Deveno said:
Yes, it is true for the ring $K[X]$, where $K$ is a field. Can you PROVE it?
It should be obvious that if d = gcd(f,g) we have:
$(f(x),g(x)) \subset (d(x))$ (why? see hint #2...)
so it only remains to show that $d(x) \in (f(x),g(x))$. If only we knew that:
$d(x) = a(x)f(x) + b(x)g(x)$ for some polynomials a and b...
Does this imply every ideal in $K[X]$ is principal...?
Suppose that $m(x)$ is irreducible...what ideals could possibly contain it...?
To show $$ (d(x)) = (f(x), g(x)) $$ ... ... ... (1)
where $$ d(x) = $$ gcd $$ (f(x), g(x)) $$
Firstly, by definition of a gcd, we have
$$ d(x)|f(x) $$ and $$ d(x)|g(x) $$
so $$ f(x) = a(x)d(x) $$ and $$ g(x) = b(x)d(x) $$
Now to show that
$$ (f(x), g(x)) \subseteq (d(x)) $$ ... ... ... (2)
Let $$ h(x) \in (f(x), g(x)) $$
Then $$ h(x) = s(x)f(x) + t(x)g(x) $$ (definition of $$ (f(x), g(x)) $$
$$ \Longrightarrow h(x) = s(x)a(x)d(x) + t(x)b(x)d(x) $$
$$ \Longrightarrow h(x) = (s(x)a(x) + t(x)b(x))d(x) $$
$$ \Longrightarrow h(x) = c(x)d(x) $$
$$ \Longrightarrow h(x) \in d(x) $$
$$ \Longrightarrow (f(x), g(x)) \subseteq (d(x)) $$
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Now to show $$ (d(x)) \subseteq (f(x), g(x)) $$ ... ... ... (3)
Note first that since K[x] is a Euclidean Domain and since d(x) is the gcd(f(x),g(x))
we have by Dummit and Foote, Theorem 4 page 275 that
$$ d(x) = a(x)f(x) + b(x)g(x) $$
Now to show (3) let $$ h(x) \in (d(x)) $$
Then $$ h(x) = s(x)d(x) $$ for some $$ s(x) \in K[X] $$
$$ \Longrightarrow h(x) = s(x)[a(x)f(x) + b(x)g(x)] $$
$$ \Longrightarrow h(x) = s(x)a(x)f(x) + s(x)b(x)g(x) $$
$$ \Longrightarrow h(x) = l(x)f(x) + m(x)g(x) $$
$$ \Longrightarrow h(x) \in (f(x),g(x)) $$
$$ \Longrightarrow (d(x) \subseteq (f(x),g(x)) $$
Thus we have shown that (2), (3) are the case and therefore $$ (d(x)) = (f(x), g(x)) $$
Can you please confirm that the above proof is correct ...
Now, I need to reflect further on the original problem ...
... but I am still not sure how to proceed ... ...
Peter