If algebraic $a$ has degree $n$, so does $-a$

Question

I feel like the best way to move forward is to use a contradiction proof.

Since $a$ is algebraic, and is of degree $n$, it has a minimal polynomial of degree $n$, so we can write $$f(a)=\sum_{k=0}^n{c_ka^k}=0$$ Now suppose $-a$ has degree less than $n$, say $n-1$. Then there exists a unique polynomial $g(x)$ such that $$g(-a)=\sum_{k=0}^{n-1}{d_k(-a)^k}=0$$ Adding a nonzero term to both sides, $$g(-a)+d_n(-a)^n=\sum_{k=0}^{n}{d_k(-a)^k}=\sum_{k=0}^{n}{d_k(-1)^ka^k}$$ Letting $c_k=d_k(-1)^k$ $$g(-a)+d_n(-a)^n=\sum_{k=0}^{n}{c_ka^k}=f(a)=0$$ But $g(-a)=0$. Thus $$g(-a)+d_n(-a)^n=f(a) \Rightarrow d_n(-a)^n=0$$ However, $d_n(-a)^n\neq 0$ by construction which leads to a contradiction. Therefore, $-a$ has degree greater than or equal to $n$. Using the same style argument for a new polynomial $h(x)$ we arrive at a similar contradiction and therefore, $-a$ has degree $n$ also.

I also need to show this for algebraic numbers $a^{-1}$ and $a-1$. Is this the method best used? And is the proof correct (outside the handwaving degree $h > n$)?

Answer 1

It's much simpler to let $p(x)$ be the minimal polynomial for $-a$. Then $p(-x)$ is a rational polynomial satisfied by $a$, hence must have degree at least $\deg a$. Conversely $q(x)$, the minimal polynomial for $a$ has that $q(-x)$ is a polynomial satisfied by $-a$, hence

$$\deg(-a)\le\deg a\le\deg(-a)$$

so equality is everywhere.

Answer 2

A simpler way is to convert this to field extension.

Let $F$ be the field. Since $-a,a^{-1}\in F(a)$ we have $F(-a),F(a^{-1})\subset F(a)$ so $[F(-a):F],[F(a^{-1}):F]\leq [F(a):F]$. Hence degree of $-a,a^{-1}$ each are no more than that of $a$.

How about equality? Notice that the above argument also work with $a$ being $-a$ instead, or $a^{-1}$ instead. Hence you have the opposite inequality.

Note: also work for $a-1$ too, very much for the same reason. In fact, work also for $2a,a+\frac{1}{2}$ or any other such combination.

Answer 3

Hint: For $a-1$, let $p(x)$ be a minimal polynomial for $a$. Argue that $p(x+1)$ is a minimal polynomial for $a-1$.

For $a^{-1}$, let $p(x)=a_0x^n+a_1x^{n-1}+\cdots +a_n$ be a minimal polynomial for $a$. Argue that $a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0$ is a minimal polynomial for $a^{-1}$. Just divide through by $x^n$.