Consider the following ideal $(2+x,x^2+5)$ in $\mathbb{Z}[x]$. Then I showed that $(2+x,x^2+5)=(9,2+x)$.
But I am not able to do the same for ideals $(1-4x,x^2+5)$ and $(1+2x,x^2+5)$. Is there some method?
EDIT: My goal is to show that these ideals can be written as $(c,ax+b)$ for some appropriate integers $a,b,c$.
Would be very grateful for help!
On
Note that $\langle 1-4x,x^2+5\rangle=\langle x+20,81\rangle$. This is because $$x^2+5=(x-20)\cdot(x+20)+5\cdot 81\,,$$ $$1-4x=(-4)\cdot (x+20)+1\cdot 81\,,$$ $$x+20=4\cdot(x^2+5)+x\cdot(1-4x)\,,$$ and $$81=16\cdot(x^2+5)+(1+4x)\cdot(1-4x)\,.$$
Similarly, $\langle 1-2x,x^2+5\rangle=\langle x+10,21\rangle$. This is because $$x^2+5=(x-10)\cdot(x+10)+5\cdot 21\,,$$ $$1-2x=(-2)\cdot(x+10)+1\cdot 21\,,$$ $$x+10=2\cdot(x^2+5)+x\cdot(1-2x)\,,$$ and $$21=4\cdot(x^2+5)+(1+2x)\cdot(1-2x)\,.$$
Hint $\bmod I=(x^2\!+\!5,\,1\!-\!4x)\!:\,\ \color{#c00}{4x\equiv 1}\,$ so $\,0\equiv 4(x^2\!+\!5)\equiv (\color{#c00}{4x})x\!+\!20\equiv x\!+\!20$
Equivalently $I$ contains $\,4(x^2\!+\!5)+x(1\!-\!4x) = 20\!+\!x$
Since $\,x\!+\!20\in I\,$ we can further reduce mod $\,x\!+\!20\,$ too, i.e. use $\,\color{#0a0}{x\equiv -20}\,$ to evaluate / reduce all other generators via $\,f(\color{#0a0}x)\equiv f(\color{#0a0}{-20}),\ $ i.e. we have $$({x\!-\!a},\, f,\,g) = (x\!-\!a,\, f\bmod{x\!-\!\color{#0a0}a},\,g\bmod{x\!-\!\color{#0a0}a}) = (x\!-\!a,f(\color{#0a0}a),g(\color{#0a0}a))\qquad$$
just like in the euclidean algorithm $\, (a,b) = (a,\,b\bmod a).\ $ In summary it is simply:
$$(1\!-\!4x,\,x^2\!+\!5) = (x\!+\!20,\,\underbrace{1\!-\!4x}_{\large f(x)},\,\underbrace{x^2\!+\!5}_{\large g(x)}) = (x\!+\!20,\!\!\underbrace{81}_{\!\!\!\!\large f(-20)},\!\underbrace{405}_{\large g(-20)}\!\!) = (x\!+\!20,81)$$
Corollary $\ f\bmod I\, =\, f(-20)\bmod 81,\ $ so $\ f\in I\iff 81\mid f(-20)$