Let L={x in RR:x is algebraic over QQ} be the subfield of RR consisting by all reals that are algebraic over QQ. Let K be finite field extension of L such that K sube CC. By the primitive element theorem K=L(a). Denote by f(x) the minimal polynomial of a over L. Prove f(x) has no real roots.

Jefferson Booth

Jefferson Booth

Answered question

2022-11-08

Let L = { x R : x  is algebraic over  Q } be the subfield of R consisting by all reals that are algebraic over Q . Let K be finite field extension of L such that K C . By the primitive element theorem K = L ( a ). Denote by f ( x ) the minimal polynomial of a over L. Prove f(x) has no real roots.

Answer & Explanation

trivialaxxf

trivialaxxf

Beginner2022-11-09Added 21 answers

Assuming K / L is a finite non-trivial extension ( K L ) then clearly a C R
If the minimal polynomial f ( x ) of a over L had a real root w then
L ( w ) / L , L / Q are both algebraic L ( w ) / Q is algebraic
and thus w L ( x w ) f ( x ) in L [ x ] a is a root of
g ( x ) = f ( x ) x w L [ x ] a n d deg f ( x ) x w < deg f ( x )
which of course cannot be.

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