While self studying algebra from Thomas Hungerford I have following question in Theorem 3.8 on page 286.
It's image:
Hiw to rigorously prove that if f splits over K then $\sigma f $ over L?
Similarly, How to prove rigorously that if f has an irreducible factor g in F then $\sigma g $ would be irreducible in L[x]?
For both, I thought of assuming that $\sigma f $ doesn't splits over L and $\sigma g $ is reducible in G and trying to find a contradiction using that $\sigma$ : K $\to$ L is an isomorphism.
But I am not sure how to obtain that it will imply (intuitively I think following must be true) f doesn't splits over K and g is reducible ie how to use map $\sigma$ for it.
Can you please give a rigorous proof for these two?
I shall be really thankful.
If $f$ splits over $K$, then $$ f=a\prod_i(x-a_i), $$ with $a,a_i\in K$. Applying $\sigma$ we have $$ \sigma f=\sigma(a)\prod_i(x-\sigma(a_i)), $$ with $\sigma(a),\sigma(a_i)\in L$ and so $\sigma(f)$ splits over $L$.
Suppose on the contrary that $\sigma g=h_1\cdot h_2$, with $h_1,h_2\in L[x]$ then, because $L\cong K$, $$ g=\sigma^{-1}(\sigma g)=\sigma^{-1}(h_1)\sigma^{-1}(h_2) $$ would be not irreducible in $K[x]$, in particular would be not irreducible in $L[x]$, because $K\subseteq L$.