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Show that the ideal generated by two polynomials is actually equivalent to the ideal generated by a single polynomial
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I've been asked to show that the ideal $I = langle x^3 - 1, x^2 - 4x + 3rangle$ in $mathbbC[x]$ is equivalent to $I = langle p(x)rangle$ for some polynomial $p(x)$.
Now, I'm not quite sure what the ideal generated by two polynomials looks like, but I know that $langle h(x)rangle = f(x)h(x) $, so I took a guess and said
$$
I = q_1(x)(x^3 - 1) + q_2(x)(x^2 - 4x + 3)
$$
I know that $x^3 - 1$ and $x^2-4x+3$ both have a common factor of $x - 1$, so this set is
$$
I = q_1(x), q_2(x)in mathbbC[x]
$$
but this is where I'm stuck. Am I going in the right direction? Is what I've written above equivalent to $langle x - 1rangle$? If so, how do you rigorously show that $ q_1(x), q_2(x)in mathbbC[x] = mathbbC[x]$?
ideals polynomial-rings
asked Jul 16 at 18:43
user3002473
3,20942138
 |Â
show 1 more comment
up vote
6
down vote
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I've been asked to show that the ideal $I = langle x^3 - 1, x^2 - 4x + 3rangle$ in $mathbbC[x]$ is equivalent to $I = langle p(x)rangle$ for some polynomial $p(x)$.
Now, I'm not quite sure what the ideal generated by two polynomials looks like, but I know that $langle h(x)rangle = f(x)h(x) $, so I took a guess and said
$$
I = q_1(x)(x^3 - 1) + q_2(x)(x^2 - 4x + 3)
$$
I know that $x^3 - 1$ and $x^2-4x+3$ both have a common factor of $x - 1$, so this set is
$$
I = q_1(x), q_2(x)in mathbbC[x]
$$
but this is where I'm stuck. Am I going in the right direction? Is what I've written above equivalent to $langle x - 1rangle$? If so, how do you rigorously show that $ q_1(x), q_2(x)in mathbbC[x] = mathbbC[x]$?
ideals polynomial-rings
asked Jul 16 at 18:43
user3002473
3,20942138
Apply the Euclidean algorithm to get the gcd of $x^3-1$ and $x^2-4x+3$. You would get that the gcd is $x-1$, so your guess is right.
– Batominovski
Jul 16 at 18:50
@Batominovski I have already done that. I guess I worded it poorly, I meant to say I found the maximal common factor.
– user3002473
Jul 16 at 18:50
Then, what is the problem here? If $x-1$ is in $I$, then any multiple of $x-1$ is in $I$. If there were anything else in $I$, $1$ would be in $I$, but that is not possible as $1$ does not vanish at $x=1$.
– Batominovski
Jul 16 at 18:51
It's stated there, I'm not sure how to show that what I've written there is actually $langle x - 1 rangle$. Why should it be? How do I know that $q_1(x) (x^2 + x + 1) + q_2(x)(x - 3)$ generates all polynomials in $mathbbC[x]$?
– user3002473
Jul 16 at 18:52
1
@Batominovski Would you mind writing your thoughts up in an answer, and being very detailed? I must not be seeing something, but I can't tell what. I have my own definitions of various things like ring ideals and polynomial rings, but I don't see how to circumvent the last part of my attempt like you said.
– user3002473
Jul 16 at 18:59
 |Â
show 1 more comment
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I've been asked to show that the ideal $I = langle x^3 - 1, x^2 - 4x + 3rangle$ in $mathbbC[x]$ is equivalent to $I = langle p(x)rangle$ for some polynomial $p(x)$.
