# 数学代写|交换代数代写commutative algebra代考|MATH2322

#### Doug I. Jones

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## 数学代写|交换代数代写commutative algebra代考|Generic Case

What do we call the generic case, regarding a projective module with $n$ generators? We consider the ring
$$\mathbf{G}n=\mathbb{Z}\left[\left(f{i, j}\right){i, j \in[1 . . n]}\right] / \mathcal{G}_n,$$ where the $f{i, j}$ ‘s are indeterminates, $F$ is the matrix $\left(f_{i, j}\right){i, j \in \llbracket 1 . . n \rrbracket}$ and $\mathcal{G}_n$ is the ideal defined by the $n^2$ relations obtained when writing $F^2=F$. With coefficients in this ring $\mathbf{G}_n$, we have the matrix $F$ whose image in $\mathbf{G}_n^n$ is what deserves to be called the generic projective module with $n$ generators. Let us reuse the notations of Theorem 1.7 in this particular case. Saying that $r_h r_k=0$ in $\mathbf{G}_n$ (for $0 \leqslant h \neq k \leqslant n$ ) signifies that we have a membership $$r_h(F) r_k(F) \in \mathcal{G}_n$$ in the ring $\mathbb{Z}\left[\left(f{i, j}\right)_{i, j \in[1 . . n \rrbracket}\right]$. This implies an algebraic identity that allows us to express this membership. This algebraic identity is naturally valid in all the commutative rings. It is therefore clear that if the membership (*) is satisfied in the generic case, it implies $r_h r_k=0$ for any projection matrix over an arbitrary commutative ring.
The same holds for the equalities $r_h u=0$ when $u$ is a minor of order $h+1$.
In short: if Theorem 1.7 is satisfied in the generic case, it is satisfied in every case. As is often the case, we therefore observe that important theorems of commutative algebra do nothing other than affirm the existence of certain particular types of algebraic identities.

## 数学代写|交换代数代写commutative algebra代考|The Semiring of Ranks

Recall that we say that a polynomial $R(X)=r_0+r_1 X+\cdots+r_n X^n$ is multiplicative when $R(1)=1$ and $R(X Y)=R(X) R(Y)$. It amounts to the same to say that the $r_i$ ‘s form a fundamental system of orthogonal idempotents, or that $R(X)$ is the rank polynomial of a finitely generated projective module.
2.1 Notations We denote by $\mathrm{H}0^{+}(\mathbf{A})$ the set of isomorphism classes of quasi-free modules over $\mathbf{A}$, and $[P]{\mathrm{H}0^{+}(\mathbf{A})}$ (or $[P]{\mathbf{A}}$, or even $\left.[P]\right)$ the class of such a module $P$ in $\mathrm{H}0^{+}(\mathbf{A})$. The set $\mathrm{H}_0^{+}(\mathbf{A})$ is equipped with a semiring structure ${ }^1$ for the inherited laws of $\oplus$ and $\otimes:[P \oplus Q]=[P]+[Q]$ and $[P \otimes Q]=[P] \cdot[Q]$. For an idempotent $e$ we will also write $[e]$ instead of $[e \mathbf{A}]$, when the context is clear. The neutral element for the multiplication is [1]. Every quasi-free module $P$ is isomorphic to a unique module ${ }^2$ $$\left(r_1 \mathbf{A}\right) \oplus\left(r_2 \mathbf{A}\right)^2 \oplus \cdots \oplus\left(r_n \mathbf{A}\right)^n,$$ where the $r_i$ ‘s are orthogonal idempotents, because then $\mathrm{e}_i(P)=r_i$. We therefore have $=[P]=\sum{k=1}^n k\left[r_k\right]$ and its rank polynomial is
$$\mathrm{R}P(X)=r_0+r_1 X+\cdots+r_n X^n$$ with $r_0=1-\left(r_1+\cdots+r_n\right)$. But while $\mathrm{R}{P \oplus Q}=\mathrm{R}P \mathrm{R}_Q$, we have $[P \oplus Q]=[P]+[Q]$ : this assures the passage from the multiplicative notation to the additive notation. Thus the “logarithm in base $X$ ” of the multiplicative polynomial $r_0+r_1 X+\cdots+r_n X^n$ is defined as the element $\sum{k=1}^n k\left[r_k\right]$ of $\mathrm{H}0^{+}(\mathbf{A})$. 2.2 Definition If $M$ is a finitely generated projective $\mathbf{A}$-module we call (generalized) rank and we denote by $\operatorname{rk}{\mathbf{A}}(M)$ or $\operatorname{rk}(M)$ the unique element of $\mathrm{H}_0^{+}(\mathbf{A})$ which has the same rank polynomial.

# 交换代数代考

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