Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 7 / 3

Choose $a_1,\dots ,a_d\in \mathfrak{m}$ whose classes form a basis of the cotangent space of a local ring $\mathfrak{m}/{\mathfrak{m}}^2$, and put $J=(a_1,\dots ,a_d)$. Spanning gives $\mathfrak{m}=J+{\mathfrak{m}}^2$. Thus the quotient module $N=\mathfrak{m}/J$ satisfies $N=\mathfrak{m}N$. It is a finitely generated module, since $A$ is a Noetherian ring.

Here is the needed form of the Nakayama lemma, with proof. Choose generators $n_1,\dots ,n_r$ of $N$ and write $n_i={\sum }_j{c}_{ij}{n}_j$ with $c_{ij}\in \mathfrak{m}$. For the column vector $n$, $(I-C)n=0$. Multiply by the adjugate to obtain $\det (I-C)n=0$. Since $\det (I-C)\equiv 1(\mathrm{mod}\mathfrak{m})$, this determinant is a unit in the local ring $A$, so every $n_i=0$. Therefore $N=0$, proving $\mathfrak{m}=J$. Conversely, any generators of $\mathfrak{m}$ must span its cotangent space, so at least $d$ are needed. HenceHere $\mu (M)$ denotes the minimal number of generators of a module $M$.

For the additional minimum, the maximal ideal $\mathfrak{m}$ is itself $\mathfrak{m}$-primary, so it supplies an eligible ideal generated by $d$ elements. ConsequentlyThe minimum here is over primary ideals. A bound $\mu (Q)\le d$ for every $\mathfrak{m}$-primary ideal would be false. Indeed, for $A=k[[u,v]]$ and $\mathfrak{m}=(u,v)$, $d=2$ but $Q={\mathfrak{m}}^2=(u^2,uv,v^2)$ is primary and needs three generators. To verify primaryness, if $ab\in Q$ and $a\notin Q$, then $b$ cannot be a unit, so $b\in \mathfrak{m}$ and $b^2\in Q$. The three displayed monomials form a basis of $Q/\mathfrak{m}Q={\mathfrak{m}}^2/{\mathfrak{m}}^3$, proving the lower bound of three, while they already generate $Q$. This illustrates that primary ideals can need more generators than the maximal ideal.

A formal power series $f={\sum }_{n\ge 0}{a}_n{x}^n$ is a unit exactly when its constant coefficient is a unit in $A$. If $fg=1$, constant coefficients give $a_0{b}_0=1$. Conversely, for invertible $a_0$, setThen every positive-degree coefficient of $fg$ is zero and its constant coefficient is one, so $g=\sum b_n{x}^n$ is an inverse. This is a formal coefficient calculation and requires no convergence.

Since $A$ is a local ring, its nonunits are exactly $\mathfrak{m}$. The nonunits of $B$ are consequently the series with $a_0\in \mathfrak{m}$, which are exactly $I=\mathfrak{m}B+xB$: such a series is $a_0+xg$, and conversely an element of $I$ has constant coefficient in $\mathfrak{m}$. The map $B\to k$ taking the constant coefficient modulo $\mathfrak{m}$ is surjective with kernel $I$, so $I$ is a maximal ideal. Every proper ideal consists of nonunits and hence lies in $I$, proving that $I$ is the unique maximal ideal. ThusThis proves the description of formal power series over a local ring.

To compute the cotangent space of a formal power series ring, expandDefine the $k$-linear mapChanging a lift $a$ by ${\mathfrak{m}}^2$ or a lift $b$ by $\mathfrak{m}$ changes the image by $I^2$, so the map is well-defined. Every series in $I$ is congruent modulo $x^{2}B\subseteq I^2$ to its first two coefficients, proving surjectivity. If $a+bx\in I^2$, its constant coefficient lies in ${\mathfrak{m}}^2$ and its linear coefficient lies in $\mathfrak{m}$, so both input classes are zero. This proves injectivity. ThereforeFinally, part (a) gives generators $a_1,\dots ,a_d$ of $\mathfrak{m}$, so $I=(a_1,\dots ,a_d,x)$. Use the Krull height theorem in its lecture form: in a Noetherian ring, a prime minimal over an ideal generated by $r$ elements has height at most $r$. The problem allows Noetherianity of $B$ to be assumed. Apply the theorem to $I$ itself, whose height is the Krull dimension of the local ring $B$. Hence $\boxed{\dim B=\mathrm{ht}(I)\le d+1}$.

Since $A$ is a regular local ring, $\dim A=d$ is finite. Choose a chain of prime ideals of length $d$,Such a chain exists because the supremum defining this finite integer Krull dimension is attained; a maximal-length chain in a local ring ends at its maximal ideal.

For each $i$, the ideal ${\mathfrak{p}}_i$ is finitely generated because $A$ is Noetherian. This ensures that its extension ${\mathfrak{p}}_{i}B$ consists precisely of series all of whose coefficients lie in ${\mathfrak{p}}_i$: if ${\mathfrak{p}}_i=(b_1,\dots ,b_r)$, write each coefficient as ${\sum }_j{b}_j{c}_{jn}$ and gather them into the finitely many series ${\sum }_n{c}_{jn}{x}^n$. Thus coefficient reduction givesThe coefficient ring $A/{\mathfrak{p}}_i$ is an integral domain, and so is its formal power series ring: the first nonzero coefficient of a product is the product of the first nonzero coefficients of its factors. Hence each ${\mathfrak{p}}_{i}B$ is prime. Also ${\mathfrak{p}}_{i}B\cap A={\mathfrak{p}}_i$, so the strict inclusions persist upon extension. Finally, $x\notin \mathfrak{m}B$, since its image in $B/\mathfrak{m}B\cong k[[x]]$ is nonzero. We obtain the prime chainof length $d+1$. Therefore $\dim B\ge d+1$. Together with part (b),Its embedding dimension is also $d+1$, by part (b), so $B$ is itself a regular local ring. This proves that formal power series preserve regular local rings.