$k$ is a field, $f, g \in k[x]$, $h = \mathsf{lcm}(f,g)$, show that $(h) = (f) \cap (g)$

Question

This comes from Rotman, A-3.63. $k[x]$ is the polynomial ring over field $k$, $(f), (g)$, etc. are principle ideals. If $h = \mathsf{lcm}(f,g)$, then there are $s, t \in k[x]$ such that

$$h = sf = tg$$

Of course any element in $(f) \cap (g)$ is in both $(f)$ and $(g)$, and for this to be true, it must essentially be 'composed' of multiples of $f$ and $g$ and therefore $h$, as lcm, will generate an ideal that contains all such 'compositions'. But this reasoning is vague at best, what is the proof?

Are you familiar with the property of the lcm which says that if $p \in k[x]$ is a multiple of $f$ and $p$ is also multiple of $g$ then $p$ is a multiple of $h = \text{lcm}(f,g)$? — Dylan, Dec 03 '21 at 00:13
Yeah, it's the connection between ideals and division I was missing. — Matt Phillips, Dec 03 '21 at 00:22

score 0 · Accepted Answer · answered Dec 03 '21 at 00:14

0

Since $f,g$ divide $h$, $h\in (f)\cap (g)$, $(h)\subseteq (f)\cap (g)$.

On the other hand, if $r\in (f)\cap (g)$, $f, g$ divide $r$, hence $h$ divides $r$, so $r\in (h)$.

Thus $(h) = (f)\cap (g)$.

answered Dec 03 '21 at 00:14

markvs

19,653

Aha. The fact that if $x \in (p)$, then $p$ divides $x$ for any $x$, is what I was missing. Thanks. – Matt Phillips Dec 03 '21 at 00:20
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$k$ is a field, $f, g \in k[x]$, $h = \mathsf{lcm}(f,g)$, show that $(h) = (f) \cap (g)$

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