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

$\alpha = 0, \tag 9$


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

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

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


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