Now, I'm not quite sure what the ideal generated by two polynomials looks like, but I know that $langle h(x)rangle = f(x)h(x) $, so I took a guess and said
$$
I = q_1(x)(x^3 - 1) + q_2(x)(x^2 - 4x + 3)
$$
I know that $x^3 - 1$ and $x^2-4x+3$ both have a common factor of $x - 1$, so this set is
$$
I = q_1(x), q_2(x)in mathbbC[x]
$$
but this is where I'm stuck. Am I going in the right direction? Is what I've written above equivalent to $langle x - 1rangle$? If so, how do you rigorously show that $ q_1(x), q_2(x)in mathbbC[x] = mathbbC[x]$?
ideals polynomial-rings
asked Jul 16 at 18:43
user3002473
3,20942138
I've been asked to show that the ideal $I = langle x^3 - 1, x^2 - 4x + 3rangle$ in $mathbbC[x]$ is equivalent to $I = langle p(x)rangle$ for some polynomial $p(x)$.
Now, I'm not quite sure what the ideal generated by two polynomials looks like, but I know that $langle h(x)rangle = f(x)h(x) $, so I took a guess and said
$$
I = q_1(x)(x^3 - 1) + q_2(x)(x^2 - 4x + 3)
$$
I know that $x^3 - 1$ and $x^2-4x+3$ both have a common factor of $x - 1$, so this set is
$$
I = q_1(x), q_2(x)in mathbbC[x]
$$
but this is where I'm stuck. Am I going in the right direction? Is what I've written above equivalent to $langle x - 1rangle$? If so, how do you rigorously show that $ q_1(x), q_2(x)in mathbbC[x] = mathbbC[x]$?
ideals polynomial-rings
asked Jul 16 at 18:43
user3002473
3,20942138
asked Jul 16 at 18:43
user3002473
3,20942138
asked Jul 16 at 18:43
user3002473
3,20942138
asked Jul 16 at 18:43
user3002473
3,20942138
3,20942138
Apply the Euclidean algorithm to get the gcd of $x^3-1$ and $x^2-4x+3$. You would get that the gcd is $x-1$, so your guess is right.
– Batominovski
Jul 16 at 18:50
@Batominovski I have already done that. I guess I worded it poorly, I meant to say I found the maximal common factor.
– user3002473
Jul 16 at 18:50
Then, what is the problem here? If $x-1$ is in $I$, then any multiple of $x-1$ is in $I$. If there were anything else in $I$, $1$ would be in $I$, but that is not possible as $1$ does not vanish at $x=1$.
– Batominovski
Jul 16 at 18:51
It's stated there, I'm not sure how to show that what I've written there is actually $langle x - 1 rangle$. Why should it be? How do I know that $q_1(x) (x^2 + x + 1) + q_2(x)(x - 3)$ generates all polynomials in $mathbbC[x]$?
– user3002473
Jul 16 at 18:52
1
@Batominovski Would you mind writing your thoughts up in an answer, and being very detailed? I must not be seeing something, but I can't tell what. I have my own definitions of various things like ring ideals and polynomial rings, but I don't see how to circumvent the last part of my attempt like you said.
– user3002473
Jul 16 at 18:59
 |Â
show 1 more comment
Apply the Euclidean algorithm to get the gcd of $x^3-1$ and $x^2-4x+3$. You would get that the gcd is $x-1$, so your guess is right.
– Batominovski
Jul 16 at 18:50
@Batominovski I have already done that. I guess I worded it poorly, I meant to say I found the maximal common factor.
– user3002473
Jul 16 at 18:50
Then, what is the problem here? If $x-1$ is in $I$, then any multiple of $x-1$ is in $I$. If there were anything else in $I$, $1$ would be in $I$, but that is not possible as $1$ does not vanish at $x=1$.
– Batominovski
Jul 16 at 18:51
It's stated there, I'm not sure how to show that what I've written there is actually $langle x - 1 rangle$. Why should it be? How do I know that $q_1(x) (x^2 + x + 1) + q_2(x)(x - 3)$ generates all polynomials in $mathbbC[x]$?
– user3002473
Jul 16 at 18:52
1
@Batominovski Would you mind writing your thoughts up in an answer, and being very detailed? I must not be seeing something, but I can't tell what. I have my own definitions of various things like ring ideals and polynomial rings, but I don't see how to circumvent the last part of my attempt like you said.
