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I've been asked to show that the ideal $I = \langle x^3 - 1, x^2 - 4x + 3\rangle$ in $\mathbb{C}[x]$ 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 $\langle h(x)\rangle = \{ f(x)h(x)\ |\ f(x)\in R[x]\}$, so I took a guess and said $$ I = \{q_1(x)(x^3 - 1) + q_2(x)(x^2 - 4x + 3)\ |\ q_1(x), q_2(x)\in \mathbb{C}[x]\} $$ I know that $x^3 - 1$ and $x^2-4x+3$ both have a common factor of $x - 1$, so this set is $$ I = \{(q_1(x)(x^2 + x + 1) + q_2(x)(x - 3))(x - 1)\ |\ q_1(x), q_2(x)\in \mathbb{C}[x]\} $$ but this is where I'm stuck. Am I going in the right direction? Is what I've written above equivalent to $\langle x - 1\rangle$? If so, how do you rigorously show that $\{q_1(x)(x^2 + x + 1) + q_2(x)(x - 3)\ |\ q_1(x), q_2(x)\in \mathbb{C}[x]\} = \mathbb{C}[x]$?

user3002473
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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 '18 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 '18 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 '18 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 $\mathbb{C}[x]$? – user3002473 Jul 16 '18 at 18:52
  • I think you were overthinking. You can simply prove what you want by looking at the definition of ring ideals to circumvent the last part of your attempt. – Batominovski Jul 16 '18 at 18:54
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    @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 '18 at 18:59

3 Answers3

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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 $p(x),q(x)\in\mathbb{C}[x]$. Thus, any $f(x)\in \mathbb{C}[x]$ 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.

Batominovski
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  • 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 '18 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 '18 at 19:09
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Hint:

Euclidean division gives

$$x^3-1 = (x^2-4x+3)(x+4) + 13(x-1)$$

so for $(x-1)q(x) \in \langle x-1\rangle$ we have$$(x-1)q(x) = \frac1{13}(x^3-1)q(x) - \frac1{13}(x^2-4x+3)(x+4)q(x) \in I$$

mechanodroid
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  • 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 '18 at 18:57
  • This is what I meant when I said you were overthinking, @user3002473. – Batominovski Jul 16 '18 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 $\langle x-1\rangle \subseteq I$. – mechanodroid Jul 16 '18 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 '18 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 $I=\langle x-1\rangle$.

mathcounterexamples.net
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