Graded Integral domain A property of a $\mathbb{Z}$-graded integral domain arose in lecture, and I for the life of me cannot see why this is the case.
Consider $$
R=\bigoplus_{i\in \mathbb{Z}}R_i
$$
Here $R$ is a $\mathbb{Z}$-graded integral domain. We considered the subring $R_0$. (Of course showing that $R_0$ is not trivial, but I was able to manage that). It was stated that $R_0$ is algebraically closed in $R$. After looking through definitions, it is still not clear to me as to why this is the case. Here being algebraically closed means that $R_0$ is equal to its algebraic closure, and the algebraic closure of $R_0$ in $R$ is the set of elements $a \in R$ such that $a$ is algebraic over $R_0$. Lastly the notion of $a \in R$ being algebraic over $R_0$ means that there exists a polynomial, say $g$, with coefficients in $R_0$ such that $g(a) = 0$. Any help would be appreciated.

Basically the idea is that if $a$ has some non-zero part in a degree other than $0$, then taking powers will shift this away from degree $0$ and so a non-zero polynomial cannot have $a$ as a zero. More precisely:
Let $0\ne g\in R_0[X]$ be a polynomial with coefficients in $R_0$ and write $g=b_nX^n+b_{n-1}X^{n-1}+\dots+b_0$ with $b_n\ne0$. Let $a\in R$ and $a_i$ be the $i$-th homogeneous component of $a$ (so that $a=\sum_{i\in \Bbb Z}a_i$). Assume that $a\notin R_0$. Assume that $a$ has some non-zero component in positive degree and let $m$ be the largest positive integers for which $a_m\ne0$ (if $a$ has no non-zero component in positive degree we take $m$ to be the minimal integer with $a_m\ne0$). Now if we plug $a$ into $g$ and write out all the homogeneous terms, we see that in the expression $g(a)$ there is exactly one term of degree $nm$, namely $b_na_m^n$. As the other non-zero $a_i$ have lower degree there is no other term of degree $nm$. By assumption $R$ is an integral domain, so we have $b_na_m^n\ne0$. In particular this means that $g(a)$ cannot be zero as it isn't zero in the $nm$-th component. We showed that no $a\in R\setminus R_0$ can satisfy a non-trivial polynomial equation over $R_0$, i.e. $R_0$ is algebraically closed in $R$.
In fact, we see that we can weaken the integral domain hypothesis: we only need that $R$ is reduced and torsionfree as an $R_0$-module.
