# $a$ and $1+a^{-1}$ have same degree over $F$ if $a$ is algebraic over $F$

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

Not really sure where to start for this one.

I know that I have to show that $[F(a):F]=[F(1+a^{-1}):F]$, and I assume that being told that $a$ is algebraic over $F$ gives me some info, but I'm not sure how to use this info.

$a$ algebraic over $F$ means that $a$ is a zero of some non-zero polynomial with coefficients in $F$.

So there exists some minimal polynomial of the form $p(x)=c_0+c_1 x+\ldots+ c_{n-1} x^{n-1}$ for $c_i\in F$ such that $p(a)=0$. We also know that $F(a)$ is isomorphic to $F[x]/\langle p(x)\rangle$.

Guidance how to approach this problem would be appreciated. Thank you.

It is obvious that $F(1+a^{-1})\subseteq F(a)$. And since $$\frac1{(1+a^{-1})-1}=a$$ we have that $F(a)\subseteq F(1+a^{-1})$.

Thus $F(a)=F(1+a^{-1})$ They have the same index because they are the same extension.

• I was thinking about this, but how are we using the information that $a$ is algebraic over $F$ here? Apr 18, 2015 at 13:52
• In fact, if $a$ is trascendental, both extensions have infinite degree. Apr 18, 2015 at 13:53
• Ah, so it being algebraic over $F$ ensures finiteness. I see Apr 18, 2015 at 13:53
• Actually, I thought that finite implied algebraic but algebraic didn't necessarily imply finiteness. Apr 18, 2015 at 14:00
• If $a$ is algebraic there will be a minimal polynomial $P$ such that $P(a)=0$ and the degree of the extension is the degree of $P$. If $a$ is trascendental, there won't be such a polynomial, and the set $\{a^j, j\in\Bbb Z_{\geq 0}\}$ will be linearly independent over $F$, so $[F(a):F]=\infty$ Apr 18, 2015 at 22:35

Note that $F(a)=F(a^{-1})$. Let $m(x)$ be the minimal polynomial of $a^{-1}$ over $F$. Then $m(x-1)$ is irreducible over $F$ and $m(x-1)$ is the minimal polynomial of $1+a^{-1}$ over $F$ and $[F(1+a^{-1}):F]=\deg{m(x-1)}=\deg{m(x)}=[F(a^{-1}):F]=[F(a):F]$.
Let $f(x)$ be the minimal polynomial of $a$. Since $f(x)$ is irreducible over $F$, the constant term of $f(x)$ is nonzero and the reciprocal polynomial $f^*(x)=x^n f(1/x)$ of $f(x)$ is also irreducible over $F$ (See here) and $\deg{f^*(x)}=n$, where $n=\deg{f(x)}$. Note that $f^*(a^{-1})=(a^{-1})^n f(a)=0$. Hence, $g(x)=f^*(x)$ is the minimal polynomial of $a^{-1}$. Furthermore, $g(x-1)$ is irreducible over $F$ and $g(x-1)$ is the minimal polynomial of $1+a^{-1}$.