Determine the maximal ideals of $mathbb R^2$ by noting $mathbb R^2 cong mathbb R[x]/(x^2-1)$
$begingroup$
This is a letter of an exericse in Artin Algebra and has been asked and answered here as well as by Brian Bi here and Takumi Murayama here. I have a different approach.
An earlier exercise is to prove $mathbb R^2 cong mathbb R[x]/(x^2-1)$. One way to do that is with the surjective homomorphism $varphi:mathbb R[x] to mathbb R^2$ defined by $varphi(x)=(1,-1)$. This solution is in the answer here and in the answer by Takumi Murayama.
The maximal ideals in $mathbb R[x]/(x^2-1)$ include the principal ideals $([x-1+(x^2-1)])$ and $([x+1+(x^2-1)])$.
The corresponding maximal ideals in $mathbb R^2$ are principal ideals generated by
$$overlinevarphi([x-1+(x^2-1)])=(overlinevarphi circ pi) (x-1) = varphi(x-1)=(0,-2)$$
and
$$overlinevarphi([x+1+(x^2-1)])=(overlinevarphi circ pi) (x+1) = varphi(x+1)=(2,0)$$
where $overline varphi$ is the isomorphism from $R[x]/(x^2-1) to mathbb R^2$.
The principal ideal generated by $(0,pm 2)$ is the same as the one generated by $(0,1)$. Also, the one by $(pm 2,0)$ is the same as the one by $(1,0)$.
We know that $((0,1))=0times mathbb R, ((1,0))=mathbb R times 0$.
After proving that $([x-1+(x^2-1)])$ and $([x+1+(x^2-1)])$ are the only maximal ideals in the quotient ring, we can say $((0,1))=0times mathbb R, ((1,0))=mathbb R times 0$ are the only maximal ideals in $mathbb R^2$.
Is this correct?
To recap, I answered the maximal ideals of $mathbb R[x]/(x^2-1)$ instead of for $mathbb R^2$. Then I converted the ideals from $mathbb R[x]/(x^2-1)$ to $mathbb R^2$.
abstract-algebra proof-verification ring-theory ideals maximal-and-prime-ideals
edited Dec 26 '18 at 21:15
user26857
39.3k124183
$endgroup$
add a comment |
$begingroup$
This is a letter of an exericse in Artin Algebra and has been asked and answered here as well as by Brian Bi here and Takumi Murayama here. I have a different approach.
An earlier exercise is to prove $mathbb R^2 cong mathbb R[x]/(x^2-1)$. One way to do that is with the surjective homomorphism $varphi:mathbb R[x] to mathbb R^2$ defined by $varphi(x)=(1,-1)$. This solution is in the answer here and in the answer by Takumi Murayama.
The maximal ideals in $mathbb R[x]/(x^2-1)$ include the principal ideals $([x-1+(x^2-1)])$ and $([x+1+(x^2-1)])$.
The corresponding maximal ideals in $mathbb R^2$ are principal ideals generated by
$$overlinevarphi([x-1+(x^2-1)])=(overlinevarphi circ pi) (x-1) = varphi(x-1)=(0,-2)$$
and
$$overlinevarphi([x+1+(x^2-1)])=(overlinevarphi circ pi) (x+1) = varphi(x+1)=(2,0)$$
where $overline varphi$ is the isomorphism from $R[x]/(x^2-1) to mathbb R^2$.
The principal ideal generated by $(0,pm 2)$ is the same as the one generated by $(0,1)$. Also, the one by $(pm 2,0)$ is the same as the one by $(1,0)$.
We know that $((0,1))=0times mathbb R, ((1,0))=mathbb R times 0$.
After proving that $([x-1+(x^2-1)])$ and $([x+1+(x^2-1)])$ are the only maximal ideals in the quotient ring, we can say $((0,1))=0times mathbb R, ((1,0))=mathbb R times 0$ are the only maximal ideals in $mathbb R^2$.
Is this correct?
To recap, I answered the maximal ideals of $mathbb R[x]/(x^2-1)$ instead of for $mathbb R^2$. Then I converted the ideals from $mathbb R[x]/(x^2-1)$ to $mathbb R^2$.
abstract-algebra proof-verification ring-theory ideals maximal-and-prime-ideals
edited Dec 26 '18 at 21:15
user26857
39.3k124183
$endgroup$
add a comment |
$begingroup$
This is a letter of an exericse in Artin Algebra and has been asked and answered here as well as by Brian Bi here and Takumi Murayama here. I have a different approach.
