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Show that for an element a in a field F

f(x) and f(x+a) have the samsame splitting field.

i want to get sure about my attempt :

without loss of generality suppose deg f is at least 1

then in the splitting field S of f

f(x)=(x-r).... in S

Then in S f(x+a)=(x+a-r)...

so the splitting ffield of f(x+a) is a subset of S. ssimilarly the reverse inclusion. hence etc.

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Hint: $f(x)$ is irreducible over some field $F \iff f(x+c)$ is irreducible over the same field for any $c \in F$.

(Prove this)

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Don't you mean $c\in F$? – Nicholas Stull Apr 21 '14 at 2:40
I do! Thanks for catching that. – Kaj Hansen Apr 21 '14 at 2:41

If $r_1,r_2,\ldots, r_n$ are the roots of $f(x)$, then $r_1-a, r_2-a,\ldots, r_n-a$ are the roots of $f(x+a)$. The splitting field is that extension field of $F$ generated by all the $r_i$'s. By adding/subtracting a fixed constant from $F$ to these $r_i$'s the generated field is not changed.

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