Question

If each n_t is prime in Exercise 18, show that $\mathbf{\le }$ may be replaced by =.

Short answer

In the step 2 of the solution, it is shown that the$\le$may be replaced by =.

Step-by-step solution

Definition of splititing field.

Assume that the K is an extension of the field F. In other words it can be said that K is an algebraic extension of F if the every element of K is algebraic over F.

Proof:

Assume that $n_1,n_2....n_t$are distinct primes.

Now, $\left[Q\left(\sqrt{n_1},...\sqrt{n_t}\right)\right]=2^t$.

Now using the inductive hypothesis.

$\sqrt{n_t},\sqrt{n_{t+1}},\sqrt{n_{t+1}}\in \left(\sqrt{n_1},...\sqrt{n_{t-1}}\right)$and $gcd\left(q_i,q_j\right)=1$

Suppose $\sqrt{n_{t+1}}\in Q\left(\sqrt{n_1},...\sqrt{n_t}\right)$

Then there are$x,y\in Q\left(\sqrt{n_1},...\sqrt{n_{t-1}}\right)$ so that
$\sqrt{n_{t+1}}=x+y\times \sqrt{n_t}$

Inductive hypothesis provides the ratio,
$Q\left(\sqrt{n_1},...\sqrt{n_t}\right):Q\left(\sqrt{n_1},...\sqrt{n_{t-1}}\right)=2$

Here localid="1659428327347" $x\ne 0$because otherwise $n_t/n_{t+1}$and also $x\ne 0$because $n_{t+1}\in Q\left(\sqrt{n_1},...\sqrt{n_{t-1}}\right)$

This provides,

$n_{t+1}=x^2+y^2{n}_t+2xy\sqrt{n_t}$

On solving the above equation,

$2xy\sqrt{n_t}=n_t-x^2-y^2{n}_t$

Since$2xy\sqrt{n_t}\notin \left(\sqrt{n_t},...\sqrt{n_{t-1}}\right)$

This is a contraction situation. So,

$\sqrt{n_{k+1}}\notin Q\left(\sqrt{n_1},...\sqrt{n_t}\right)$

Above equation implies that,

$$\begin{gathered} \left[Q\left(\sqrt{n_1},...\sqrt{n_{t-1}}\right):Q\left(\sqrt{n_1},...\sqrt{n_t}\right)\right]=2or, \\ \left[Q\left(\sqrt{n_1},...\sqrt{n_{t-1}}\right):Q\right]=2 \end{gathered}$$

Therefore result is true for the value of t = 1 .

Now let us assume that the result is true for t = k ,

$\left[Q\left(\sqrt{n_1},...\sqrt{n_k}\right):Q\right]=2^k$

Now to show the assumption is true for t = k + 1

$Q\left(\sqrt{n_1},...\sqrt{n_{t-1}}\right)=Q\left(\sqrt{n_1},...\sqrt{n_k}\right)\sqrt{n_{t+1}}$

Thus from the above expression the chain of inclusions is,

$Q\subset Q\left(\sqrt{n_1},...\sqrt{n_{k+1}}\right)=Q\left(\sqrt{n_1},....\sqrt{n_k}\right)\sqrt{n_{t+1}}$

Thus,

$\begin{array}{rcl}Q\left(\sqrt{n_1},...\sqrt{n_{k+1}}\right)& =& 2-2^k\\ & =& 2^{k+1}\end{array}$

Therefore the above holds fort = k + 1

$\left[Q\left(\sqrt{n_1},...\sqrt{n_t}\right):Q\right]=2^t$

Therefore proved $\le$can be replaced by =.