Ideals Generated by polynomials So I am currently studying a course in commutative algebra and the main object that we are looking at are ideals generated by polynomials in n variables. But the one thing I don't understand when working with these ideals is when we reduce the generating set to something much simpler. For e.g.
Consider the Ideal   $I$ = $<x^2-4x + 3, x^2 +x -2>$, then since $x -1$ is a common  factor of both the polynomials in the generating set we deduce that I is infact $<x-1>$. So my question is what is the criteria that applies when we are reducing the generating set to something much simpler.
Based on what I understand, I am guessing in the above example that since every polynomial is divisible by $x-1$ we can say the ideal is generated by $x-1$ (wouldn't this result in the loss of any elements?). But I am not entirely convinced by my reasoning and would prefer to hear it from someone who understands this stuff better. 
Also using the same reasoning as above can we then say that the ideal $I$ = $<x^3 - x^2 + x>$ = $<x>$ ?
 A: The fact that $x-1$ divides both of the generators of $I$ only means that $I \subseteq (x-1)$. For the other inclusion, try subtracting one of the generators from the other.
Subsequently, it is also true that $(x^3-x^2+x) \subseteq (x)$, but the other inclusion is false (the LHS does not contain polynomials of degree $1$, say).
A: $I=\langle(x-1)(x-3),(x-1)(x+2)\rangle=(x-1)\langle x-3,x+2\rangle =(x-1)A$,  since $x-3$ and $x+2$  are coprime.
You can't apply the same reasoning to $\langle x^3-x^2+x\rangle$  and $\langle x\rangle$.
A: The tuple notation is used both for gcds and ideals because they share many of the same laws, e.g. those below that are familiar from wide use in the Euclidean algorithm and related results:
$$\quad a(b,c)\, =\, (ab,ac)\qquad \rm [Distributive\ Law]\qquad $$
$$ (a,b)\, =\, (a,b')\ \ \ {\rm if}\ \ \ b\equiv b'\!\!\!\pmod a\qquad $$
Applied to your example $ $ (where, $ $ in your case, $\ f(x) = x\!+\!2\in\Bbb Z[x])$
$$\begin{align}((x\!-\!1)(x\!-\!3),(x\!-\!1)f(x))\, =&\,\ (x\!-\!1)\,(x\!-\!3,\,f(x))\\
=&\,\ (x\!-\!1)\,(x\!-\!3,\,f(3))\ \ {\rm by\ \ mod}\ x\!-\!3\!:\ x\equiv 3\Rightarrow f(x)\equiv f(3)\\
=&\,\ (x\!-\!1)\ \  {\rm when}\ \ f(3)\ \ \ \text{is a unit (invertible)}
\end{align}$$
