aug 13 at 0:12
Alex Wertheim
15.7k22848
answered Aug 13 at 0:12
Alex Wertheim
15.7k22848
15.7k22848
add a comment |Â
add a comment |Â
up vote
3
down vote
Hints:
$Rightarrow$: If $R$ is a field, let $sin S$, and consider multiplication by $s$ in $S$. Check this is an injective $R$-linear map. What can you conclude, knowing $S$ is a finite dimensional $R$-vector space?
$Leftarrow$: If $S$ is a field, consider $rin R$; you know $r^-1in S$, hence it is a root of a monic polynomial in $R[X]$. Deduce from this polynomial equation that $r^-1$ is a polynomial in $r$, hence it belongs to $R$.
answered Aug 13 at 0:16
Bernard
111k635103
add a comment |Â
up vote
3
down vote
Hints:
$Rightarrow$: If $R$ is a field, let $sin S$, and consider multiplication by $s$ in $S$. Check this is an injective $R$-linear map. What can you conclude, knowing $S$ is a finite dimensional $R$-vector space?
$Leftarrow$: If $S$ is a field, consider $rin R$; you know $r^-1in S$, hence it is a root of a monic polynomial in $R[X]$. Deduce from this polynomial equation that $r^-1$ is a polynomial in $r$, hence it belongs to $R$.
answered Aug 13 at 0:16
Bernard
111k635103
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Hints:
$Rightarrow$: If $R$ is a field, let $sin S$, and consider multiplication by $s$ in $S$. Check this is an injective $R$-linear map. What can you conclude, knowing $S$ is a finite dimensional $R$-vector space?
$Leftarrow$: If $S$ is a field, consider $rin R$; you know $r^-1in S$, hence it is a root of a monic polynomial in $R[X]$. Deduce from this polynomial equation that $r^-1$ is a polynomial in $r$, hence it belongs to $R$.
answered Aug 13 at 0:16
Bernard
111k635103
Hints:
$Rightarrow$: If $R$ is a field, let $sin S$, and consider multiplication by $s$ in $S$. Check this is an injective $R$-linear map. What can you conclude, knowing $S$ is a finite dimensional $R$-vector space?
$Leftarrow$: If $S$ is a field, consider $rin R$; you know $r^-1in S$, hence it is a root of a monic polynomial in $R[X]$. Deduce from this polynomial equation that $r^-1$ is a polynomial in $r$, hence it belongs to $R$.
answered Aug 13 at 0:16
Bernard
111k635103
answered Aug 13 at 0:16
Bernard
111k635103
answered Aug 13 at 0:16
Bernard
111k635103
answered Aug 13 at 0:16
Bernard
111k635103
111k635103
add a comment |Â
add a comment |Â
See Atiyah–Macdonald, Prop 5.1 and 5.7.
– lhf
Aug 13 at 0:11
This is one of the most duplicated ring-theory questions on the site that I know of.
– rschwieb
Aug 13 at 20:12