# Let $K$ be an extension of a field $F$ such that $[K:F] = 13$.Suppose $a$ $∈$ $K-F$.What is the value of $[F(a):F]$?

Since 13 is prime and we can write $$[K:F]$$ $$=$$ $$[K:F(a)]$$ $$[F(a):F]$$ , i think answer should be 1 or 13.

So it is $$13$$ since $$F(a)$$ is not $$F$$

We note that the given $$a \in K \setminus F$$ implies

$$a \notin F; \tag 0$$

but clearly

$$F \subset F(a) \subset K; \tag 1$$

thus we have in general

$$[K:F] = [K:F(a)][F(a):F]; \tag 2$$

given that

$$[K:F] = 13, \tag 3$$

(2) becomes

$$[K:F(a)][F(a):F] = 13; \tag 4$$

since $$13$$ is prime we thus find

$$[F(a):F] \in \{1, 13 \}; \tag 5$$

now if

$$[F(a):F] = 1, \tag 6$$

it follows that $$a \in F(a)$$ and $$1 \in F$$ are linearly dependent over $$F$$, so there exist

$$\alpha, \beta \in F, \tag 7$$

not both zero, with

$$\alpha a + \beta = 0; \tag 8$$

if now

$$\alpha = 0, \tag 9$$

then

$$\beta = 0, \tag{10}$$

which contradicts our assumption that at least one of $$\alpha$$, $$\beta$$ does not vanish; therefore,

$$\alpha \ne 0 \Longrightarrow a = \alpha^{-1}\beta \in F \Rightarrow \Leftarrow a \notin F; \tag{11}$$

thus (6) is impossible; the only option other than $$[F(a):F] = 1$$ allowed by (5) is then

$$[F(a):F] = 13. \tag{12}$$