Suppose that $a$ is algebraic over a field $F$. Show that $a$ and $1+a^{-1}$ have the same degree over $F$.

Question

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Suppose that $a$ is algebraic over a field $F$. Show that $a$ and $1+a^{-1}$ have the same degree over $F$.

My idea is proof that $F(a)=F(1+a^{-1})$ but i dont know how? or other idea?

Answer 1

Note that $a \neq 0$, otherwise $a^{-1}$ is not defined.

Let $f(x) = \displaystyle\sum_{i=0}^n c_i x^i$. Suppose that $f(a)= \displaystyle\sum_{i=0}^n c_i a^i = 0$. Then, dividing by $a^n$, which we can since it is not zero, $\displaystyle\sum_{i=0}^n c_i a^{n-i} = 0$. So, if $g(x) = \displaystyle\sum_{i=0}^n c_{n-i} x^i$, then $g(a^{-1}) = 0$. Now, define $h(x) = g(x-1)$, then $h(x)$ has degree $n$, and $h(a^{-1}+1) = g(a^{-1}) = 0$. Thus,$ \deg 1+a^{-1} \leq \deg a$. I leave you to see the opposite direction in a similar manner, using the fact that $a = \frac{1}{(1+a^{-1}) - 1}$. Basically, we have constructed a degree preserving correspondence between polynomials satisfied by $a$ and by $1+a^{-1}$. Hence, the degree of the minimal polynomial of both are the same, which is the same as the degree of the field extensions adjoining them.

For example (only to illustrate the procedure, these are not irreducible polynomials) , suppose that $a=2$ satisfies $f(x) = x^2-5x+6$. Then, by what we have done, $\frac{1}{2}$ satisfies the polynomial $g(x) = (6x^2-5x+1)$, and hence $1 + \frac {1}{2}$ would satisfy the polynomial $g(x-1) = 6(x-1)^2-5(x-1)+1 = 6x^2-17x+12$, which is true. This has the same degree as $f(x)$.

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Alternately, we have $1+a^{-1} \in F(a)$, and $a = \frac{1}{(1+a^{-1}) - 1} \in F(a^{-1}+1)$. Hence, $F(a) = F(1+a^{-1})$, and now we can conclude from the fact that $[F(x) : F]$ is the degree of the minimal polynomial of $x$ over $F$.