# Homogeneous Ideal in Laurent Polynomial Ring

This question is exercise 13.2 from Matsumura, Commutative Ring Theory.

Let $R = R_0 + R_1 + \cdots$ be a graded ring, $I$ an ideal of $R$ and $t$ an indeterminate over $R$. Set $R' = R[t, t^{-1}]$ and consider $R'$ as a graded ring where $t$ has degree $0$. Then an ideal $I$ of $R$ is homogeneous if and only if $T_t(IR') = IR'$ (where $T_t$ is defined as here).

This is what I tried so far,

Write $R' = R_0[t,t^{-1}] + R_1[t,t^{-1}] + \cdots$.

If $I$ is homogeneous then let $f = f_0+f_1+\cdots + f_m \in I \Rightarrow \text{ all } \ f_i \in \ I$. Let $r' = r_0'+ r_1'+\cdots + r_n'$ where $r_i \in R_i[t,t^{-1}]$ where not all $f_i$ are $0$.

Now $T_t(fr') = (f)(r_0'+ r_1't+\cdots + r_n't^n) = (f_0+f_1+\cdots + f_m )(r_0'+ r_1't+\cdots + r_n't^n)$.

As all $f_i \in I$ clearly $T_t(fr') \in IR'$, so $T_t(fr') \subset IR'$. We can check injectivity by expanding the right side of above equation, if it zero then all coefficients of powers of $t$ are zero, hence $r' = 0$.

And inverse can be defined by $T_{t^{-1}}$. Hence $T_t(IR') = IR'$.

Am I correct till here?

How to start the converse, please give me some hints.

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A hint for the reverse direction: every $f\in IR'$ can be written as $\sum_i f_i t^i$, for $f_i\in R$. What can you say about $f_i$? Now, apply this to $T_t(g)$, for $g\in I$. – David Moews Jan 31 '13 at 7:51
@DavidMoews got it thanks, is my above work correct? – Ram Jan 31 '13 at 8:00
The equation showing the action of $T_t$ is not right. But the idea should be to expand the $f$ that $T_t$ is acting on into homogeneous pieces $f_i$ and show that each $f_i\in I R'$, and so then $T_t(f)\in IR'.$ – David Moews Jan 31 '13 at 8:08

If $I$ is homogeneous in $R$, then $IR'$ is homogeneous in $R'$ (the inclusion map $R\to R'$ is graded). This shows that $IR'=\bigoplus_{n\ge 0} I_n[t,t^{-1}]$, where $I_n=I\cap R_n$. In order to prove that $T_t(IR')\subseteq IR'$ pick up an element $x\in I_n[t,t^{-1}]$, $x=at^j$ with $a\in I_n$. Then $T_t(x)=at^{j+1}\in I_n[t,t^{-1}]$. Similarly one can show $T_{t^{-1}}(IR')\subseteq IR'$. Now use that $T_t\circ T_{t^{-1}}=T_{t^{-1}}\circ T_t=\operatorname{Id}_R$: $IR'=T_t(T_{t^{-1}}(IR'))\subseteq T_t(IR')$, hence $IR'=T_t(IR')$.
Conversely, let $x\in I$, $x=x_0+x_1+\cdots+x_n$. Then $T_t(x)=x_0+x_1t+\cdots+x_nt^n$. Since $T_t(IR')\subseteq IR'$ we get $T_t(x)\in IR'$. But $IR'=I[t,t^{-1}]$, and thus we can write $T_t(x)=\sum_{j\in\mathbb Z}a_jt^j$ with $a_j\in I$. In particular, we get $x_i\in I$ for all $i$.