Determine the maximal ideals of $mathbb R^2$ by determining **all** its ideals.
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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 had a different approach here and have yet another approach.
$mathbb R$ is a field so its only ideals are $(1)$ and $(0)$. Thus, by Structure of ideals in the product of two rings all the ideals of $mathbb R^2$ are $$(0)times(0),(1)times(1),(0)times(1),(1)times(0)$$. Then it's obvious $$(0)times(1),(1)times(0)$$ are the maximal ideals.
Is this correct also?
abstract-algebra proof-verification ring-theory ideals maximal-and-prime-ideals
$endgroup$
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 had a different approach here and have yet another approach.
$mathbb R$ is a field so its only ideals are $(1)$ and $(0)$. Thus, by Structure of ideals in the product of two rings all the ideals of $mathbb R^2$ are $$(0)times(0),(1)times(1),(0)times(1),(1)times(0)$$. Then it's obvious $$(0)times(1),(1)times(0)$$ are the maximal ideals.
Is this correct also?
abstract-algebra proof-verification ring-theory ideals maximal-and-prime-ideals
$endgroup$
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 had a different approach here and have yet another approach.
$mathbb R$ is a field so its only ideals are $(1)$ and $(0)$. Thus, by Structure of ideals in the product of two rings all the ideals of $mathbb R^2$ are $$(0)times(0),(1)times(1),(0)times(1),(1)times(0)$$. Then it's obvious $$(0)times(1),(1)times(0)$$ are the maximal ideals.
Is this correct also?
abstract-algebra proof-verification ring-theory ideals maximal-and-prime-ideals
$endgroup$
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 had a different approach here and have yet another approach.
$mathbb R$ is a field so its only ideals are $(1)$ and $(0)$. Thus, by Structure of ideals in the product of two rings all the ideals of $mathbb R^2$ are $$(0)times(0),(1)times(1),(0)times(1),(1)times(0)$$. Then it's obvious $$(0)times(1),(1)times(0)$$ are the maximal ideals.
Is this correct also?
abstract-algebra proof-verification ring-theory ideals maximal-and-prime-ideals
abstract-algebra proof-verification ring-theory ideals maximal-and-prime-ideals
asked Dec 29 '18 at 11:57
user198044
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1 Answer
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Yes, since $(1) times (1) = mathbb{R} times mathbb{R}$ is the whole ring and $(0) times (0)$ is contained in both $(0) times (1)$ and $(1) times (0)$. It is also clear that $(0) times (1) not subset (1) times (0)$ and vice versa.
answered Dec 29 '18 at 12:02
0x5390x539
1,450518
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$begingroup$
Thank you 0x539! I have to wait for 6-8 minutes to accept an answer.
$endgroup$
– user198044
Dec 29 '18 at 12:04
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1 Answer
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1 Answer
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active
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active
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votes
$begingroup$
Yes, since $(1) times (1) = mathbb{R} times mathbb{R}$ is the whole ring and $(0) times (0)$ is contained in both $(0) times (1)$ and $(1) times (0)$. It is also clear that $(0) times (1) not subset (1) times (0)$ and vice versa.
answered Dec 29 '18 at 12:02
0x5390x539
1,450518
$endgroup$
$begingroup$
Thank you 0x539! I have to wait for 6-8 minutes to accept an answer.
$endgroup$
– user198044
Dec 29 '18 at 12:04
add a comment |
$begingroup$
Yes, since $(1) times (1) = mathbb{R} times mathbb{R}$ is the whole ring and $(0) times (0)$ is contained in both $(0) times (1)$ and $(1) times (0)$. It is also clear that $(0) times (1) not subset (1) times (0)$ and vice versa.
answered Dec 29 '18 at 12:02
0x5390x539
1,450518
$endgroup$
$begingroup$
Thank you 0x539! I have to wait for 6-8 minutes to accept an answer.
$endgroup$
– user198044
Dec 29 '18 at 12:04
add a comment |
$begingroup$
Yes, since $(1) times (1) = mathbb{R} times mathbb{R}$ is the whole ring and $(0) times (0)$ is contained in both $(0) times (1)$ and $(1) times (0)$. It is also clear that $(0) times (1) not subset (1) times (0)$ and vice versa.
answered Dec 29 '18 at 12:02
0x5390x539
1,450518
$endgroup$
Yes, since $(1) times (1) = mathbb{R} times mathbb{R}$ is the whole ring and $(0) times (0)$ is contained in both $(0) times (1)$ and $(1) times (0)$. It is also clear that $(0) times (1) not subset (1) times (0)$ and vice versa.
answered Dec 29 '18 at 12:02
0x5390x539
1,450518
answered Dec 29 '18 at 12:02
0x5390x539
1,450518
answered Dec 29 '18 at 12:02
0x5390x539
1,450518
answered Dec 29 '18 at 12:02
0x5390x539
1,450518
1,450518
$begingroup$
Thank you 0x539! I have to wait for 6-8 minutes to accept an answer.
$endgroup$
– user198044
Dec 29 '18 at 12:04
add a comment |
$begingroup$
Thank you 0x539! I have to wait for 6-8 minutes to accept an answer.
$endgroup$
– user198044
Dec 29 '18 at 12:04
$begingroup$
Thank you 0x539! I have to wait for 6-8 minutes to accept an answer.
$endgroup$
– user198044
Dec 29 '18 at 12:04
$begingroup$
Thank you 0x539! I have to wait for 6-8 minutes to accept an answer.
$endgroup$
– user198044
Dec 29 '18 at 12:04
add a comment |
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