Tom Marly here noted that
if R is a graded field then $R$ is concentrated in degree 0, i.e., $R=R_0$ and $R_n =0$ for all $n \neq 0$.
Is this proposition mentioned in any book or paper ?? please, I need to document this proposition.
Thank you.
Asked
Active
Viewed 604 times
1
Njl
- 31
-
Who is Tom Marly ? – Jean Marie Dec 18 '17 at 22:15
-
@JeanMarie I put the link, thank u – Njl Dec 18 '17 at 22:22
-
It's Tom Marley, from University of Nebraska. – Zach Teitler Dec 18 '17 at 23:05
-
@ZachTeitler yes but this is "Note" not paper " my university except only books and papers in references". – Njl Dec 18 '17 at 23:10
-
The same question already asked here. Please see the answer. – Ritam Halder Aug 03 '24 at 12:32
2 Answers
4
If $R$ is a $\mathbf Z$-graded field then it is easy to see that $k := R_0$ is a field. When $R_0$ is a field there are two possibilities:
$R = k$
$R \cong k[x, x^{-1}]$
For suppose that $R \ne k$ and let $\pi \in R$ be homogeneous of minimal positive degree. If $x$ is an indeterminate over $k$ with $\deg x = \deg \pi$, then we obtain a graded map
$$ \phi : k[x,x^{-1}] \to R, $$
by sending $x$ to $\pi$. The kernel of $\phi$ is a graded ideal of degree $0$ (i.e. $\ker \phi \subseteq k$) thus $\ker \phi = 0$. This shows that $\phi$ is injective.
To see that $\phi$ is surjective, let $\alpha \in R \setminus \operatorname{Im} \phi$ be of minimal positive degree. Since $\pi$ has minimal positive degree in $R$, $\deg \pi^{-1} \alpha \ge 0$. Since $\alpha$ has minimal positive degree in $R \setminus \operatorname{Im} \phi$ it follows that $\deg \pi^{-1}\alpha = 0$. But then $\pi^{-1}\alpha \in k$ so $\alpha \in k[\pi, \pi^{-1}] = \operatorname{Im } \phi$.
Therefore, when $R_0$ is a field, either $R = R_0$ or $R$ is a Laurent polynomial ring.
Sera Gunn
- 27,981
4
A reference for this fact with a complete proof is Remark 1.3.10 in C. Nastasescu, F. Van Oystaeyen, Methods of Graded Rings, Springer 2004.
Fred Rohrer
- 531