– user3002473
Jul 16 at 18:59
Apply the Euclidean algorithm to get the gcd of $x^3-1$ and $x^2-4x+3$. You would get that the gcd is $x-1$, so your guess is right.
– Batominovski
Jul 16 at 18:50
Apply the Euclidean algorithm to get the gcd of $x^3-1$ and $x^2-4x+3$. You would get that the gcd is $x-1$, so your guess is right.
– Batominovski
Jul 16 at 18:50
@Batominovski I have already done that. I guess I worded it poorly, I meant to say I found the maximal common factor.
– user3002473
Jul 16 at 18:50
@Batominovski I have already done that. I guess I worded it poorly, I meant to say I found the maximal common factor.
– user3002473
Jul 16 at 18:50
Then, what is the problem here? If $x-1$ is in $I$, then any multiple of $x-1$ is in $I$. If there were anything else in $I$, $1$ would be in $I$, but that is not possible as $1$ does not vanish at $x=1$.
– Batominovski
Jul 16 at 18:51
Then, what is the problem here? If $x-1$ is in $I$, then any multiple of $x-1$ is in $I$. If there were anything else in $I$, $1$ would be in $I$, but that is not possible as $1$ does not vanish at $x=1$.
– Batominovski
Jul 16 at 18:51
It's stated there, I'm not sure how to show that what I've written there is actually $langle x - 1 rangle$. Why should it be? How do I know that $q_1(x) (x^2 + x + 1) + q_2(x)(x - 3)$ generates all polynomials in $mathbbC[x]$?
– user3002473
Jul 16 at 18:52
It's stated there, I'm not sure how to show that what I've written there is actually $langle x - 1 rangle$. Why should it be? How do I know that $q_1(x) (x^2 + x + 1) + q_2(x)(x - 3)$ generates all polynomials in $mathbbC[x]$?
– user3002473
Jul 16 at 18:52
1
1
@Batominovski Would you mind writing your thoughts up in an answer, and being very detailed? I must not be seeing something, but I can't tell what. I have my own definitions of various things like ring ideals and polynomial rings, but I don't see how to circumvent the last part of my attempt like you said.
– user3002473
Jul 16 at 18:59
@Batominovski Would you mind writing your thoughts up in an answer, and being very detailed? I must not be seeing something, but I can't tell what. I have my own definitions of various things like ring ideals and polynomial rings, but I don't see how to circumvent the last part of my attempt like you said.
– user3002473
Jul 16 at 18:59
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
Well, there is no need to complete the last part of OP's attempt, but I am answering anyhow. The OP can try again to show that the greatest common divisor of $x^2+x+1$ and $x-3$ is $1$. That is,
$$1=(x^2+x+1),p(x)+(x-3),q(x)$$
for some $p(x),q(x)inmathbbC[x]$. Thus, any $f(x)in mathbbC[x]$ is in the ideal generated by $x^2+x+1$ and $x-3$ as
$$f(x)=f(x)cdot 1=(x^2+x+1),p(x),f(x)+(x-3),q(x),f(x),.$$
But this is practically the same thing that mechanodroid did, and can be skipped entirely.
answered Jul 16 at 19:02
Batominovski
23.2k22777
Ohhhhhhhh see, this is what I was curious about, and now that I've seen it I realize why mechanodroid's answer works more efficiently. Thanks for answering.
– user3002473
Jul 16 at 19:06
I think what I was missing was the idea that the gcd of $f(x)$ and $g(x)$ can be written as $s(x)f(x) + t(x)g(x)$ for some polynomials $s(x)$ and $t(x)$.
– user3002473
Jul 16 at 19:09
add a comment |Â
up vote
2
down vote
Hint:
Euclidean division gives
$$x^3-1 = (x^2-4x+3)(x+4) + 13(x-1)$$
so for $(x-1)q(x) in langle x-1rangle$ we have$$(x-1)q(x) = frac113(x^3-1)q(x) - frac113(x^2-4x+3)(x+4)q(x) in I$$
answered Jul 16 at 18:54
mechanodroid
22.3k52041
Ah I see what you're doing here. Interesting. I'm clearly very confused about something, I haven't figured out where my understanding is failing, and I don't think this answer addresses that. However, I do like the direction you've taken. I certainly wouldn't have seen it myself.
