Suppose $a$, $b$ and $c$ are distinct real numbers. Let $f_1 = (x - a)(x - b) $, $f_2 = (x - b)(x - c)$ and $ f_3 = (x - a)(x - c) $. Then $\langle f_1, f_2, f_3\rangle = 1$.
How to approach this problem ? Any hints ?
Suppose $a$, $b$ and $c$ are distinct real numbers. Let $f_1 = (x - a)(x - b) $, $f_2 = (x - b)(x - c)$ and $ f_3 = (x - a)(x - c) $. Then $\langle f_1, f_2, f_3\rangle = 1$.
How to approach this problem ? Any hints ?
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Hint: It's fairly easy to show that $1$ is in this ideal with elementary work (there are fancier reasons, but it's not necessary).
If you look at $f_1-f_2$ and $f_1-f_3$, you get two linear polynomials. By an appropriate constant factor, you can make these have the same leading coefficient. Then, by subtracting, you get a nonzero constant, which can be scaled to $1$.
A fancier answer: $\mathbb{C}[x]$ is a PID, so $\langle f_1,f_2,f_3\rangle=\langle g\rangle$. But then any roots of $g$ will also be roots of $f_1$, $f_2$, and $f_3$. Since $f_1$, $f_2$, and $f_3$ do not have any roots in common, $g$ cannot have any roots (over $\mathbb{C}$). Hence $g$ is a unit.