Question

If each ${\text{n}}_{\text{t}}$ is prime in Exercise $\text{18}$, show that $\le$ 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, $\left[Q\left(\sqrt{n_1},\mathrm{...}\sqrt{n_t}\right)\right]=2^t$ .

Now using the inductive hypothesis.

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

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

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

$\sqrt{n_{t+1}}=x+y\times \sqrt{n_t}$

Inductive hypothesis provides the ratio,

$Q\left(\sqrt{n_1},\mathrm{....}\sqrt{n_t}\right):Q\left(\sqrt{n_1},\mathrm{....}\sqrt{n_{t-1}}\right)=2$

Here $x\ne 0$ because otherwise $n_t/n_{t+1}$ and also $x\ne 0$ because

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

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 $2xy\sqrt{n_t}\notin \left(\sqrt{n_1},\mathrm{....}\sqrt{n_{t-1}}\right)$

This is a contraction situation. So,

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

Above equation implies that,

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

$\left[Q\left(\sqrt{n_1},\mathrm{....}\sqrt{n_{t-1}}\right):Q\right]=2$

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

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

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

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

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

Thus from the above expression the chain of inclusions is,

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

Thus,

$\begin{array}{c}Q\left(\sqrt{n_1},\mathrm{....}\sqrt{n_{k+1}}\right)=2\cdot 2^k\\ =2^{k+1}\end{array}$

Therefore the above holds for $t=k+1$

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

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