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The mod-$p$ Iwasawa algebra is the completed group algebrawhere $U$ runs through open normal subgroups and the transition maps come from quotient maps. This is the mod-$p$ algebra, distinguished from the integral Iwasawa algebra ${\mathbb{Z}}_p[[K]]$.

Use the stated complete separated negative filtered ring structure, writing $F_{-n}={\mathfrak{m}}^n$ and $\nu (x)=\max \{n:x\in F_{-n}\}$ for a nonzero element. In the usual odd-prime congruence-group situation,The dimension-three assertion records the three independent entries of a trace-zero matrix; the argument below only needs the given graded polynomial-algebra property. Completeness and separatedness are with respect to this filtration and follow from the inverse-limit topology for the standard filtration. If $x,y\ne 0$, their leading symbols are nonzero, so their product in the graded domain is nonzero. Therefore $\nu (xy)=\nu (x)+\nu (y)$ and $xy\ne 0$. This proves the domain assertion by filtered-graded transfer for complete rings.

For a left ideal $I$, its leading symbols form a homogeneous left ideal $\mathrm{gr}I$ of the polynomial ring. The Hilbert basis theorem gives finitely many homogeneous generators; choose lifts $a_1,\dots ,a_s\in I$. Given $x\in I$, express its leading symbol using these generators, lift the homogeneous coefficients, and subtract the resulting $\sum r_j{a}_j$. The remainder is still in $I$ and has strictly larger valuation. Repeat. At a stage where the remainder has valuation $n$, the coefficient contribution for $a_j$ has valuation at least $n-\nu (a_j)$. Thus each accumulated coefficient is a Cauchy series, converging to an element of ${\Omega }_K$. The remainder tends to zero, and separatedness gives $x={\sum }_j{r}_j{a}_j$. Hence $I$ is algebraically generated by these finitely many lifts; no prior assumption that $I$ is closed was used. This proves $\boxed{{\Omega }_K\text{ is left Noetherian and a domain}}$ under the stated filtration hypothesis. The same proof for right ideals also gives right Noetherianity.

For the classical fractions, a left Noetherian domain satisfies the left Ore condition. Here is the common-multiple argument. If two nonzero principal left ideals $Ra,Rb$ had zero intersection, the ideals $Ra{b}^i$, $i\ge 0$, would form a direct sum: from a finite relation, its first term lies in $Ra\cap Rb$, and then cancellation of the nonzero $b$ proves the assertion inductively. Their partial sums give an infinite ascending chain, impossible in a left Noetherian ring. Thus $Ra\cap Rb\ne 0$, giving common left multiples for every two nonzero denominators. The right-handed argument also applies. Ore localization at all nonzero elements therefore produces the classical division ring of fractions, with expressions $s^{-1}a$ or $b{s}^{-1}$; every nonzero fraction is invertible.

There is a genuine prime restriction behind the auxiliary hypothesis. For $p=2$, the matrix $z=-I$ is a nonidentity element of $K$ of order two. In ${\mathbb{F}}_2[[K]]$, the nonzero element $z-1$ satisfies $(z-1{)}^2=z^2-2z+1=0$. It is nonzero already in a finite quotient separating $z$ from one. Thus the actual $p=2$ algebra is not a domain, and cannot have the supplied separated graded-domain filtration. The intended conclusion applies for odd $p$; at $p=2$ it instead applies to a suitable deeper torsion-free congruence subgroup, such as the kernel modulo four.