Show that if K⊆F is an extension of degree k, every element α∈F has a minimal polynomial over K of degree &lt;=k. Find theminimal polymials of complex numers over R explicitly to verify this fact

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Show that if K⊆F is an extension of degree k, every element α∈F has a minimal polynomial over K of degree &amp;lt;=k. Find theminimal polymials of complex numers over R explicitly to verify this fact

nitruraviX · Accepted Answer

To  show that the degree of the minimal polynomial of any complex number over R is of degree at most 2.  This is the problem statement.  Now, C=R(i)={a+ib:a,b∈R}.  So, deg(CR)=2, (with I, i as a basis).  To show that if z∈C, then z satisfies a polynomial f(z) with coefficients in R with deg(f)&amp;lt;=2  Proof of the statement. To start with any z satisfies a polynomial equation of degree less than or equal to 2.  Let z=a+ib,a,b∈R  ⇒(z−a)−ib⇒(z−a)2=−b2⇒z2−2az+(a2+b2)=0  Completing the proof  If z=a (b=0) is real, the minimal polynomial is z-a, deg=1  If z=a+ib, with b≠0, then z∉R, and the minimal polynomial is f(z)=z2−2az+(a2+b2), (deg=2).

Show that if K sube F is an extension of degree k, every element alpha in F has a minimal polynomial over K of degree <=k. Find theminimal polymials of complex numers over RR explicitly to verify this fact

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