Determine the maximal ideals of $mathbb{R}^2$.
Determine the maximal ideals of $mathbb{R}^2$.
Well for any real number that is not divisible by another number other than 1 and itself generates a maximal ideal for $mathbb{R}$. Is that right? Would it be the same as for $mathbb{R}^2$?
abstract-algebra ring-theory
edited Dec 26 '18 at 21:09
user26857
39.2k123983
asked Oct 20 '13 at 5:09
sarah
2111834
add a comment |
Determine the maximal ideals of $mathbb{R}^2$.
Well for any real number that is not divisible by another number other than 1 and itself generates a maximal ideal for $mathbb{R}$. Is that right? Would it be the same as for $mathbb{R}^2$?
abstract-algebra ring-theory
edited Dec 26 '18 at 21:09
user26857
39.2k123983
asked Oct 20 '13 at 5:09
sarah
2111834
I'm kind of confused by what you're asking. $mathbb R$ is a field, so it has only one proper ideal (the zero ideal). You can use that to try to find the ideals of $mathbb R^2$.
– Ian Coley
Oct 20 '13 at 5:12
No, that is not right: there doesn't exist "any real number that is not divisible by another number other than $1$ and itself." Fields have two ideals: $0$ and the whole field.
– anon
Oct 20 '13 at 5:14
I guess OP means the product ring structure where addition and multiplication are defined componentwise.
– user43208
Oct 20 '13 at 5:29
I wonder how this question was answered and accepted by the OP without even knowing what product is the OP taking for $;Bbb R^2;$...unless (s)he meant actually the external direct product $;Bbb RtimesBbb R;$ of the real field with itself. Mistery...
– DonAntonio
Oct 20 '13 at 10:55
add a comment |
Determine the maximal ideals of $mathbb{R}^2$.
Well for any real number that is not divisible by another number other than 1 and itself generates a maximal ideal for $mathbb{R}$. Is that right? Would it be the same as for $mathbb{R}^2$?
abstract-algebra ring-theory
edited Dec 26 '18 at 21:09
user26857
39.2k123983
asked Oct 20 '13 at 5:09
sarah
2111834
Determine the maximal ideals of $mathbb{R}^2$.
Well for any real number that is not divisible by another number other than 1 and itself generates a maximal ideal for $mathbb{R}$. Is that right? Would it be the same as for $mathbb{R}^2$?
abstract-algebra ring-theory
abstract-algebra ring-theory
edited Dec 26 '18 at 21:09
user26857
39.2k123983
asked Oct 20 '13 at 5:09
sarah
2111834
edited Dec 26 '18 at 21:09
user26857
39.2k123983
asked Oct 20 '13 at 5:09
sarah
2111834
edited Dec 26 '18 at 21:09
user26857
39.2k123983
edited Dec 26 '18 at 21:09
user26857
39.2k123983
edited Dec 26 '18 at 21:09
user26857
39.2k123983
39.2k123983
asked Oct 20 '13 at 5:09
sarah
2111834
asked Oct 20 '13 at 5:09
sarah
2111834
asked Oct 20 '13 at 5:09
sarah
2111834
2111834
I'm kind of confused by what you're asking. $mathbb R$ is a field, so it has only one proper ideal (the zero ideal). You can use that to try to find the ideals of $mathbb R^2$.
– Ian Coley
Oct 20 '13 at 5:12
No, that is not right: there doesn't exist "any real number that is not divisible by another number other than $1$ and itself." Fields have two ideals: $0$ and the whole field.
– anon
Oct 20 '13 at 5:14
I guess OP means the product ring structure where addition and multiplication are defined componentwise.
– user43208
Oct 20 '13 at 5:29
I wonder how this question was answered and accepted by the OP without even knowing what product is the OP taking for $;Bbb R^2;$...unless (s)he meant actually the external direct product $;Bbb RtimesBbb R;$ of the real field with itself. Mistery...