– user3002473
Jul 16 at 18:57
This is what I meant when I said you were overthinking, @user3002473.
– Batominovski
Jul 16 at 18:59
@user3002473 As Batomintovski earlier said, the ideal will be generated with the gcd of $x^3-1$ and $x^2-4x+3$. You calculated the gcd $x-1$ using factorization. The Euclidean algorithm in general also gives polynomials $g, h$ such that $g(x)(x^3-1) + h(x)(x^2-4x+1) = x-1$ which are useful to show $langle x-1rangle subseteq I$.
– mechanodroid
Jul 16 at 19:01
@mechanodroid Ok, I guess I'm having too fixed a mindset, expecting to see a certain path to the answer, when there are perfectly reasonable alternatives like this. Thanks for explaining.
– user3002473
Jul 16 at 19:03
add a comment |Â
up vote
2
down vote
This is a general result. The ring of polynomials over a field is a principal ideal domain. This is the case of $mathbb C[x]$. And in a principal ideal domain, every ideal is principal. As $I$ is an ideal, it is principal, which means generated by a single element which is the $gcd$ of $(x^3-1, x^2-4x+3)=x-1$.
You can retrieve this result directly. An element of $I$ can be divided by $x-1$ as both polynomials are multiple of $x-1$. Conversely, as $$x-1=1/13[(x^3-1)-(x^2-4x+3)(x+4)]$$ every multiple of $x-1$ is an element of $I=langle x-1rangle$.
answered Jul 16 at 19:08
mathcounterexamples.net
24.2k21653
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Well, there is no need to complete the last part of OP's attempt, but I am answering anyhow. The OP can try again to show that the greatest common divisor of $x^2+x+1$ and $x-3$ is $1$. That is,
$$1=(x^2+x+1),p(x)+(x-3),q(x)$$
for some $p(x),q(x)inmathbbC[x]$. Thus, any $f(x)in mathbbC[x]$ is in the ideal generated by $x^2+x+1$ and $x-3$ as
$$f(x)=f(x)cdot 1=(x^2+x+1),p(x),f(x)+(x-3),q(x),f(x),.$$
But this is practically the same thing that mechanodroid did, and can be skipped entirely.
answered Jul 16 at 19:02
Batominovski
23.2k22777
Ohhhhhhhh see, this is what I was curious about, and now that I've seen it I realize why mechanodroid's answer works more efficiently. Thanks for answering.
– user3002473
Jul 16 at 19:06
I think what I was missing was the idea that the gcd of $f(x)$ and $g(x)$ can be written as $s(x)f(x) + t(x)g(x)$ for some polynomials $s(x)$ and $t(x)$.
– user3002473
Jul 16 at 19:09
add a comment |Â
up vote
3
down vote
accepted
Well, there is no need to complete the last part of OP's attempt, but I am answering anyhow. The OP can try again to show that the greatest common divisor of $x^2+x+1$ and $x-3$ is $1$. That is,
$$1=(x^2+x+1),p(x)+(x-3),q(x)$$
for some $p(x),q(x)inmathbbC[x]$. Thus, any $f(x)in mathbbC[x]$ is in the ideal generated by $x^2+x+1$ and $x-3$ as
$$f(x)=f(x)cdot 1=(x^2+x+1),p(x),f(x)+(x-3),q(x),f(x),.$$
But this is practically the same thing that mechanodroid did, and can be skipped entirely.
answered Jul 16 at 19:02
Batominovski
23.2k22777
Ohhhhhhhh see, this is what I was curious about, and now that I've seen it I realize why mechanodroid's answer works more efficiently. Thanks for answering.
