Determine maximal ideals of $mathbb R[x]/(x^2)$
$begingroup$
Artin Algebra Chapter 11
For (b), a solution can be found here, which I think is the same as Takumi Murayama here. I have a question about the solution of Brian Bi here.
- What does it mean that "Obviously $I$ can't contain any nonzero constant without being the entire ring." ?
My confusion is that the elements of $R$ are of this form $$[p(x)+(x^2)]$$ rather than real numbers. So constants in $R$ are $[p(x)+(x^2)]$ for $p(x)=x^2+k$ for some $k in mathbb R$?
I think it means that for $a in mathbb R$, if $[a+(x^2)] in I$, then $I = R$. But I think we can also have $p(x) in mathbb R[x]$ such that $[p(x)+(x^2)]=[b+(x^2)]$ for some $b in mathbb R$ such as $p(x)=x^2-17$ and $b=-17$. If I'm right, then is this correct?
$$[a+(x^2)] in I implies [a+(x^2)][frac1a+(x^2)] = [1+(x^2)] in I implies I = R$$
- Why exactly do we have
$$forall c in I setminus (alpha), exists a,b in R: c = aalpha + b, a ne 0 ne b?$$
I think I understand intuitively, for finitely generated $I=(alpha, beta_1, ..., beta_n)$, we have $$c=a alpha + sum_{i=1}^{n} b_i beta_i$$
where at least one of the $b_i$'s is not zero, but
Why is $a ne 0$?
What if $I$ is not finitely generated?
abstract-algebra ideals maximal-and-prime-ideals finitely-generated
$endgroup$
add a comment |
$begingroup$
Artin Algebra Chapter 11
For (b), a solution can be found here, which I think is the same as Takumi Murayama here. I have a question about the solution of Brian Bi here.
- What does it mean that "Obviously $I$ can't contain any nonzero constant without being the entire ring." ?
My confusion is that the elements of $R$ are of this form $$[p(x)+(x^2)]$$ rather than real numbers. So constants in $R$ are $[p(x)+(x^2)]$ for $p(x)=x^2+k$ for some $k in mathbb R$?
I think it means that for $a in mathbb R$, if $[a+(x^2)] in I$, then $I = R$. But I think we can also have $p(x) in mathbb R[x]$ such that $[p(x)+(x^2)]=[b+(x^2)]$ for some $b in mathbb R$ such as $p(x)=x^2-17$ and $b=-17$. If I'm right, then is this correct?
$$[a+(x^2)] in I implies [a+(x^2)][frac1a+(x^2)] = [1+(x^2)] in I implies I = R$$
- Why exactly do we have
$$forall c in I setminus (alpha), exists a,b in R: c = aalpha + b, a ne 0 ne b?$$
I think I understand intuitively, for finitely generated $I=(alpha, beta_1, ..., beta_n)$, we have $$c=a alpha + sum_{i=1}^{n} b_i beta_i$$
where at least one of the $b_i$'s is not zero, but
Why is $a ne 0$?
What if $I$ is not finitely generated?
abstract-algebra ideals maximal-and-prime-ideals finitely-generated
$endgroup$
add a comment |
$begingroup$
Artin Algebra Chapter 11
For (b), a solution can be found here, which I think is the same as Takumi Murayama here. I have a question about the solution of Brian Bi here.
- What does it mean that "Obviously $I$ can't contain any nonzero constant without being the entire ring." ?
My confusion is that the elements of $R$ are of this form $$[p(x)+(x^2)]$$ rather than real numbers. So constants in $R$ are $[p(x)+(x^2)]$ for $p(x)=x^2+k$ for some $k in mathbb R$?
I think it means that for $a in mathbb R$, if $[a+(x^2)] in I$, then $I = R$. But I think we can also have $p(x) in mathbb R[x]$ such that $[p(x)+(x^2)]=[b+(x^2)]$ for some $b in mathbb R$ such as $p(x)=x^2-17$ and $b=-17$. If I'm right, then is this correct?
$$[a+(x^2)] in I implies [a+(x^2)][frac1a+(x^2)] = [1+(x^2)] in I implies I = R$$
- Why exactly do we have
$$forall c in I setminus (alpha), exists a,b in R: c = aalpha + b, a ne 0 ne b?$$
I think I understand intuitively, for finitely generated $I=(alpha, beta_1, ..., beta_n)$, we have $$c=a alpha + sum_{i=1}^{n} b_i beta_i$$
where at least one of the $b_i$'s is not zero, but
Why is $a ne 0$?
What if $I$ is not finitely generated?
abstract-algebra ideals maximal-and-prime-ideals finitely-generated
$endgroup$
Artin Algebra Chapter 11
For (b), a solution can be found here, which I think is the same as Takumi Murayama here. I have a question about the solution of Brian Bi here.
- What does it mean that "Obviously $I$ can't contain any nonzero constant without being the entire ring." ?
