Ring theory: Ideals being equal Question: Prove directly, without gcd computations, the following equalities of ideals.
(i) $(5, 7) = (1)$ in $\Bbb{Z}$ (of course (1) = $\Bbb Z$). 
(ii) $(15, 9) = (3)$ in $Z$.
(iii) $(X^3 −1,X^2 −1) = (X −1)  \text{ in } \Bbb{Q} [X]$
Attempted solution: 
Try to show that $(1)$ is a subset of $(5,7)$ and that $(5,7)$ is a subset of $(1)$
First thing is $(5,7) = 5x + 7y$ with $x,y$ integers. Then by the closure under addition of the integers $(5,7)$ is a subset of $(1)$ as $(1)$ is the integers
Next to show $(5,7)$ is a subset of $(1)$ I try to say if $x$ belongs to integers, $x=5q + 7p, 1=5q + 7p$, then $x=x . 1= x(5q + 7 p)$ which is $5q' +7p'$, which shows that if $x$ is an integer then it's also in $(5,7)$
 A: For (i), note that $3 \times 7 - 4 \times 5 = 1$, whence $1 \in (5, 7)$, so that $(5, 7) = Z$;
for (ii), note that $2 \times 15 - 3 \times 9 = 3$, whence $3 \in (15, 9)$, whence $(3) \subset (15, 9)$, and clearly $(15, 9) \subset (3)$, so that $(15, 9) = (3)$;
for (iii), observe that $(x^3 - 1) - x(x^2 -1) = x - 1$, whence $x -1 \in (x^3 -1, x^2 - 1)$, and that $x^3  - 1 = (x - 1)(x^2 + x + 1)$, $x^2 -1 = (x -1)(x + 1)$, whence $x^3 - 1, x^2 - 1 \in (x - 1)$, whence $(x^3 - 1, x^2 - 1) = (x - 1)$.
Call it a wrap, it's in the can, Mr. Hitchcock!
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Which also means:  have some good fun with math this weekend!
Hope this helps, ALL of it!  Cheers,
and as always,
Fiat Lux!!!
A: Hint:  Your rings are PIDs. Think about GCDs
A: For (i) you have that $1 = 5x + 7y$ for some $x,y \in \Bbb{Z}$. This can be done using the Euclidean algorithm. One solution is $x = 3$, $y=-2$. Thus any integer $z$ can be written as $z = z\cdot1 = z\cdot(5\cdot3 - 14\cdot2)$, and we have $\Bbb{Z} = (1) = (5,7)$. 
For (ii), we use basically the same process, but show that $3$ can be written as $15x+9y$, since both $15$ and $9$ can obviously be written as a multiple of $3$. Euclidean algorithm once again gives $x=2,y=-3$. 
For (iii), we have that $X^3-1 = (X-1)(X^2+X+1)$, and $X^2-1 = (X-1)(X+1)$, thus $(X^3-1,X^2-1) \subset (X-1)$. For the other direction we want some $f,g \in \Bbb{Q}[X]$ such that $(X^3-1)f + (X^2-1)g = X-1$. We use the same Euclidean algorithm, but use polynomial division instead of integer division and get $f = 1$ and $g=-X$. This gives $(X-1) \subset (X^3 -1,X^2-1)$. 
