I have to prove that if $a$ is algebraic number then so is $\alpha+1$. I've tried to construct the polynomial for $\alpha+1$ using the polynomial for $\alpha$ but that didn't lead me to anything. Let the $W(x)=a_n x^n+a_{n-1} x^{n-1} +\dots + a_1x+a_0$ be a polynomial and $W(\alpha)=0$. Consider $$W(\alpha+1)=a_n (\alpha+1)^n+a_{n-1} (\alpha+1)^{n-1} +\dots + a_1(\alpha+1)+a_0$$ $$W(\alpha+1)=a_n(\alpha^n+{n\choose 1}a^{n-1}+\dots+{n\choose n-1}\alpha +1)+\\a_{n-1}(\alpha^{n-1}+{n-1\choose 1}a^{n-2}+\dots+{n-1\choose n-2}\alpha +1)+\dots +\\a_2\alpha+a_1+a_0$$ Then taking first element of every bracket I can obtain $W(\alpha)$ which is $0$. Now I'm left with the rest of the stuff and don't know where to go from there. Is this strategy any good? If not - what would be a proper way to prove this. If I'm doing it correctly -what's next?
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6$\begingroup$ If $W(x)$ is a polynomial with root $\alpha$ then $W(x-1)$ is a polynomial with root $\alpha+1$. $\endgroup$– Peter ForemanJun 17, 2020 at 21:27
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If you are to substitute $x \mapsto x - 1$ in your characteristic polynomial $W(x)$ then you obtain a polynomial $$W'(x) = a_n (x-1)^n + \dots a_1 (x -1) + a_0.$$ It should be clear that $W'(\alpha + 1) = W(\alpha) = 0$. If you want to find the coefficients $a'_k$ of $W'(x)$ you will have to use binomial expansion similar to the way you have already done to calculate them.