An earlier exercise is to prove $mathbb R^2 cong mathbb R[x]/(x^2-1)$. One way to do that is with the surjective homomorphism $varphi:mathbb R[x] to mathbb R^2$ defined by $varphi(x)=(1,-1)$. This solution is in the answer here and in the answer by Takumi Murayama.
The maximal ideals in $mathbb R[x]/(x^2-1)$ include the principal ideals $([x-1+(x^2-1)])$ and $([x+1+(x^2-1)])$.
The corresponding maximal ideals in $mathbb R^2$ are principal ideals generated by
$$overlinevarphi([x-1+(x^2-1)])=(overlinevarphi circ pi) (x-1) = varphi(x-1)=(0,-2)$$
and
$$overlinevarphi([x+1+(x^2-1)])=(overlinevarphi circ pi) (x+1) = varphi(x+1)=(2,0)$$
where $overline varphi$ is the isomorphism from $R[x]/(x^2-1) to mathbb R^2$.
The principal ideal generated by $(0,pm 2)$ is the same as the one generated by $(0,1)$. Also, the one by $(pm 2,0)$ is the same as the one by $(1,0)$.
We know that $((0,1))=0times mathbb R, ((1,0))=mathbb R times 0$.
After proving that $([x-1+(x^2-1)])$ and $([x+1+(x^2-1)])$ are the only maximal ideals in the quotient ring, we can say $((0,1))=0times mathbb R, ((1,0))=mathbb R times 0$ are the only maximal ideals in $mathbb R^2$.
Is this correct?
To recap, I answered the maximal ideals of $mathbb R[x]/(x^2-1)$ instead of for $mathbb R^2$. Then I converted the ideals from $mathbb R[x]/(x^2-1)$ to $mathbb R^2$.
abstract-algebra proof-verification ring-theory ideals maximal-and-prime-ideals
edited Dec 26 '18 at 21:15
user26857
39.3k124183
$endgroup$
This is a letter of an exericse in Artin Algebra and has been asked and answered here as well as by Brian Bi here and Takumi Murayama here. I have a different approach.
An earlier exercise is to prove $mathbb R^2 cong mathbb R[x]/(x^2-1)$. One way to do that is with the surjective homomorphism $varphi:mathbb R[x] to mathbb R^2$ defined by $varphi(x)=(1,-1)$. This solution is in the answer here and in the answer by Takumi Murayama.
The maximal ideals in $mathbb R[x]/(x^2-1)$ include the principal ideals $([x-1+(x^2-1)])$ and $([x+1+(x^2-1)])$.
The corresponding maximal ideals in $mathbb R^2$ are principal ideals generated by
$$overlinevarphi([x-1+(x^2-1)])=(overlinevarphi circ pi) (x-1) = varphi(x-1)=(0,-2)$$
and
$$overlinevarphi([x+1+(x^2-1)])=(overlinevarphi circ pi) (x+1) = varphi(x+1)=(2,0)$$
where $overline varphi$ is the isomorphism from $R[x]/(x^2-1) to mathbb R^2$.
The principal ideal generated by $(0,pm 2)$ is the same as the one generated by $(0,1)$. Also, the one by $(pm 2,0)$ is the same as the one by $(1,0)$.
We know that $((0,1))=0times mathbb R, ((1,0))=mathbb R times 0$.
After proving that $([x-1+(x^2-1)])$ and $([x+1+(x^2-1)])$ are the only maximal ideals in the quotient ring, we can say $((0,1))=0times mathbb R, ((1,0))=mathbb R times 0$ are the only maximal ideals in $mathbb R^2$.
Is this correct?
To recap, I answered the maximal ideals of $mathbb R[x]/(x^2-1)$ instead of for $mathbb R^2$. Then I converted the ideals from $mathbb R[x]/(x^2-1)$ to $mathbb R^2$.
