Let $R=\bigoplus_{n \ge 0} R_n$ be a graded integral domain. If $R$ is a PID, then is $R_0$ a field ?
Since $R$ is Noetherian, I know that $R_0$ is Noetherian and $R$ is a finitely generated $R_0$-algebra. But apart from that I am unable to conclude anything about $R_0$.
Please help.
No: $R$ could be any PID that is not a field, concentrated in degree $0$.
However, if you assume that $R_n\neq 0$ for some $n>0$, then the answer is yes. Let $I=\bigoplus_{n>0} R_n$; then $I$ is a principal ideal so it is generated by some element $x$ (which by hypothesis is nonzero). Now let $a\in R_0$ be any nonzero element. The ideal $(a,x)$ must be principal. But since $a$ can only be a multiple of elements of degree $0$, $(a,x)$ must be generated by some element of degree $0$. Since every element of degree $0$ in $(a,x)$ is a multiple of $a$, this means $a$ generates $(a,x)$.
In other words, $a$ divides $x$, and we have $x=ay$ for some $y$. But now observe that $y\in I$, so $y$ is a multiple of $x$. If $y=bx$, we then have $bay=y$ so $ba=1$. Thus $a$ is a unit. Since $a\in R_0$ was an arbitrary nonzero element, this implies $R_0$ is a field.
(It is in fact then easy to conclude moreover that $R$ is a polynomial ring $R_0[x]$.)