Let a be a complex number that is algebraic over Q. Show that a is algebraic over Q.

Step 1&nbsp;The polynomial is an example:h(x)=x2−a&nbsp;Clearly,&nbsp;h(a)=0&nbsp;We thus have the following:[Q(a):Q(a)]≤deg(h)&nbsp;Let f(x) be the minimal polynomial in Q[x]&nbsp;Step 2&nbsp;Now by tower rule, we have:&nbsp;[Q(a):Q]=[Q(a):Q(a)][Q(a):Q]≤degh(x).deg&nbsp;f(x)&nbsp;Hence,&nbsp;[Q(a):Q]&nbsp;is a finite extension and so algebraic.&nbsp;Step 3&nbsp;Answer:&nbsp;[Q(a):Q]&nbsp;is algebraic.

If a is a zero of f(x)∈Q[x] then a is a zero of f(x2)∈Q[x].

Step 1&nbsp;The polynomial is an example:h(x)=x2−a&nbsp;Clearly,&nbsp;h(a)=0&nbsp;We thus have the following:[Q(a):Q(a)]≤deg(h)&nbsp;Let f(x) be the minimal polynomial in Q[x]&nbsp;Step 2&nbsp;Now by tower rule, we have:&nbsp;[Q(a):Q]=[Q(a):Q(a)][Q(a):Q]≤degh(x).deg&nbsp;f(x)&nbsp;Hence,&nbsp;[Q(a):Q]&nbsp;is a finite extension and so algebraic.&nbsp;Step 3&nbsp;Answer:&nbsp;[Q(a):Q]&nbsp;is algebraic.

If a is a zero of f(x)∈Q[x] then a is a zero of f(x2)∈Q[x].

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High School
Algebra
Algebra I
Algebra foundations

veksetz

Answered question

2021-12-10

Let a be a complex number that is algebraic over Q. Show that

\sqrt{a}

is algebraic over Q.

Answer & Explanation

esfloravaou

Beginner2021-12-11Added 43 answers

Step 1
The polynomial is an example:
$h (x) = x^{2} - a$
Clearly, $h (\sqrt{a}) = 0$
We thus have the following:
$[Q (\sqrt{a}) : Q (a)] \leq d e g (h)$
Let f(x) be the minimal polynomial in Q[x]
Step 2
Now by tower rule, we have:
$[Q (\sqrt{a}) : Q] = [Q (\sqrt{a}) : Q (a)] [Q (a) : Q] \leq d e g h (x) . d e g f (x)$
Hence, $[Q (\sqrt{a}) : Q]$
is a finite extension and so algebraic.
Step 3
Answer: $[Q (\sqrt{a}) : Q]$ is algebraic.

Barbara Meeker

Beginner2021-12-12Added 38 answers

If a is a zero of

f (x) \in Q [x]

then

\sqrt{a}

is a zero of

f (x^{2}) \in Q [x]

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