– user3002473
Jul 16 at 19:06
I think what I was missing was the idea that the gcd of $f(x)$ and $g(x)$ can be written as $s(x)f(x) + t(x)g(x)$ for some polynomials $s(x)$ and $t(x)$.
– user3002473
Jul 16 at 19:09
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Well, there is no need to complete the last part of OP's attempt, but I am answering anyhow. The OP can try again to show that the greatest common divisor of $x^2+x+1$ and $x-3$ is $1$. That is,
$$1=(x^2+x+1),p(x)+(x-3),q(x)$$
for some $p(x),q(x)inmathbbC[x]$. Thus, any $f(x)in mathbbC[x]$ is in the ideal generated by $x^2+x+1$ and $x-3$ as
$$f(x)=f(x)cdot 1=(x^2+x+1),p(x),f(x)+(x-3),q(x),f(x),.$$
But this is practically the same thing that mechanodroid did, and can be skipped entirely.
answered Jul 16 at 19:02
Batominovski
23.2k22777
Well, there is no need to complete the last part of OP's attempt, but I am answering anyhow. The OP can try again to show that the greatest common divisor of $x^2+x+1$ and $x-3$ is $1$. That is,
$$1=(x^2+x+1),p(x)+(x-3),q(x)$$
for some $p(x),q(x)inmathbbC[x]$. Thus, any $f(x)in mathbbC[x]$ is in the ideal generated by $x^2+x+1$ and $x-3$ as
$$f(x)=f(x)cdot 1=(x^2+x+1),p(x),f(x)+(x-3),q(x),f(x),.$$
But this is practically the same thing that mechanodroid did, and can be skipped entirely.
answered Jul 16 at 19:02
Batominovski
23.2k22777
edited Jul 17 at 7:22
answered Jul 16 at 19:02
Batominovski
23.2k22777
answered Jul 16 at 19:02
Batominovski
23.2k22777
answered Jul 16 at 19:02
Batominovski
23.2k22777
23.2k22777
Ohhhhhhhh see, this is what I was curious about, and now that I've seen it I realize why mechanodroid's answer works more efficiently. Thanks for answering.
– user3002473
Jul 16 at 19:06
I think what I was missing was the idea that the gcd of $f(x)$ and $g(x)$ can be written as $s(x)f(x) + t(x)g(x)$ for some polynomials $s(x)$ and $t(x)$.
– user3002473
Jul 16 at 19:09
add a comment |Â
Ohhhhhhhh see, this is what I was curious about, and now that I've seen it I realize why mechanodroid's answer works more efficiently. Thanks for answering.
– user3002473
Jul 16 at 19:06
I think what I was missing was the idea that the gcd of $f(x)$ and $g(x)$ can be written as $s(x)f(x) + t(x)g(x)$ for some polynomials $s(x)$ and $t(x)$.
– user3002473
Jul 16 at 19:09
Ohhhhhhhh see, this is what I was curious about, and now that I've seen it I realize why mechanodroid's answer works more efficiently. Thanks for answering.
– user3002473
Jul 16 at 19:06
Ohhhhhhhh see, this is what I was curious about, and now that I've seen it I realize why mechanodroid's answer works more efficiently. Thanks for answering.
– user3002473
Jul 16 at 19:06
I think what I was missing was the idea that the gcd of $f(x)$ and $g(x)$ can be written as $s(x)f(x) + t(x)g(x)$ for some polynomials $s(x)$ and $t(x)$.
– user3002473
Jul 16 at 19:09
I think what I was missing was the idea that the gcd of $f(x)$ and $g(x)$ can be written as $s(x)f(x) + t(x)g(x)$ for some polynomials $s(x)$ and $t(x)$.
– user3002473
Jul 16 at 19:09
add a comment |Â
up vote
2
down vote
Hint:
Euclidean division gives
$$x^3-1 = (x^2-4x+3)(x+4) + 13(x-1)$$
so for $(x-1)q(x) in langle x-1rangle$ we have$$(x-1)q(x) = frac113(x^3-1)q(x) - frac113(x^2-4x+3)(x+4)q(x) in I$$
answered Jul 16 at 18:54
mechanodroid
22.3k52041
Ah I see what you're doing here. Interesting. I'm clearly very confused about something, I haven't figured out where my understanding is failing, and I don't think this answer addresses that. However, I do like the direction you've taken. I certainly wouldn't have seen it myself.
