Find the degrees of field extensions

Question

Let $F\leq E$ be a field extension and $a\in E$ is a transcendental element. Put $K=F(a^2+a+1)\leq F(a).$ Find the degrees of $|F(a):K|$ and $|K:F|$. Could someone please give me any hints on how to do that ?

Answer 1

First I claim that $[F(a):F(a^2+a+1)]=2$. Since $a$ is a root of the polynomial $$ p(X)= X^2 + X + 1 - (a^2+a+1) \in F(a^2+a+1)[X] $$ we see that $[F(a):F(a^2+a+1)] \le 2$. On the other hand, if it were $1$ then there would be polynomials $q(X),p(X) \in F[X]$, $p(X)\neq 0$, such that $a = \frac{q(a^2+a+1)}{p(a^2+a+1)}$ (as every element in $F(a^2+a+1)$ is of this form, because $F(a^2+a+1)$ is the fraction field of $F[a^2+a+1]$ and $a^2+a+1$ is transcendental over $F$). But then $a$ would be a root of $q(X^2+X+1)- X\cdot p(X^2+X+1)\in F[X]$ (which is not the zero-polynomial, since $\deg q(X^2+X+1)$ is even and $\deg(X\cdot p(X^2+X+1))$ is odd), contradicting the fact that $a$ is transcendental over $F$. We conclude $[F(a):F(a^2+a+1)] = 2$.

On the other hand, $[F(a^2+a+1): F] = \infty$ follows from the fact that $a^2+a+1$ is transcendental over $F$ (if there were a polynomial $p(X)\in F[X]\setminus\{0\}$ and $p(a^2+a+1)=0$, then $a$ is a root of $p(X^2+X+1)\in F[X]\setminus\{0\}$ and hence algebraic, contradicting the fact that $a$ is transcendental over $F$).