Is a subdomain of an Euclidean domain an Euclidean domain? Is every subdomain of an Euclidean domain also an Euclidean domain?
I'm having some trouble grasping some of the concepts, so could someone please help with this question?
Any help would be appreciated!
 A: Firstly let's think of a non-Euclidean domain - $\mathbb{Z}[X]$.
Note that every Euclidean domain is a pid (principal ideal domain). For a proof I suggest this textbook (Theorem 18.21). To check that $\mathbb{Z}[X]$ is a non-Euclidean domain consider the
ideal $<X,2>$, and conclude it's not principal. 
Now $\mathbb{Z}[X]$ will be the subdomain, so we should look for
something that contains it and that is an Euclidean domain. $\mathbb{Q}[X]$
is field (in particular it is a integral domain) so it has $\mathbb{Z}[X]$
as its subdomain. Also note that there is a division algorithm for rings
of polynomials with coefficients in a field, and its valuation is the
degree of the polynomial. Alternatively, we can think of the field
of fractions of $\mathbb{Z}[X]$ and take $\mathbb{Z}[X]$ as its subdomain. 
A: Not at all. Technically speaking, all fields can be considered Euclidean domains, and every domain is a subring of a field.
If your definition of an Euclidean ring does not allow fields, then just add a variable: 
Whenver $D$ is any domain whatsoever, we have its field of fractions $k$. Then
$$D \subseteq k \subseteq k[X]$$
and $k[X]$ is an Euclidean domain.
A: Maybe your confusion lies in thinking that having an euclidean function is a universal property, when it is actually a $\forall \exists$-property.
So if $\varphi$ is an euclidean function on the ring $R$, and $R'$ is a subring, for $(a,b) \in  R' \times (R' - \{0\})$, there are $q,r$ in $\mathbf{R}$ such that $a = bq + r$ with $r=0$ or $\varphi(r) < \varphi(b)$.
Since both $q$ and $r$ are only known to be in $R$, one can't conclude that they are in $R'$ by knowing that $a$ and $b$ are.
