How are euclidean rings and norms related? How could I link norms with euclidean rings?
Why is the norm 1 for invertible elements?
Why are there arbitrarily constant in fields?
Why are local PIDs euclidean (a more intuitive approach, since my text book doesn't explain this very well)?
Thankyou
 A: This is the definition of Euclidian rings I know: It is a commutative integral domain with unity($1$) that we will denote $E$, where there exists a function $\phi$$:$ $E$ $\rightarrow$ $\Bbb Z$ satisfying
(1) If $a,b$ $\in$ $E$, and $a,b$ are nonzero, $a$ $\mid$ $b$, then $\phi(a)$ $\leq$ $\phi(b)$
(2) For $a,b$ $\in$ $E$, $a\neq 0$, there are elements $q,r$ $\in$ $E$ such that $b = aq + r$ and $\phi(r)$ $\lt$ $\phi(a)$
Such a function $\phi$ is what is what I mean by a norm. $\phi(0)$ has to be less than any other norm, because for any $a \neq 0$,we have that $a = aq + r$ for some $r,q \in E$ with $\phi(r)$ $\lt$ $\phi(a)$ which means that $a - aq$ $=$ $a(1-q)$ $=$ $r$, which means that $a \mid r$, which in turn means that $\phi(a)$ $\leq$ $\phi(r)$; a contradiction, unless $r = 0$. We have, in summary, $\phi(r) = \phi(0) $ $\lt$ $\phi(a)$.
That the norm of an invertible element (unit) is the same as the norm of $1$ in an Euclidian ring (domain):
Call the Euclidian ring $E$. Let the norm be $\phi$, let $a$ be invertible in $E$. Then for every element $\qquad$ $\,$ $b$ $\neq$ $0$ $\in$ $E$, $\phi$$(a)$ $\leq$ $\phi$$(b)$, since  a unit divides every element in any ring with $1$. Especially, $\phi (a)$ $\leq$ $\phi(1)$. But $1$ is also a unit [$1\ast1$ $=$ $1$], so  $\phi(1)$ $\leq$ $\phi(a)$. Therefore, $\phi(a)$ = $\phi(1)$ , since two numbers can't both be greater than the other number.
Constant norm in a field:
In a field you could define the norm to be $1$ for each element except $0$, since every nonzero element in a field is a unit and therefore each one of them divides each other. By defining the norm of $0$ to be the number zero, you are assured of having a remainder with norm less than the norm of the element you "divide" with. Of course, you can let the norm of all the nonzero elements of the field be any constant $c$, as long as the norm of $0$ is less than this constant.
Local PIDs are euclidean
Before I go through a proof of this result, there are some things I will assume you know, and if you don't know these things, I expect you to look them up.
First of all, for a commutative ring $R$ with unity, every ideal in $R$ not equal to $R$ is contained in a maximal ideal. This is a consequence of Zorn's lemma. This in turn, means that every non-unit is contained in a maximal ideal, because the ideal generated by this non-unit is unequal to $R$; otherwise the non-unit would divide $1$. An element not contained in any maximal ideal must therefore be a unit.
I expect you to know what an $R$-module is, what it means for it to be finitely generated, and most importantly, this;
Nakayama's lemma: If $M$ is a finitely generated $R$-module, and $A$ is an ideal contained in the Jacobson-radical $\Bbb J$ of $R$, then $AM = M \implies M=0$. [The Jacobson-radical of $R$ is the intersection of all the maximal ideals in $R$].
Now for the good stuff:
A local PID is a PID with only one maximal ideal. Therefore, its Jacobson-radical is equal to this one ideal. 
The fact that a field is a local PID I suspect you can deduce. $\{0\}$ is the maximal ideal and its generator is $0$.
I already provided a norm on fields. To show condition (2), let $a,b \in R$ ($R$ a field), $a \neq 0$. $b = a(a^{-1}b) + 0$. Now, $\phi(a) = 1, \phi(0) = 0$, so 
$\phi(0) \lt \phi(a)$.
Let $R$ be a local PID. Let $pR$ be the only maximal ideal in $R$. For $a \neq 0$, define $\phi:R \to \Bbb N$ by $\phi(a) =n$ where $n$ is the first $n$ such that $a \notin p^nR$, and let $\phi(0) = 0$.
Are you missing something? No. We will have to prove that such an n exists for each nonzero element in $R$.
Let $M =\bigcap_{n=1}^{\infty}p^nR$. This is the intersection of all ideals $p^nR$, where $n \in \Bbb N$. Now, $pM \subseteq M$, because $M$ is an ideal, and $M \subseteq pM$, because $pM =\bigcap_{n=2}^{\infty}p^nR \supseteq \bigcap_{n=1}^{\infty}p^nR$.
In effect, $pM = M$, which means that $(pR)M = M$, since $RM = M$, since $1 \in R$. But $M$ is finitely generated, because every ideal in a PID is finitely generated (of the form $cR$). Because the ideal $M$ can be viewed as a module, and $pR$ is the Jacobson-radical, and therefore contained in itself, we have by Nakayama that $M = 0$.
What do we do with this? Well, $M = 0$ means that only $0$ is in every ideal of the form $p^nR$. So if $a \neq 0$, then $a$ can't be in $p^nR$ for all $n$. There must be a least such n by the well-ordering principle. We have confirmed that $\phi$ is well-defined.
Do we have an Euclidean norm though? Let $a$ be a unit. Then $a \notin pR$, so $\phi(a) = 1$, and for $b \neq 0$ we clearly have $\phi(b) \geq \phi(a)$.
Anyway, let $a \neq 0, b \neq 0$. Further, let $\phi(a) = m+1$, and let $\phi(b) = n+1$. Then $a = p^mr$ for $m \in \Bbb N$ (or $m = 0$)and $r \in R$. We have $b= p^ns$ for $n \in \Bbb N$ (or $n = 0$) and $s \in R$. Let $p^0$ mean $1$.
$a \mid b$ means $p^mr \mid p^ns$. Then $m \gt n$ gives us $p^{m-n}r \mid s$ $\implies$ $b = p^np^{m-n}rk$ for some $k \in R$, which is basically saying that $b = p^mrk$, but $b$ can't be in $p^mR$, because $p^mR \subseteq p^{n+1}R$. Therefore, $\phi(a) \leq \phi(b)$
Property (2): Let $a, b \in R, a \neq 0$. Let $\phi(a) = n$. We have $a = p^{n-1}u$, where $u$ is a unit. If $u$ wasn't, it would have to be in pR [in a local ring, every element outside the maximal ideal is a unit], say $u = pr$, and we get $a = p^{n-1}pr = p^nr$; a contradiction.
So now, if $\phi(b) \lt n$, consider $b = a\cdot0 + b$. This is in accordance with (2). If $\phi(b) \geq n$, then $b = p^mx$ for some $x$ in $R$ and $m \in \Bbb N \cup 0$, where $m \geq n-1$, which means that $b$ is also of the form $p^{n-1}y$. We see that $b = au^{-1}y + 0 = p^{n-1}u \cdot u^{-1}y = p^{n-1}y = b$.
Of course, $\phi(0) \lt \phi(a)$ when $a \neq 0$.
It seems to me that property (2) is the only thing that requires there to be only one maximal ideal. The maximal ideal in a field is the $0$-ideal, generated by $0$, that is, let $p=0$ and write $0 = 0 \cdot R$. Now, any nonzero elements are not in this ideal, so their norm would all be $1$.
