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## The set of all formal Laurent series is a ring

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.2 Exercise 7.2.5

Let $F$ be a field and define the ring $F((x))$ of formal Laurent series with coefficients from $F$ as the set of all formal series of the form $\sum_{n \geq N} a_nx^n$ where $a_n \in F$ and $N \in \mathbb{Z}$.

(1) Prove that $F((x))$ is a field. (Define appropriate operators.)
(2) Define the map $v : F((x))^\times \rightarrow \mathbb{Z}$ by $v(\alpha) = \min \{ k \ |\ a_k \neq 0 \}$. Prove that $v$ is a discrete valuation on $F((x))$ whose discrete valuation ring is $F[[x]]$. (See §7.1 #26 and §7.2 #3.)

Solution: Note that we can think of the elements of $F((x))$ as functions $\alpha : \mathbb{Z} \rightarrow F$ with the property that there exists a minimal element $k \in \mathbb{Z}$ such that $\alpha(k) \neq 0$, together with the zero function. With this interpretation in mind, the notation $\alpha = \sum_{n \geq N} a_n x^n$ means that $a_n = 0$ for all $n < N$; note that $N$ is not necessarily maximal with this property. That is, if $a_n = 0$ for all $N \leq n < M$, then we also say $\alpha = \sum_{n \geq M} a_n x^n$. With this in mind, we define addition and multiplication on $F((x))$ as follows. $$\left( \displaystyle\sum_{n \geq N} a_n x^n \right) + \left( \displaystyle\sum_{n \geq M} b_n x^n \right) = \displaystyle\sum_{n \geq \min(N,M)} (a_n+b_n)x^n$$ $$\left( \displaystyle\sum_{n \geq N} a_n x^n \right) \cdot \left( \displaystyle\sum_{n \geq M} b_n x^n \right) = \displaystyle\sum_{n \geq N+M} \left( \displaystyle\sum_{i+j=n} a_ib_j \right) x^n$$ For the definition of multiplication, note that for n an integer there are only finitely many pairs $(i,j)$ such that $i+j = 0$, $i \geq N$, and $j \geq M$.

First we show that $F((x))$ is a ring, is commutative, has a 1, and that every nonzero element has a multiplicative inverse. Let $\alpha = \sum_{n \geq N} a_n x^n$, $\beta = \sum_{n \geq M} b_n x^n$, and $\gamma = \sum_{n \geq Q} c_n x^n$.

(+ is associative) We have\begin{align*}(\alpha + \beta) + \gamma =&\ \left( \left( \displaystyle\sum_{n \geq N} a_n x^n \right) + \left( \displaystyle\sum_{n \geq M} b_n x^n \right) \right) + \left( \displaystyle\sum_{n \geq Q} c_n x^n \right)\\
=&\ \left( \displaystyle\sum_{n \geq \min(N,M)} (a_n+b_n) x^n \right) + \left( \displaystyle\sum_{n \geq Q} c_n x^n \right)\\
=&\ \displaystyle\sum_{n \geq \min(\min(N,M),Q)} ((a_n + b_n) + c_n) x^n\\
=&\ \displaystyle\sum_{n \geq \min(N,\min(M,Q))} (a_n + (b_n + c_n)) x^n\\
=&\ \left( \displaystyle\sum_{n \geq N} a_n x^n \right) + \left( \displaystyle\sum_{n \geq \min(M,Q)} (b_n + c_n) x^n \right)\\
=&\ \left( \displaystyle\sum_{n \geq N} a_n x^n \right) + \left( \left( \displaystyle\sum_{n \geq M} b_n x^n \right) + \left( \displaystyle\sum_{n \geq Q} c_n x^n \right) \right)\\
=&\ \alpha + (\beta + \gamma)\end{align*}so that addition is associative.

(Zero exists) Consider $0 = \sum_{n \geq N} 0 \cdot x^n$; note that our choice of the “beginning” index $N$ is arbitrary, so we may choose it as conveniently as possible. Then \begin{align*}\alpha + 0 =&\ \left( \displaystyle\sum_{n \geq N} a_n x^n \right) + \left( \displaystyle\sum_{n \geq N} 0 \cdot x^n \right)\\
=&\ \displaystyle\sum_{n \geq \min(N,N)} (a_n + 0) x^n\\
=&\ \left( \displaystyle\sum_{n \geq N} a_n x^n \right) = \alpha\\
=&\ \displaystyle\sum_{n \geq \min(N,N)} (0+a_n) x^n\\
=&\ \left( \displaystyle\sum_{n \geq N} 0 \cdot x^n \right) + \left( \displaystyle\sum_{n \geq N} a_n x^n \right)\\
=&\ 0 + \alpha\end{align*}so that 0 is an additive identity.