My confusion is that the elements of $R$ are of this form $$[p(x)+(x^2)]$$ rather than real numbers. So constants in $R$ are $[p(x)+(x^2)]$ for $p(x)=x^2+k$ for some $k in mathbb R$?
I think it means that for $a in mathbb R$, if $[a+(x^2)] in I$, then $I = R$. But I think we can also have $p(x) in mathbb R[x]$ such that $[p(x)+(x^2)]=[b+(x^2)]$ for some $b in mathbb R$ such as $p(x)=x^2-17$ and $b=-17$. If I'm right, then is this correct?
$$[a+(x^2)] in I implies [a+(x^2)][frac1a+(x^2)] = [1+(x^2)] in I implies I = R$$
- Why exactly do we have
$$forall c in I setminus (alpha), exists a,b in R: c = aalpha + b, a ne 0 ne b?$$
I think I understand intuitively, for finitely generated $I=(alpha, beta_1, ..., beta_n)$, we have $$c=a alpha + sum_{i=1}^{n} b_i beta_i$$
where at least one of the $b_i$'s is not zero, but
Why is $a ne 0$?
What if $I$ is not finitely generated?
abstract-algebra ideals maximal-and-prime-ideals finitely-generated
abstract-algebra ideals maximal-and-prime-ideals finitely-generated
asked Dec 22 '18 at 11:43
user198044
add a comment |
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1 Answer
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$begingroup$
The simplest way to approach these questions is by using the isomorphism theorem, in the form
The ideals of $mathbb R[x]/(f(x))$ correspond to the ideals of $mathbb R[x]$ that contain $(f(x))$.
Then use that $mathbb R[x]$ is a PID to find the ideals of $mathbb R[x]$ that contain $(f(x))$.
answered Dec 22 '18 at 12:08
lhflhf
166k10171396
$endgroup$
$begingroup$
That's the solution of Takumi Murayama. It's probably simpler, but my question is about a different solution.
$endgroup$
– user198044
Dec 25 '18 at 20:33
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The simplest way to approach these questions is by using the isomorphism theorem, in the form
The ideals of $mathbb R[x]/(f(x))$ correspond to the ideals of $mathbb R[x]$ that contain $(f(x))$.
Then use that $mathbb R[x]$ is a PID to find the ideals of $mathbb R[x]$ that contain $(f(x))$.
answered Dec 22 '18 at 12:08
lhflhf
166k10171396
$endgroup$
$begingroup$
That's the solution of Takumi Murayama. It's probably simpler, but my question is about a different solution.
$endgroup$
– user198044
Dec 25 '18 at 20:33
add a comment |
$begingroup$
The simplest way to approach these questions is by using the isomorphism theorem, in the form
The ideals of $mathbb R[x]/(f(x))$ correspond to the ideals of $mathbb R[x]$ that contain $(f(x))$.
Then use that $mathbb R[x]$ is a PID to find the ideals of $mathbb R[x]$ that contain $(f(x))$.
answered Dec 22 '18 at 12:08
lhflhf
166k10171396
$endgroup$
$begingroup$
That's the solution of Takumi Murayama. It's probably simpler, but my question is about a different solution.
$endgroup$
– user198044
Dec 25 '18 at 20:33
add a comment |
$begingroup$
The simplest way to approach these questions is by using the isomorphism theorem, in the form
The ideals of $mathbb R[x]/(f(x))$ correspond to the ideals of $mathbb R[x]$ that contain $(f(x))$.
Then use that $mathbb R[x]$ is a PID to find the ideals of $mathbb R[x]$ that contain $(f(x))$.
answered Dec 22 '18 at 12:08
lhflhf
166k10171396
$endgroup$
The simplest way to approach these questions is by using the isomorphism theorem, in the form
The ideals of $mathbb R[x]/(f(x))$ correspond to the ideals of $mathbb R[x]$ that contain $(f(x))$.
Then use that $mathbb R[x]$ is a PID to find the ideals of $mathbb R[x]$ that contain $(f(x))$.
answered Dec 22 '18 at 12:08
lhflhf
166k10171396
answered Dec 22 '18 at 12:08
lhflhf
166k10171396
answered Dec 22 '18 at 12:08
lhflhf
166k10171396
answered Dec 22 '18 at 12:08
lhflhf
166k10171396
166k10171396
$begingroup$
That's the solution of Takumi Murayama. It's probably simpler, but my question is about a different solution.
$endgroup$
– user198044
Dec 25 '18 at 20:33
add a comment |
$begingroup$
That's the solution of Takumi Murayama. It's probably simpler, but my question is about a different solution.
$endgroup$
– user198044
Dec 25 '18 at 20:33
$begingroup$
That's the solution of Takumi Murayama. It's probably simpler, but my question is about a different solution.
$endgroup$
– user198044
Dec 25 '18 at 20:33
$begingroup$
That's the solution of Takumi Murayama. It's probably simpler, but my question is about a different solution.
$endgroup$
– user198044
Dec 25 '18 at 20:33
add a comment |
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