– DonAntonio
Oct 20 '13 at 10:55
add a comment |
I'm kind of confused by what you're asking. $mathbb R$ is a field, so it has only one proper ideal (the zero ideal). You can use that to try to find the ideals of $mathbb R^2$.
– Ian Coley
Oct 20 '13 at 5:12
No, that is not right: there doesn't exist "any real number that is not divisible by another number other than $1$ and itself." Fields have two ideals: $0$ and the whole field.
– anon
Oct 20 '13 at 5:14
I guess OP means the product ring structure where addition and multiplication are defined componentwise.
– user43208
Oct 20 '13 at 5:29
I wonder how this question was answered and accepted by the OP without even knowing what product is the OP taking for $;Bbb R^2;$...unless (s)he meant actually the external direct product $;Bbb RtimesBbb R;$ of the real field with itself. Mistery...
– DonAntonio
Oct 20 '13 at 10:55
I'm kind of confused by what you're asking. $mathbb R$ is a field, so it has only one proper ideal (the zero ideal). You can use that to try to find the ideals of $mathbb R^2$.
– Ian Coley
Oct 20 '13 at 5:12
I'm kind of confused by what you're asking. $mathbb R$ is a field, so it has only one proper ideal (the zero ideal). You can use that to try to find the ideals of $mathbb R^2$.
– Ian Coley
Oct 20 '13 at 5:12
No, that is not right: there doesn't exist "any real number that is not divisible by another number other than $1$ and itself." Fields have two ideals: $0$ and the whole field.
– anon
Oct 20 '13 at 5:14
No, that is not right: there doesn't exist "any real number that is not divisible by another number other than $1$ and itself." Fields have two ideals: $0$ and the whole field.
– anon
Oct 20 '13 at 5:14
I guess OP means the product ring structure where addition and multiplication are defined componentwise.
– user43208
Oct 20 '13 at 5:29
I guess OP means the product ring structure where addition and multiplication are defined componentwise.
– user43208
Oct 20 '13 at 5:29
I wonder how this question was answered and accepted by the OP without even knowing what product is the OP taking for $;Bbb R^2;$...unless (s)he meant actually the external direct product $;Bbb RtimesBbb R;$ of the real field with itself. Mistery...
– DonAntonio
Oct 20 '13 at 10:55
I wonder how this question was answered and accepted by the OP without even knowing what product is the OP taking for $;Bbb R^2;$...unless (s)he meant actually the external direct product $;Bbb RtimesBbb R;$ of the real field with itself. Mistery...
– DonAntonio
Oct 20 '13 at 10:55
add a comment |
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Not quite, since there aren't any primes in $mathbb{R}$ - every non-zero element has a multiplicative inverse. Any non-trivial ideal in $mathbb{R}$ is, in fact, $mathbb{R}$. For if $r in I$ and $r ne 0$, then
$$frac{1}{r} I subseteq I implies frac{1}{r} r in I implies 1 in I$$
And any ideal containing $1$ is the entire ring. So $0$ is actually the unique maximal ideal.
On the other hand, suppose that $(a, b) in I$ for some ideal $I subseteq mathbb{R} times mathbb{R}$. Consider some cases:
If $a$ and $b$ are both non-zero, show that $0 ne I = mathbb{R} times mathbb{R}$ by constructing something analogous to $1/r$.
If every element of $I$ looks like $(0, b)$, conclude that $I = 0 times mathbb{R}$ by thinking about the element $(0, frac{1}{b})$.
Likewise for $(a, 0)$.
add a comment |
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Not quite, since there aren't any primes in $mathbb{R}$ - every non-zero element has a multiplicative inverse. Any non-trivial ideal in $mathbb{R}$ is, in fact, $mathbb{R}$. For if $r in I$ and $r ne 0$, then
$$frac{1}{r} I subseteq I implies frac{1}{r} r in I implies 1 in I$$
And any ideal containing $1$ is the entire ring. So $0$ is actually the unique maximal ideal.
On the other hand, suppose that $(a, b) in I$ for some ideal $I subseteq mathbb{R} times mathbb{R}$. Consider some cases:
If $a$ and $b$ are both non-zero, show that $0 ne I = mathbb{R} times mathbb{R}$ by constructing something analogous to $1/r$.
