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If $R$ is a commutative Euclidean domain, then the ring $R((x))$ of formal Laurent series will also be an Euclidean domain. (Here we can assume that $R((x))$ is a commutative integral domain.)

Additional information, I have:

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

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 $j \geq M.$

Then we can check all the properties.

Sara Tancredi
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    Someone told me to use the order function instead of the degree. I couldn't get that. – Sara Tancredi Nov 21 '13 at 05:46
  • Dear Sara: What do you mean by you couldn't get that? You didn't understand what they meant, or it didn't work? – Bruno Joyal Nov 21 '13 at 06:03
  • Yeah I didn't understand what that mean. – Sara Tancredi Nov 21 '13 at 06:05
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    The order of $f=a_nx^n + a_{n+1}x^{n+1} + \dots$ is $v(f) = n$, if $a_n\neq 0$. If you want something which behaves like an absolute value, you should take $|f| = e^{-v(f)}$. – Bruno Joyal Nov 21 '13 at 06:09
  • Alright. But how will we prove it. Will we show the euclidean algorithm exists for this ring of formal laurent series? – Sara Tancredi Nov 21 '13 at 06:13
  • If $R$ is euclidean with degree function $\deg : R \to \mathbb{N}$, you have to construct a suitable degree function $R((x)) \to \mathbb{N}$. My guess is that the degree of the smallest coefficient could work. Or the minimum of the degrees of all coefficients. – Martin Brandenburg Nov 21 '13 at 08:25

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Hint. Let $\varphi$ be an euclidean algorithm on $R$. For an element $f\in R((x))$ let $r(f)\in R$ be the first non-zero coefficient of $f$. Set $\psi(f)=\varphi(r(f))$. Prove that $\psi$ is an euclidean algorithm on $R((x))$.