- #1
mathmari
Gold Member
MHB
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Hey! ![Eek! :eek: :eek:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Let $F$ be a field, $f\in F[x]$ be non-constant and $K$ the splitting field of $f$ over $F$.
Let $g\in F[x]$ be a non-constant polynomial, that divides $f$.
I want to show that $g$ can be splitted into linear factors of $K[x]$. Is $K$ the splitting field of $g$ over $F$ ?
Since $g\in F[x]$ be a non-constant polynomial, that divides $f$, we have that $f(x)=g(x)h(x), h(x)\in F[x]$.
Since $K$ is the splitting field of $f$, $f$ can be decomposed into linear factors of $K[x]$.
So, $f(x)=\prod_i (x-a_i)^i$. Since $f(x)=g(x)h(x)$, we have that $\prod_i (x-a_i)^i=g(x)h(x)$.
Do we conclude from that that the product of some of these linear factors is equal to $g(x)$ ? (Wondering)
Let $F$ be a field, $f\in F[x]$ be non-constant and $K$ the splitting field of $f$ over $F$.
Let $g\in F[x]$ be a non-constant polynomial, that divides $f$.
I want to show that $g$ can be splitted into linear factors of $K[x]$. Is $K$ the splitting field of $g$ over $F$ ?
Since $g\in F[x]$ be a non-constant polynomial, that divides $f$, we have that $f(x)=g(x)h(x), h(x)\in F[x]$.
Since $K$ is the splitting field of $f$, $f$ can be decomposed into linear factors of $K[x]$.
So, $f(x)=\prod_i (x-a_i)^i$. Since $f(x)=g(x)h(x)$, we have that $\prod_i (x-a_i)^i=g(x)h(x)$.
Do we conclude from that that the product of some of these linear factors is equal to $g(x)$ ? (Wondering)