Show that the ideal generated by two polynomials is actually equivalent to the ideal generated by a single polynomial 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 $\{q_1(x)(x^2 + x + 1) + q_2(x)(x - 3)\ |\ q_1(x), q_2(x)\in \mathbb{C}[x]\} = \mathbb{C}[x]$?
 A: 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.
A: 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$$(x-1)q(x) = \frac1{13}(x^3-1)q(x) - \frac1{13}(x^2-4x+3)(x+4)q(x) \in I$$
A: 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 $I=\langle x-1\rangle$.