If every element of $I$ looks like $(0, b)$, conclude that $I = 0 times mathbb{R}$ by thinking about the element $(0, frac{1}{b})$.
Likewise for $(a, 0)$.
add a comment |
Not quite, since there aren't any primes in $mathbb{R}$ - every non-zero element has a multiplicative inverse. Any non-trivial ideal in $mathbb{R}$ is, in fact, $mathbb{R}$. For if $r in I$ and $r ne 0$, then
$$frac{1}{r} I subseteq I implies frac{1}{r} r in I implies 1 in I$$
And any ideal containing $1$ is the entire ring. So $0$ is actually the unique maximal ideal.
On the other hand, suppose that $(a, b) in I$ for some ideal $I subseteq mathbb{R} times mathbb{R}$. Consider some cases:
If $a$ and $b$ are both non-zero, show that $0 ne I = mathbb{R} times mathbb{R}$ by constructing something analogous to $1/r$.
If every element of $I$ looks like $(0, b)$, conclude that $I = 0 times mathbb{R}$ by thinking about the element $(0, frac{1}{b})$.
Likewise for $(a, 0)$.
add a comment |
Not quite, since there aren't any primes in $mathbb{R}$ - every non-zero element has a multiplicative inverse. Any non-trivial ideal in $mathbb{R}$ is, in fact, $mathbb{R}$. For if $r in I$ and $r ne 0$, then
$$frac{1}{r} I subseteq I implies frac{1}{r} r in I implies 1 in I$$
And any ideal containing $1$ is the entire ring. So $0$ is actually the unique maximal ideal.
On the other hand, suppose that $(a, b) in I$ for some ideal $I subseteq mathbb{R} times mathbb{R}$. Consider some cases:
If $a$ and $b$ are both non-zero, show that $0 ne I = mathbb{R} times mathbb{R}$ by constructing something analogous to $1/r$.
If every element of $I$ looks like $(0, b)$, conclude that $I = 0 times mathbb{R}$ by thinking about the element $(0, frac{1}{b})$.
Likewise for $(a, 0)$.
Not quite, since there aren't any primes in $mathbb{R}$ - every non-zero element has a multiplicative inverse. Any non-trivial ideal in $mathbb{R}$ is, in fact, $mathbb{R}$. For if $r in I$ and $r ne 0$, then
$$frac{1}{r} I subseteq I implies frac{1}{r} r in I implies 1 in I$$
And any ideal containing $1$ is the entire ring. So $0$ is actually the unique maximal ideal.
On the other hand, suppose that $(a, b) in I$ for some ideal $I subseteq mathbb{R} times mathbb{R}$. Consider some cases:
If $a$ and $b$ are both non-zero, show that $0 ne I = mathbb{R} times mathbb{R}$ by constructing something analogous to $1/r$.
If every element of $I$ looks like $(0, b)$, conclude that $I = 0 times mathbb{R}$ by thinking about the element $(0, frac{1}{b})$.
Likewise for $(a, 0)$.
answered Oct 20 '13 at 5:14
user61527
add a comment |
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I'm kind of confused by what you're asking. $mathbb R$ is a field, so it has only one proper ideal (the zero ideal). You can use that to try to find the ideals of $mathbb R^2$.
– Ian Coley
Oct 20 '13 at 5:12
No, that is not right: there doesn't exist "any real number that is not divisible by another number other than $1$ and itself." Fields have two ideals: $0$ and the whole field.
– anon
Oct 20 '13 at 5:14
I guess OP means the product ring structure where addition and multiplication are defined componentwise.
– user43208
Oct 20 '13 at 5:29
I wonder how this question was answered and accepted by the OP without even knowing what product is the OP taking for $;Bbb R^2;$...unless (s)he meant actually the external direct product $;Bbb RtimesBbb R;$ of the real field with itself. Mistery...
– DonAntonio
Oct 20 '13 at 10:55