– user3002473
Jul 16 at 18:57
This is what I meant when I said you were overthinking, @user3002473.
– Batominovski
Jul 16 at 18:59
@user3002473 As Batomintovski earlier said, the ideal will be generated with the gcd of $x^3-1$ and $x^2-4x+3$. You calculated the gcd $x-1$ using factorization. The Euclidean algorithm in general also gives polynomials $g, h$ such that $g(x)(x^3-1) + h(x)(x^2-4x+1) = x-1$ which are useful to show $langle x-1rangle subseteq I$.
– mechanodroid
Jul 16 at 19:01
@mechanodroid Ok, I guess I'm having too fixed a mindset, expecting to see a certain path to the answer, when there are perfectly reasonable alternatives like this. Thanks for explaining.
– user3002473
Jul 16 at 19:03
add a comment |Â
up vote
2
down vote
Hint:
Euclidean division gives
$$x^3-1 = (x^2-4x+3)(x+4) + 13(x-1)$$
so for $(x-1)q(x) in langle x-1rangle$ we have$$(x-1)q(x) = frac113(x^3-1)q(x) - frac113(x^2-4x+3)(x+4)q(x) in I$$
answered Jul 16 at 18:54
mechanodroid
22.3k52041
Ah I see what you're doing here. Interesting. I'm clearly very confused about something, I haven't figured out where my understanding is failing, and I don't think this answer addresses that. However, I do like the direction you've taken. I certainly wouldn't have seen it myself.
– user3002473
Jul 16 at 18:57
This is what I meant when I said you were overthinking, @user3002473.
– Batominovski
Jul 16 at 18:59
@user3002473 As Batomintovski earlier said, the ideal will be generated with the gcd of $x^3-1$ and $x^2-4x+3$. You calculated the gcd $x-1$ using factorization. The Euclidean algorithm in general also gives polynomials $g, h$ such that $g(x)(x^3-1) + h(x)(x^2-4x+1) = x-1$ which are useful to show $langle x-1rangle subseteq I$.
– mechanodroid
Jul 16 at 19:01
@mechanodroid Ok, I guess I'm having too fixed a mindset, expecting to see a certain path to the answer, when there are perfectly reasonable alternatives like this. Thanks for explaining.
– user3002473
Jul 16 at 19:03
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint:
Euclidean division gives
$$x^3-1 = (x^2-4x+3)(x+4) + 13(x-1)$$
so for $(x-1)q(x) in langle x-1rangle$ we have$$(x-1)q(x) = frac113(x^3-1)q(x) - frac113(x^2-4x+3)(x+4)q(x) in I$$
answered Jul 16 at 18:54
mechanodroid
22.3k52041
Hint:
Euclidean division gives
$$x^3-1 = (x^2-4x+3)(x+4) + 13(x-1)$$
so for $(x-1)q(x) in langle x-1rangle$ we have$$(x-1)q(x) = frac113(x^3-1)q(x) - frac113(x^2-4x+3)(x+4)q(x) in I$$
answered Jul 16 at 18:54
mechanodroid
22.3k52041
answered Jul 16 at 18:54
mechanodroid
22.3k52041
answered Jul 16 at 18:54
mechanodroid
22.3k52041
answered Jul 16 at 18:54
mechanodroid
22.3k52041
22.3k52041
Ah I see what you're doing here. Interesting. I'm clearly very confused about something, I haven't figured out where my understanding is failing, and I don't think this answer addresses that. However, I do like the direction you've taken. I certainly wouldn't have seen it myself.
– user3002473
Jul 16 at 18:57
This is what I meant when I said you were overthinking, @user3002473.