abstract-algebra proof-verification ring-theory ideals maximal-and-prime-ideals
abstract-algebra proof-verification ring-theory ideals maximal-and-prime-ideals
edited Dec 26 '18 at 21:15
user26857
39.3k124183
edited Dec 26 '18 at 21:15
user26857
39.3k124183
edited Dec 26 '18 at 21:15
user26857
39.3k124183
edited Dec 26 '18 at 21:15
user26857
39.3k124183
edited Dec 26 '18 at 21:15
user26857
39.3k124183
39.3k124183
asked Dec 22 '18 at 9:03
user198044
add a comment |
add a comment |
1 Answer
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$begingroup$
Yes, your argument is fine (modulo that you still need to show that $(x+1)$ and $(x-1)$ are the only maximal ideals of $Bbb R[x]/(x^2 - 1)$). If you have an isomorphism $f : Rto S,$ then the maximal ideals of $S$ are precisely the images via $f$ of maximal ideals of $R.$
I only have two comments: you can probably remove some of the steps and some of the notation, but if you really want to spell everything out, that's fine. Secondly, there's a minor typo in step 2: $overline{varphi}circpi(x+1) = varphi(x + 1),$ not $varphi(x-1).$
Another comment not related to the correctness of the proof: as in your first link, it is also not too difficult to argue directly that if you have a finite product of fields $A = k_1timesdotstimes k_n,$ the maximal ideals are precisely the ideals $$mathfrak{m}_i := {(x_1,dots, x_n)in Amid x_i = 0},$$
and you could generalize without too much trouble to a direct product of finitely many commutative rings. For infinite products of fields, the answer is stranger, as you might expect.
answered Dec 22 '18 at 17:10
StahlStahl
16.7k43455
$endgroup$
$begingroup$
The last part sounds like the introduction to algebraic geometry, so I won't try to understand it now. I will understand it soon though. Thank you!
$endgroup$
– user198044
Dec 25 '18 at 20:28
1
$begingroup$
@JackBauer it's not, it's a basic fact of rings. Ideals in a (finite) product are products of ideals
$endgroup$
– leibnewtz
Dec 25 '18 at 21:22
$begingroup$
@leibnewtz I assume they were talking about the infinite product result (which falls more under weird results involving set theory and other foundations related stuff I think)
$endgroup$
– Stahl
Dec 25 '18 at 21:28
$begingroup$
@Stahl My impression was that they were mixing up the result you cited with the Nullstellensatz, but who knows
$endgroup$
– leibnewtz
Dec 25 '18 at 21:53
$begingroup$
@leibnewtz Oh, is it this one? If it's in Dummit Foote but not Artin, that explains why people say to do the exercises in Dummit Foote but read Artin.
$endgroup$
– user198044
Dec 29 '18 at 11:59
add a comment |
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$begingroup$
Yes, your argument is fine (modulo that you still need to show that $(x+1)$ and $(x-1)$ are the only maximal ideals of $Bbb R[x]/(x^2 - 1)$). If you have an isomorphism $f : Rto S,$ then the maximal ideals of $S$ are precisely the images via $f$ of maximal ideals of $R.$
I only have two comments: you can probably remove some of the steps and some of the notation, but if you really want to spell everything out, that's fine. Secondly, there's a minor typo in step 2: $overline{varphi}circpi(x+1) = varphi(x + 1),$ not $varphi(x-1).$
Another comment not related to the correctness of the proof: as in your first link, it is also not too difficult to argue directly that if you have a finite product of fields $A = k_1timesdotstimes k_n,$ the maximal ideals are precisely the ideals $$mathfrak{m}_i := {(x_1,dots, x_n)in Amid x_i = 0},$$
and you could generalize without too much trouble to a direct product of finitely many commutative rings. For infinite products of fields, the answer is stranger, as you might expect.
answered Dec 22 '18 at 17:10
StahlStahl
16.7k43455
$endgroup$
$begingroup$
The last part sounds like the introduction to algebraic geometry, so I won't try to understand it now. I will understand it soon though. Thank you!
$endgroup$
– user198044
Dec 25 '18 at 20:28
1
$begingroup$
@JackBauer it's not, it's a basic fact of rings. Ideals in a (finite) product are products of ideals
$endgroup$
– leibnewtz
Dec 25 '18 at 21:22
$begingroup$
@leibnewtz I assume they were talking about the infinite product result (which falls more under weird results involving set theory and other foundations related stuff I think)
$endgroup$
– Stahl
Dec 25 '18 at 21:28
$begingroup$
@Stahl My impression was that they were mixing up the result you cited with the Nullstellensatz, but who knows
$endgroup$
– leibnewtz
Dec 25 '18 at 21:53
$begingroup$
@leibnewtz Oh, is it this one? If it's in Dummit Foote but not Artin, that explains why people say to do the exercises in Dummit Foote but read Artin.