(Negatives exist) Define $\overline{\alpha} \in F((x))$ as follows: $\overline{\alpha} = \sum_{n \geq N} (-a_n) x^n$. Note then that
\begin{align*}\alpha + \overline{\alpha} =&\ \left( \displaystyle\sum_{n \geq N} a_n x^n \right) + \left( \displaystyle\sum_{n \geq N} (-a_n) x^n \right)\\
=&\ \displaystyle\sum_{n \geq \min(N,N)} (a_n - a_n) x^n\\
=&\ \displaystyle\sum_{n \geq N} 0 \cdot x^n\\
=&\ 0\end{align*}Similarly, $\overline{\alpha} + \alpha = 0$. Thus $\overline{\alpha}$ is an additive inverse for $\alpha$.

(+ is commutative) We have\begin{align*}\alpha + \beta =&\ \left( \displaystyle\sum_{n \geq N} a_n x^n \right) + \left( \displaystyle\sum_{n \geq M} b_n x^n \right)\\
=&\ \displaystyle\sum_{n \geq \min(N,M)} (a_n + b_n) x^n\\
=&\ \displaystyle\sum_{n \geq \min(M,N)} (b_n + a_n) x^n\\
=&\ \left( \displaystyle\sum_{n \geq M} b_n x^n \right) + \left( \displaystyle\sum_{n \geq N} a_n x^n \right)\\
=&\ \beta + \alpha\end{align*}so that addition is commutative.

($\cdot$ is associative) Note that\begin{align*}(\alpha\beta)\gamma =&\ \left( \left( \displaystyle\sum_{n \geq N} a_n x^n \right) \left( \displaystyle\sum_{n \geq M} b_n x^n \right) \right) \left( \displaystyle\sum_{n \geq Q} c_n x^n \right)\\
=&\ \left( \displaystyle\sum_{n \geq N+M} \left( \displaystyle\sum_{i+j = n} a_ib_j \right) x^n \right) \left( \displaystyle\sum_{n \geq Q} c_n x^n \right)\\
=&\ \displaystyle\sum_{n \geq (N+M)+Q} \left( \displaystyle\sum_{t+k = n} \left( \displaystyle\sum_{i+j = t} a_ib_j \right) c_k \right) x^n\\
=&\ \displaystyle\sum_{n \geq N+M+Q} \left( \displaystyle\sum_{t+k = n} \displaystyle\sum_{i+j = t} a_i b_j c_k \right) x^n\\
=&\ \displaystyle\sum_{n \geq N+M+Q} \left( \displaystyle\sum_{i+j+k=n} a_ib_jc_k \right) x^n\\
=&\ \displaystyle\sum_{n \geq N+M+Q} \left( \displaystyle\sum_{i+s = n} \displaystyle\sum_{j+k=s} a_ib_jc_k \right) x^n\\
=&\ \displaystyle\sum_{n \geq N+(M+Q)} \left( \displaystyle\sum_{i+s=n} a_i \left( \displaystyle\sum_{j+k=s} b_jc_k \right) \right) x^n\\
=&\ \left( \displaystyle\sum_{n \geq N} a_n x^n \right) \left( \displaystyle\sum_{n \geq M+Q} \left( \displaystyle\sum_{j+k = n} b_jc_k \right) x^n \right)\\
=&\ \left( \displaystyle\sum_{n \geq N} a_n x^n \right) \left( \left( \displaystyle\sum_{n \geq M} b_n x^n \right) \left( \displaystyle\sum_{n \geq Q} c_n x^n \right) \right)\\
=&\ \alpha(\beta\gamma)\end{align*}so that multiplication is associative.

($\cdot$ distributes over +) We have\begin{align*}\alpha(\beta + \gamma) =&\ \left( \displaystyle\sum_{n \geq N} a_n x^n \right) \left( \left( \displaystyle\sum_{n \geq M} b_n x^n \right) + \left( \displaystyle\sum_{n \geq Q} c_n x^n \right) \right)\\
=&\ \left( \displaystyle\sum_{n \geq N} a_n x^n \right) \left( \displaystyle\sum_{n \geq \min(M,Q)} (b_n + c_n) x^n \right)\\
=&\ \displaystyle\sum_{n \geq N + \min(M,Q)} \left( \displaystyle\sum_{i+j = n} a_i(b_j + c_j) \right) x^n\\
=&\ \displaystyle\sum_{n \geq \min(N+M,N+Q)} \left( \displaystyle\sum_{i+j=n} a_ib_j + a_i c_j \right) x^n\\
=&\ \displaystyle\sum_{n \geq \min(N+M,N+Q)} \left( \left( \displaystyle\sum_{i+j=n} a_ib_j \right) + \left( \displaystyle\sum_{i+j=n} a_ic_j \right) \right) x^n\\
=&\ \left( \displaystyle\sum_{n \geq N+M} \left( \displaystyle\sum_{i+j=n} a_ib_j \right) x^n \right) + \left( \displaystyle\sum_{n \geq N+Q} \left( \displaystyle\sum_{i+j=n} a_ic_j \right) x^n \right)\\
=&\ \left( \displaystyle\sum_{n \geq N} a_n x^n \right) \left( \displaystyle\sum_{n \geq M} b_n x^n \right) + \left( \displaystyle\sum_{n \geq N} a_n x^n \right) \left( \displaystyle\sum_{n \geq Q} c_n x^n \right)\\
=&\ \alpha\beta + \alpha\gamma\end{align*}
Thus multiplication distributes over addition on the left. Similarly, multiplication distributes over addition on the right. Thus $F((x))$ is a ring.