– Batominovski
Jul 16 at 18:59
@user3002473 As Batomintovski earlier said, the ideal will be generated with the gcd of $x^3-1$ and $x^2-4x+3$. You calculated the gcd $x-1$ using factorization. The Euclidean algorithm in general also gives polynomials $g, h$ such that $g(x)(x^3-1) + h(x)(x^2-4x+1) = x-1$ which are useful to show $langle x-1rangle subseteq I$.
– mechanodroid
Jul 16 at 19:01
@mechanodroid Ok, I guess I'm having too fixed a mindset, expecting to see a certain path to the answer, when there are perfectly reasonable alternatives like this. Thanks for explaining.
– user3002473
Jul 16 at 19:03
add a comment |Â
Ah I see what you're doing here. Interesting. I'm clearly very confused about something, I haven't figured out where my understanding is failing, and I don't think this answer addresses that. However, I do like the direction you've taken. I certainly wouldn't have seen it myself.
– user3002473
Jul 16 at 18:57
This is what I meant when I said you were overthinking, @user3002473.
– Batominovski
Jul 16 at 18:59
@user3002473 As Batomintovski earlier said, the ideal will be generated with the gcd of $x^3-1$ and $x^2-4x+3$. You calculated the gcd $x-1$ using factorization. The Euclidean algorithm in general also gives polynomials $g, h$ such that $g(x)(x^3-1) + h(x)(x^2-4x+1) = x-1$ which are useful to show $langle x-1rangle subseteq I$.
– mechanodroid
Jul 16 at 19:01
@mechanodroid Ok, I guess I'm having too fixed a mindset, expecting to see a certain path to the answer, when there are perfectly reasonable alternatives like this. Thanks for explaining.
– user3002473
Jul 16 at 19:03
Ah I see what you're doing here. Interesting. I'm clearly very confused about something, I haven't figured out where my understanding is failing, and I don't think this answer addresses that. However, I do like the direction you've taken. I certainly wouldn't have seen it myself.
– user3002473
Jul 16 at 18:57
Ah I see what you're doing here. Interesting. I'm clearly very confused about something, I haven't figured out where my understanding is failing, and I don't think this answer addresses that. However, I do like the direction you've taken. I certainly wouldn't have seen it myself.
– user3002473
Jul 16 at 18:57
This is what I meant when I said you were overthinking, @user3002473.
– Batominovski
Jul 16 at 18:59
This is what I meant when I said you were overthinking, @user3002473.
– Batominovski
Jul 16 at 18:59
@user3002473 As Batomintovski earlier said, the ideal will be generated with the gcd of $x^3-1$ and $x^2-4x+3$. You calculated the gcd $x-1$ using factorization. The Euclidean algorithm in general also gives polynomials $g, h$ such that $g(x)(x^3-1) + h(x)(x^2-4x+1) = x-1$ which are useful to show $langle x-1rangle subseteq I$.
– mechanodroid
Jul 16 at 19:01
@user3002473 As Batomintovski earlier said, the ideal will be generated with the gcd of $x^3-1$ and $x^2-4x+3$. You calculated the gcd $x-1$ using factorization. The Euclidean algorithm in general also gives polynomials $g, h$ such that $g(x)(x^3-1) + h(x)(x^2-4x+1) = x-1$ which are useful to show $langle x-1rangle subseteq I$.
– mechanodroid
Jul 16 at 19:01
@mechanodroid Ok, I guess I'm having too fixed a mindset, expecting to see a certain path to the answer, when there are perfectly reasonable alternatives like this. Thanks for explaining.
– user3002473
Jul 16 at 19:03
@mechanodroid Ok, I guess I'm having too fixed a mindset, expecting to see a certain path to the answer, when there are perfectly reasonable alternatives like this. Thanks for explaining.
– user3002473
Jul 16 at 19:03
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This is a general result. The ring of polynomials over a field is a principal ideal domain. This is the case of $mathbb C[x]$. And in a principal ideal domain, every ideal is principal. As $I$ is an ideal, it is principal, which means generated by a single element which is the $gcd$ of $(x^3-1, x^2-4x+3)=x-1$.