$endgroup$
– user198044
Dec 29 '18 at 11:59
add a comment |
$begingroup$
Yes, your argument is fine (modulo that you still need to show that $(x+1)$ and $(x-1)$ are the only maximal ideals of $Bbb R[x]/(x^2 - 1)$). If you have an isomorphism $f : Rto S,$ then the maximal ideals of $S$ are precisely the images via $f$ of maximal ideals of $R.$
I only have two comments: you can probably remove some of the steps and some of the notation, but if you really want to spell everything out, that's fine. Secondly, there's a minor typo in step 2: $overline{varphi}circpi(x+1) = varphi(x + 1),$ not $varphi(x-1).$
Another comment not related to the correctness of the proof: as in your first link, it is also not too difficult to argue directly that if you have a finite product of fields $A = k_1timesdotstimes k_n,$ the maximal ideals are precisely the ideals $$mathfrak{m}_i := {(x_1,dots, x_n)in Amid x_i = 0},$$
and you could generalize without too much trouble to a direct product of finitely many commutative rings. For infinite products of fields, the answer is stranger, as you might expect.
answered Dec 22 '18 at 17:10
StahlStahl
16.7k43455
$endgroup$
$begingroup$
The last part sounds like the introduction to algebraic geometry, so I won't try to understand it now. I will understand it soon though. Thank you!
$endgroup$
– user198044
Dec 25 '18 at 20:28
1
$begingroup$
@JackBauer it's not, it's a basic fact of rings. Ideals in a (finite) product are products of ideals
$endgroup$
– leibnewtz
Dec 25 '18 at 21:22
$begingroup$
@leibnewtz I assume they were talking about the infinite product result (which falls more under weird results involving set theory and other foundations related stuff I think)
$endgroup$
– Stahl
Dec 25 '18 at 21:28
$begingroup$
@Stahl My impression was that they were mixing up the result you cited with the Nullstellensatz, but who knows
$endgroup$
– leibnewtz
Dec 25 '18 at 21:53
$begingroup$
@leibnewtz Oh, is it this one? If it's in Dummit Foote but not Artin, that explains why people say to do the exercises in Dummit Foote but read Artin.
$endgroup$
– user198044
Dec 29 '18 at 11:59
add a comment |
$begingroup$
Yes, your argument is fine (modulo that you still need to show that $(x+1)$ and $(x-1)$ are the only maximal ideals of $Bbb R[x]/(x^2 - 1)$). If you have an isomorphism $f : Rto S,$ then the maximal ideals of $S$ are precisely the images via $f$ of maximal ideals of $R.$
I only have two comments: you can probably remove some of the steps and some of the notation, but if you really want to spell everything out, that's fine. Secondly, there's a minor typo in step 2: $overline{varphi}circpi(x+1) = varphi(x + 1),$ not $varphi(x-1).$
Another comment not related to the correctness of the proof: as in your first link, it is also not too difficult to argue directly that if you have a finite product of fields $A = k_1timesdotstimes k_n,$ the maximal ideals are precisely the ideals $$mathfrak{m}_i := {(x_1,dots, x_n)in Amid x_i = 0},$$
and you could generalize without too much trouble to a direct product of finitely many commutative rings. For infinite products of fields, the answer is stranger, as you might expect.
answered Dec 22 '18 at 17:10
StahlStahl
16.7k43455
$endgroup$
Yes, your argument is fine (modulo that you still need to show that $(x+1)$ and $(x-1)$ are the only maximal ideals of $Bbb R[x]/(x^2 - 1)$). If you have an isomorphism $f : Rto S,$ then the maximal ideals of $S$ are precisely the images via $f$ of maximal ideals of $R.$
I only have two comments: you can probably remove some of the steps and some of the notation, but if you really want to spell everything out, that's fine. Secondly, there's a minor typo in step 2: $overline{varphi}circpi(x+1) = varphi(x + 1),$ not $varphi(x-1).$
Another comment not related to the correctness of the proof: as in your first link, it is also not too difficult to argue directly that if you have a finite product of fields $A = k_1timesdotstimes k_n,$ the maximal ideals are precisely the ideals $$mathfrak{m}_i := {(x_1,dots, x_n)in Amid x_i = 0},$$
and you could generalize without too much trouble to a direct product of finitely many commutative rings. For infinite products of fields, the answer is stranger, as you might expect.
answered Dec 22 '18 at 17:10
StahlStahl
16.7k43455
answered Dec 22 '18 at 17:10
StahlStahl
16.7k43455
answered Dec 22 '18 at 17:10
StahlStahl
16.7k43455
answered Dec 22 '18 at 17:10
StahlStahl
16.7k43455
16.7k43455
$begingroup$
The last part sounds like the introduction to algebraic geometry, so I won't try to understand it now. I will understand it soon though. Thank you!