($\cdot$ is commutative) Note that $F$ is commutative. Thus we have\begin{align*}\alpha\beta =&\ \left( \displaystyle\sum_{n \geq N} a_n x^n \right) \left( \displaystyle\sum_{n \geq M} b_n x^n \right)\\
=&\ \displaystyle\sum_{n \geq N+M} \left( \displaystyle\sum_{i+j=n} a_ib_j \right) x^n\\
=&\ \displaystyle\sum_{n \geq M+N} \left( \displaystyle\sum_{j+i=n} b_ja_i \right) x^n\\
=&\ \left( \displaystyle\sum_{n \geq M} b_n x^n \right) \left( \displaystyle\sum_{n \geq N} a_n x^n \right)\\
=&\ \beta\alpha\end{align*}Thus $F((x))$ is commutative.

(One exists) Note that $F$ has a 1. Define $1 = \sum_{n \geq 0} e_n x^n$ by $e_0 = 1$ and $e_{i+1} = 0$. Then\begin{align*}\alpha \cdot 1 =&\ \left( \displaystyle\sum_{n \geq N} a_n x^n \right) \left( \displaystyle\sum_{n \geq 0} e_n x^n \right)\\
=&\ \displaystyle\sum_{n \geq N+0} \left( \displaystyle\sum_{i+j=n} a_ie_j \right) x^n\\
=&\ \displaystyle\sum_{n \geq N} a_n e_0 x^n\\
=&\ \displaystyle\sum_{n \geq N} a_n x^n\\
=&\ \alpha\end{align*}Similarly, $1 \cdot \alpha = \alpha$. Thus $F((x))$ has a 1.

(Inverses exist) Without loss of generality, suppose $\alpha \neq 0$ and that $N$ is minimal such that $a_N \neq 0$. Since $F$ is a field, $a_N^{-1}$ exists. Define $\delta = \sum_{n \geq -N} d_n x^n$ by $d_{-N} = a_n^{-1}$ and $$d_{k+1} = -a_N^{-1}(\sum_{i+j=k+N+1, j \leq k} a_id_j).$$ Now we have\begin{align*}\alpha\delta =&\ \left( \displaystyle\sum_{n \geq N} a_n x^n \right) \cdot \left( \displaystyle\sum_{n \geq -N} d_n x^n \right)\\
=&\ \displaystyle\sum_{n \geq N-N} \left( \displaystyle\sum_{i+j = n} a_id_j \right) x^n\\
=&\ 1\end{align*}For the last step, we handle the cases $n = 0$ and $n > 0$ separately. Thus every nonzero element $\alpha$ has an inverse.

Thus $F((x))$ is a field.

Now to show that $v$ is a valuation, we need to show three things.

$v(\alpha + \beta) \geq \min(v(\alpha), v(\beta))$: Suppose $N$ and $M$ in the definitions of $\alpha$ and $\beta$ are maximal; that is, $a_N$ and $b_M$ are nonzero. Then letting $Q = \min(N,M)$, then $a_k = 0$ for all $k < Q$. Thus $$v(\alpha + \beta) \geqslant \min(v(\alpha), v(\beta)).$$ $v$ is surjective: We can define $\theta_k \in F((x))$ by $\theta_k(n) = 1$ if $k = n$ and 0 otherwise. Clearly then $v(\theta_k) = k$, so that $v$ is surjective.

$v(\alpha\beta) = v(\alpha) + v(\beta)$: Suppose again that $N$ and $M$ are maximal. Then the $(N+M)$-th coefficient of $\alpha\beta$ is $a_Nb_M \neq 0$, since $F$ is a field, and all previous coefficients are zero by definition. Thus $v(\alpha\beta) = v(\alpha) + v(\beta)$.
Thus $v$ is a discrete valuation on $F((x))$.

Finally, we show that the discrete valuation ring of $F((x))$ is $F[[x]]$. Of course $0 \in F[[x]]$, and if $v(\alpha) \geq 0$, then without loss of generality, $\alpha = \sum_{n \geq 0} a_n x^n \in F[[x]]$. Conversely, if $\alpha \in F[[x]]$, then evidently $v(\alpha) \geq 0$ or $\alpha = 0$.