Let a be a complex number that is algebraic over Q and let r be a rational number. Show that ar is algebraic over Q.

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Joseph Lewis · Accepted Answer

Step 1  Let a be a complex number that is algebraic over Q  Step 2  Now r is a rational number, r=qp,∀p,q∈Q. Now, (ar)p=(aqp)p  =aq  aq∈Q(a)  Step 3  Then, we can say xp−aq∈Q(a)[x]  Step 4  From ar is a zero of (xp−aq), we get that ar is algebraic over Q(a)  [Q(ar):Q(a)]&amp;lt;∞  ⇒[Q(a):Q]&amp;lt;∞  [Q(ar):Q]=[Q(ar):Q(a)][Q(a):Q]&amp;lt;∞  Step 5  Therefore, we can conclude that  ar is algebraic over Q.

Samantha Brown · Accepted Answer

Let a be a complex number that is algebraic over Q  Now r is a rational number, r=qp for p,q∈Z.  (ar)p=(aqp)p  =aq  aq∈Q(a)  Then is xp−aq∈Q(a)[x]  From ar is a zero of xp−aq, we get ar is algebraic over Q(a).  [Q(ar):Q(a)]&amp;lt;∞  ⇒[Q(a):Q]&amp;lt;∞  [Q(ar):Q]=[Q(ar):Q(a)][Q(a):Q]&amp;lt;∞  Now we can conclude ar is algebraic over Q.

nick1337 · Accepted Answer

Step 1 Let a be a complex number that is algebraic over Q and let r be a rational number. We have to show that ar is algebraic over Q. Step 2 Let r=pq Where p,q,∈Z,q≠0 Now (ar)q=(ap/q)q=ap Therefore, ap∈q(a) it follows that xq−ap∈q(a)[x]

Let a be a complex number that is algebraic over

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