$endgroup$
– user198044
Dec 25 '18 at 20:28
1
$begingroup$
@JackBauer it's not, it's a basic fact of rings. Ideals in a (finite) product are products of ideals
$endgroup$
– leibnewtz
Dec 25 '18 at 21:22
$begingroup$
@leibnewtz I assume they were talking about the infinite product result (which falls more under weird results involving set theory and other foundations related stuff I think)
$endgroup$
– Stahl
Dec 25 '18 at 21:28
$begingroup$
@Stahl My impression was that they were mixing up the result you cited with the Nullstellensatz, but who knows
$endgroup$
– leibnewtz
Dec 25 '18 at 21:53
$begingroup$
@leibnewtz Oh, is it this one? If it's in Dummit Foote but not Artin, that explains why people say to do the exercises in Dummit Foote but read Artin.
$endgroup$
– user198044
Dec 29 '18 at 11:59
add a comment |
$begingroup$
The last part sounds like the introduction to algebraic geometry, so I won't try to understand it now. I will understand it soon though. Thank you!
$endgroup$
– user198044
Dec 25 '18 at 20:28
1
$begingroup$
@JackBauer it's not, it's a basic fact of rings. Ideals in a (finite) product are products of ideals
$endgroup$
– leibnewtz
Dec 25 '18 at 21:22
$begingroup$
@leibnewtz I assume they were talking about the infinite product result (which falls more under weird results involving set theory and other foundations related stuff I think)
$endgroup$
– Stahl
Dec 25 '18 at 21:28
$begingroup$
@Stahl My impression was that they were mixing up the result you cited with the Nullstellensatz, but who knows
$endgroup$
– leibnewtz
Dec 25 '18 at 21:53
$begingroup$
@leibnewtz Oh, is it this one? If it's in Dummit Foote but not Artin, that explains why people say to do the exercises in Dummit Foote but read Artin.
$endgroup$
– user198044
Dec 29 '18 at 11:59
$begingroup$
The last part sounds like the introduction to algebraic geometry, so I won't try to understand it now. I will understand it soon though. Thank you!
$endgroup$
– user198044
Dec 25 '18 at 20:28
$begingroup$
The last part sounds like the introduction to algebraic geometry, so I won't try to understand it now. I will understand it soon though. Thank you!
$endgroup$
– user198044
Dec 25 '18 at 20:28
1
1
$begingroup$
@JackBauer it's not, it's a basic fact of rings. Ideals in a (finite) product are products of ideals
$endgroup$
– leibnewtz
Dec 25 '18 at 21:22
$begingroup$
@JackBauer it's not, it's a basic fact of rings. Ideals in a (finite) product are products of ideals
$endgroup$
– leibnewtz
Dec 25 '18 at 21:22
$begingroup$
@leibnewtz I assume they were talking about the infinite product result (which falls more under weird results involving set theory and other foundations related stuff I think)
$endgroup$
– Stahl
Dec 25 '18 at 21:28
$begingroup$
@leibnewtz I assume they were talking about the infinite product result (which falls more under weird results involving set theory and other foundations related stuff I think)
$endgroup$
– Stahl
Dec 25 '18 at 21:28
$begingroup$
@Stahl My impression was that they were mixing up the result you cited with the Nullstellensatz, but who knows
$endgroup$
– leibnewtz
Dec 25 '18 at 21:53
$begingroup$
@Stahl My impression was that they were mixing up the result you cited with the Nullstellensatz, but who knows
$endgroup$
– leibnewtz
Dec 25 '18 at 21:53
$begingroup$
@leibnewtz Oh, is it this one? If it's in Dummit Foote but not Artin, that explains why people say to do the exercises in Dummit Foote but read Artin.
$endgroup$
– user198044
Dec 29 '18 at 11:59
$begingroup$
@leibnewtz Oh, is it this one? If it's in Dummit Foote but not Artin, that explains why people say to do the exercises in Dummit Foote but read Artin.
$endgroup$
– user198044
Dec 29 '18 at 11:59
add a comment |
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