Are $\langle X-1\rangle $ and $\langle X+1\rangle $ comaximal in $\Bbb Z[X]$?

Question

The GCD is $1$ so i was thinking that they are indeed comaximal. Can someone tell me if this is true?

Answer 1

They are not, because $1\notin \langle X-1,X+1\rangle$. Indeed, if you could find two polynomials $P(X),Q(X)\in \Bbb Z[X]$ such that $$(X-1)P(X)+(X+1)Q(X)=1,$$ then evaluating the above identity at $1$ would give you $2Q(1)=1$, which is impossible since $Q(1)\in \Bbb Z$.

The reason these ideals are not comaximal despite having GCD equal to $1$ is that $\mathbb{Z}[X]$ is not a PID, and in fact the ideal $\langle X-1,X+1\rangle$ is not principal.

Answer 2

It's special case $\,a=1,\ f(x) = x+1\,$ in

Theorem $\ \ (x\!-\!a,f) = (1) \iff f(a)\mid 1$

Proof $\,\ I = (x\!-\!a,f) = (x\!-\!a,\,f\bmod x\!-\!a) = (x\!-\!a,f(a)).\ $ Hence $\ f(a)\mid 1\,\Rightarrow\, 1\in I$

Conversely $\ 1\in I = (x\!-\!a,f(a))\,\Rightarrow\, (x\!-\!a)\,g + f(a)\,h = 1\,\overset{\large x\,=\,a}{\Longrightarrow}\, f(a)h(a)= 1\,\Rightarrow\, f(a)\mid 1$