An ideal generated by a polynomial What does it mean if an ideal is generated by a polynomial where the ring is Q[X]? e.g if the polynomial that generates the ideal is $x^2-5x+6$ does this mean that its scalar multiples, $nx^2-5nx+6n$ for $n \in \mathbb{Q}$ are also generators? And also are its factors i.e. $(x-3)(x-2)$ in the ideal too? So would the ideal also be generated by $(x-2)$?
 A: The ideal $I= (x^2-5x+6)$ generated by the polynomial $p(x)= x^2-5x+6$ is the subset
$$I= \{(x^2-5x+6) q(x) \mid q(x) \in \mathbb Q[x]\}.$$
In particular neither $x-2$ nor $x-3$ belong to $I$.
A: When thinking about ideals keep in mind the analogy between Vector space and subspace. If $V$ is a vector space over a field $F$ and $W$ is a subspace of $V$ means if $a,b \in W$ then $a+b \in W$ and $\lambda a\in W,\: \lambda \in F$.
Same way an ideal is defined as a subspace is closed under addition and the scalars are coming from a ring rather than a field.
Here the ring is $\mathbb{Q}[X]$. So the ideal generated by the "vector" $x^2 - 5x+6$ would be, by definition must contain all the "scalar" multiples which will be $p(x) \in \mathbb{Q}[X]$. Hence the ideal is $\{p(x)(x^2-5x+6) : p(x) \in \mathbb{Q}[X]\}$
As for the next question, in the ring $\mathbb{Q}[X]$ there are some elements called the units. They are nothing but elements whose multiplicative inverse also exist in that ring. In $\mathbb{Q}[X]$ the units are precisely the elements of $\mathbb{Q}$. An ideal doesn't change as long its generators are multiplied by units. Hence $n(x^2-5x+6)$ are also generators of the ideal for every $n\in \mathbb{Q}$.
The ideal generated by $x-2$ is "bigger" than the ideal generator $x^2 - 5x +6$ because the former contains $x-2$ but no matter with which polynomial you multiply $x^2 - 5x +6$ you'll never get $x-2$ because the degree always increases when you multiply by a non zero polynomial