You can retrieve this result directly. An element of $I$ can be divided by $x-1$ as both polynomials are multiple of $x-1$. Conversely, as $$x-1=1/13[(x^3-1)-(x^2-4x+3)(x+4)]$$ every multiple of $x-1$ is an element of $I=langle x-1rangle$.
answered Jul 16 at 19:08
mathcounterexamples.net
24.2k21653
add a comment |Â
up vote
2
down vote
This is a general result. The ring of polynomials over a field is a principal ideal domain. This is the case of $mathbb C[x]$. And in a principal ideal domain, every ideal is principal. As $I$ is an ideal, it is principal, which means generated by a single element which is the $gcd$ of $(x^3-1, x^2-4x+3)=x-1$.
You can retrieve this result directly. An element of $I$ can be divided by $x-1$ as both polynomials are multiple of $x-1$. Conversely, as $$x-1=1/13[(x^3-1)-(x^2-4x+3)(x+4)]$$ every multiple of $x-1$ is an element of $I=langle x-1rangle$.
answered Jul 16 at 19:08
mathcounterexamples.net
24.2k21653
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This is a general result. The ring of polynomials over a field is a principal ideal domain. This is the case of $mathbb C[x]$. And in a principal ideal domain, every ideal is principal. As $I$ is an ideal, it is principal, which means generated by a single element which is the $gcd$ of $(x^3-1, x^2-4x+3)=x-1$.
You can retrieve this result directly. An element of $I$ can be divided by $x-1$ as both polynomials are multiple of $x-1$. Conversely, as $$x-1=1/13[(x^3-1)-(x^2-4x+3)(x+4)]$$ every multiple of $x-1$ is an element of $I=langle x-1rangle$.
answered Jul 16 at 19:08
mathcounterexamples.net
24.2k21653
This is a general result. The ring of polynomials over a field is a principal ideal domain. This is the case of $mathbb C[x]$. And in a principal ideal domain, every ideal is principal. As $I$ is an ideal, it is principal, which means generated by a single element which is the $gcd$ of $(x^3-1, x^2-4x+3)=x-1$.
You can retrieve this result directly. An element of $I$ can be divided by $x-1$ as both polynomials are multiple of $x-1$. Conversely, as $$x-1=1/13[(x^3-1)-(x^2-4x+3)(x+4)]$$ every multiple of $x-1$ is an element of $I=langle x-1rangle$.
answered Jul 16 at 19:08
mathcounterexamples.net
24.2k21653
answered Jul 16 at 19:08
mathcounterexamples.net
24.2k21653
answered Jul 16 at 19:08
mathcounterexamples.net
24.2k21653
answered Jul 16 at 19:08
mathcounterexamples.net
24.2k21653
24.2k21653
add a comment |Â
add a comment |Â
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Apply the Euclidean algorithm to get the gcd of $x^3-1$ and $x^2-4x+3$. You would get that the gcd is $x-1$, so your guess is right.
– Batominovski
Jul 16 at 18:50
@Batominovski I have already done that. I guess I worded it poorly, I meant to say I found the maximal common factor.
– user3002473
Jul 16 at 18:50
Then, what is the problem here? If $x-1$ is in $I$, then any multiple of $x-1$ is in $I$. If there were anything else in $I$, $1$ would be in $I$, but that is not possible as $1$ does not vanish at $x=1$.
– Batominovski
Jul 16 at 18:51
It's stated there, I'm not sure how to show that what I've written there is actually $langle x - 1 rangle$. Why should it be? How do I know that $q_1(x) (x^2 + x + 1) + q_2(x)(x - 3)$ generates all polynomials in $mathbbC[x]$?
– user3002473
Jul 16 at 18:52
1
@Batominovski Would you mind writing your thoughts up in an answer, and being very detailed? I must not be seeing something, but I can't tell what. I have my own definitions of various things like ring ideals and polynomial rings, but I don't see how to circumvent the last part of my attempt like you said.
– user3002473
Jul 16 at 18:59