Exercise 5.33

Question

Let $\Gamma$ be a totally ordered abelian group. We shall show how to construct a field $K$ and a valuation $v$ of $K$ with $\Gamma$ as value group. Let $k$ be any field and let $A = k[\Gamma]$ be the group algebra of $\Gamma$ over $k$. By definition, $A$ is freely generated as a $k$-vector space by elements $x_\alpha~(\alpha \in \Gamma)$ such that $x_\alpha x_\beta = x_{\alpha+\beta}$. Show that $A$ is an integral domain.
  \par If $u = \lambda_1x_{\alpha_1} + \cdots + \lambda_nx_{\alpha_n}$ is any non-zero element of $A$, where the $\lambda_i$ are all $
e 0$ and $\alpha_1 < \cdots < \alpha_n$, define $v_0(u)$ to be $\alpha_1$. Show that the mapping $v_0 : A - \{0\} 	o \Gamma$ satisfies the conditions $(1)$ and $(2)$ of Exercise $31$.
  \par Let $K$ be the field of fractions of $A$. Show that $v_0$ can be uniquely extended to a valuation $v$ of $K$, and that the value group of $v$ is precisely $\Gamma$.

Amit Awasthi · Accepted Answer

\begin{proof}
  Suppose $uv = 0$ for nonzero $u,v \in A$. Then, $u = \lambda_1x_{\alpha_1} + \cdots + \lambda_mx_{\alpha_m}, v = \eta_1x_{\beta_1} + \cdots + \eta_nx_{\beta_n}$, where we assume without loss of generality that $\alpha_i,\beta_j$ are totally ordered by the order on $\Gamma$. Then, the least term in $uv$ with respect to this order is $\lambda_1\eta_1x_{\alpha_1+\beta_1}$, which is nonzero since $\lambda_1\eta_1 
e 0$, a contradiction. Thus, $A$ is a domain.
  \par Consider $u,v$ as above. Then, $v_0(uv) = \alpha_1+\beta_1 = v_0(u)+v_0(v)$, and $v_0(u+v) \ge \min\{\alpha_1,\beta_1\} = \min\{v_0(u),v_0(v)\}$, and so $v_0$ satisfies $(1),(2)$ in AM Exercise 5.31.
  \par We want to extend $v_0$ to a valuation $v : K \setminus \{0\} 	o \Gamma$. Such an extension must satisfy $(1)$ in AM Exercise $5.31$, i.e., $v(a/s) + v(s) = v(a)$. Thus, $v(a/s) = v(a) - v(s)$ uniquely determines $v$, and since $v_0$ maps onto $\Gamma$ and every element $-v(s) \in \Gamma$ by $\Gamma$'s additive group structure, we see the value group of $v$ is precisely $\Gamma$.
\end{proof}

Exercise 5.33

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Exercise 5.33

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