Lengths of truncated plane local rings

The classes of the monomials $x^i{y}^j$ with $i+j\le t-1$ form a $k$-basis of $R\mathrm{/}{\mathfrak{m}}^t$. Indeed every monomial of total degree $\ge t$ is a product of $t$ or more linear forms and so lies in ${\mathfrak{m}}^t$, and every element of ${\mathfrak{m}}^t$, expanded as a sum of products of elements of $\mathfrak{m}$, is a sum of monomials of degree $\ge t$; hence ${\mathfrak{m}}^t$ is exactly the $k$-span of the monomials of degree $\ge t$ and the displayed classes are a basis of the quotient. Consequently $R\mathrm{/}{\mathfrak{m}}^t$ is nonzero for $t\ge 1$ with ${\dim }_k(R\mathrm{/}{\mathfrak{m}}^t)=\mathrm{\#}\{(i,j):i+j\le t-1\}={\sum }_{d=0}^{t-1}(d+1)=\left(\genfrac{}{}{0px}{}{t+1}{2}\right)$, while $R\mathrm{/}{\mathfrak{m}}^0=R\mathrm{/}R=0$.

For $t\ge 1$ the ring $R\mathrm{/}{\mathfrak{m}}^t$ is local with unique maximal ideal $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^t$. Let $f\in R\setminus \mathfrak{m}$; writing $f=c+g$ with $c=f(0,0)\in k^{\times }$ and $g\in \mathfrak{m}$, the class of $g$ has $g^t\in {\mathfrak{m}}^t$, so the class of $f$ is $c$ times the class of $1+g\mathrm{/}c$, which is a unit with inverse the finite geometric sum ${\sum }_{i=0}^{t-1}(-g\mathrm{/}c{)}^i$. Thus every element outside $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^t$ is a unit; since a proper ideal contains no unit, every proper ideal of $R\mathrm{/}{\mathfrak{m}}^t$ is contained in $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^t$, which is therefore the unique maximal ideal.

For $t\ge 1$ the localisation map $\lambda :R\mathrm{/}{\mathfrak{m}}^t\to (R\mathrm{/}{\mathfrak{m}}^t{)}_{\mathfrak{m}\mathrm{/}{\mathfrak{m}}^t}$ is an isomorphism. It is surjective because a denominator outside $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^t$ is a unit by step 1.2, so $a\mathrm{/}s=a{s}^{-1}$; it is injective because $\lambda (a)=0$ means $ua=0$ for some $u\mathrm{\notin }\mathfrak{m}\mathrm{/}{\mathfrak{m}}^t$, and $u$ is a unit, whence $a=0$.

For $t\ge 1$, by [F3] applied to $R$, the ideal ${\mathfrak{m}}^t$ and the multiplicative set $R\setminus \mathfrak{m}$, there is a canonical isomorphism $(R\mathrm{/}{\mathfrak{m}}^t{)}_{\mathfrak{m}\mathrm{/}{\mathfrak{m}}^t}\cong O\mathrm{/}{\mathfrak{m}}^{t}O$. Composing with step 2.1 gives the required isomorphism $R\mathrm{/}{\mathfrak{m}}^t\to O\mathrm{/}{\mathfrak{m}}^{t}O$ for $t\ge 1$. For $t=0$, $R\mathrm{/}{\mathfrak{m}}^0=R\mathrm{/}R=0$ and $O\mathrm{/}{\mathfrak{m}}^{0}O=O\mathrm{/}O=0$, so the natural map is directly an isomorphism of zero rings.

Order the monomials of degree $\le t-1$ by decreasing total degree and let $M_j\subseteq R\mathrm{/}{\mathfrak{m}}^t$ be the $k$-span of the classes of the first $j$ monomials, so that $0=M_0<M_1<\cdots <M_N=R\mathrm{/}{\mathfrak{m}}^t$ with $N=\left(\genfrac{}{}{0px}{}{t+1}{2}\right)$. Multiplication by any element of $\mathfrak{m}$ raises total degree, hence sends each $M_j$ into $M_{j-1}$; therefore $\mathfrak{m}$ acts as $0$ on every quotient $M_j\mathrm{/}{M}_{j-1}$, and each quotient is a one-dimensional $k$-vector space, spanned by one monomial class of some degree $i+j=t-1,t-2,\dots ,0$. Transporting this chain through the ring isomorphism of step 3.1 gives a chain of $O$-submodules of $O\mathrm{/}{\mathfrak{m}}^{t}O$ whose successive quotients are one-dimensional $k$-vector spaces.

Each successive quotient in step 4.1 is annihilated by $\mathfrak{m}$ and is a one-dimensional $k$-vector space, hence simple as an $O$-module: an $O$-submodule would be a $k$-subspace, and there is no proper nonzero one. Therefore the transported chain is a composition series of $O\mathrm{/}{\mathfrak{m}}^{t}O$ over $O$ with $N=\left(\genfrac{}{}{0px}{}{t+1}{2}\right)$ factors, so ${\mathrm{\ell }}_O(O\mathrm{/}{\mathfrak{m}}^{t}O)=N$; since the same chain exhibits a $k$-basis, ${\dim }_k(O\mathrm{/}{\mathfrak{m}}^{t}O)=N$ as well, and the displayed factors are the one-dimensional spaces spanned by the monomials $x^i{y}^j$ with $i+j=t-1,\dots ,0$.