### 数学代写|交换代数代写commutative algebra代考|MATH3033

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## 数学代写|交换代数代写commutative algebra代考|Polynomials and finitely generated algebras

If, in addition, $\mathfrak{E}=\left{\mathrm{s}{1}, \ldots, \mathrm{s}{n}\right}$ happens to be a finite set, one sets $R[E]=R\left[\mathrm{~s}{1}, \ldots, \mathrm{s}{n}\right]$, a mnemonic déja vu of the polynomial ring in $n$ indeterminates; in this case, $R\left[s_{1}, \ldots\right.$, $\left.s_{n}\right]$ is said to he finitely generated (or of finite type) nver $R$.

The following statement is also adaptable for infinitely generated algebras, but the use in this book is mainly in the finitely generated case.

Proposition 1.2.2. Let $R \subset S$ be an R-algebra of finite type. Then there is an R-isomorphism $R\left[X_{1}, \ldots, X_{n}\right] / I \simeq S$, for a suitable ideal I of the polynomial ring $R\left[X_{1}, \ldots, X_{n}\right]$.
The proof is an immediate consequence of the universal property of the polynomial ring $R\left[X_{1}, \ldots, X_{n}\right]$ and of the first theorem of the homomorphism for rings ( $c f$. Proposition 1.1.2).

A surjective homomorphism as in Proposition 1.2.2 and its kernel are respectively called a polynomial presentation and a presentation ideal of the $R$-algebra $S$. An alternative terminology for the presentation ideal is ideal of relations. It is understood that these notions are not uniquely defined by the algebra itself as they depend on the choice of a set of generators.

A remarkable case is that of an $R$-subalgebra of the polynomial ring $R\left[X_{1}, \ldots, X_{n}\right]$. Even in the case where $R=k$ is a field, the richness of the structure of the $k$-subalgebras is anything but easily understood. At first sight, a finitely generated $k$-subalgebra of $k\left[X_{1}, \ldots, X_{n}\right]$ resembles any other integral domain of finite type over $k$. However, this resemblance is misleading since $e_{x} g_{s}$, there are cases when such an algebra may turn up to be isomorphic to the homogeneous coordinate ring of a so-called unirational projective variety.

## 数学代写|交换代数代写commutative algebra代考|The transcendence degree

In this part, one focus on integral domains of finite type over a field $k$. The results of this subsection are independent from the characteristic of the base field $k$. Since no other base ring will come up other than $k$ itself, one will denote a $k$-algebra by the letter $R$ (instead of $S$ ).

One uses freely the following notation, originally conceived by Kronecker and rigorously set by Steinitz ([148]) more than one century ago: if $K \mid k$ is a field extension, i. e., $k$ is a subfield of the field $K$, and $\mathfrak{X} \subset K$ is a subset, then $k(\mathfrak{X})$ denotes the smallest inclusionwise subfield of $K$ containing $k$ and $\mathfrak{X}$. If $\mathfrak{X}$ consists of a single element $x$ one writes $k(x)$ for short.

Recall that such an element $x$ is said to be algebraic over $k$ provided it is a root of a nonzero polynomial in $k[X]$. The extension $K \mid k$ is algebraic if all of its elements are algebraic over $k$.

Given a field extension $K \mid k$, the algebraic closure of $k$ in $K$ is the set of elements of $K$ which are algebraic over $k$. By the elementary theory of algebraic elements in a field extension, one knows that this is an intermediate (“Zwischenkörper” in the terminology of Steinitz) field between $k$ and $K$. For lack of better notation, it is usually denoted by $\bar{k}$ if the ambient field $K$ is fixed in the discussion. This construction has the formal properties of a closure operator; in particular, taking the closure of a closure does not do anything, i. e., $\overline{(\bar{k})}=\bar{k}$. One says that $k$ is algebraically closed in $K$ if $\bar{k}=k$.
In this book, one assumes the elementary theory of algebraic extensions, a topic that is part of a general algebraic training no matter how tricky parts of Galois theory maybe (specially in prime characteristic), whereas the main focus in this part is the transcendental side of field theory in its relation to the underlying ring theoretic aspects.

Thus, let $R$ stand for an integral domain of finite type over $k$. Let $K$ denote the field of fractions of $R$. The resulting inclusion $k \subset K$ makes $K$ into a finitely generated field extension $K \mid k$ : a finite set of generators of $R$ over $k$ will generate $K$ as a field extension of $k$ as well.

Given a field extension $K \mid k$, a finite subset $\mathfrak{X}=\left{x_{1}, \ldots, x_{n}\right} \subset K$ is said to be algebraically independent over $k$ if the surjective $k$-homomorphism $k\left[X_{1}, \ldots, X_{n}\right] \rightarrow k[\mathfrak{X}]$ mapping $X_{i}$ to $x_{i}(1 \leq i \leq n)$ is injective.

Though this definition sounds repetitive, as it asserts that an algebraically independent set is essentially a set of indeterminates, its role will become clear as one progresses in the theory. This notion can be extended to possibly infinite sets by requiring that every finite subset have the property.

The next notion plays for finitely generated field extensions a similar role as a vector basis does for vector spaces-in fact, both are particular cases of a more general matroid theory phenomenon, but one will refrain from bringing it up here. As finitely generated extensions include finite extensions as a special case, one must allow for the new notion to encode this flexibility. The most general statement goes like the following.

## 数学代写|交换代数代写commutative algebra代考|Basic properties of the transcendence degree

However difficult recognizing whether a certain set is algebraically independent, there are some basic steps that come to help.

Proposition 1.2.8 (Modding out irreducible polynomials). Let $B=k\left[X_{1}, \ldots, X_{n}\right]$ be $a$ polynomial ring over a field $k$ and let $f \in B$ denote a nonzero irreducible polynomial. Then $\operatorname{trdeg}_{k}(B /(f))=n-1$.

Proof. First, $\operatorname{trdeg}{k}(B)=n$ since $\left{X{1}, \ldots, X_{n}\right}$ is a transcendence basis of $B$ over $k$. Write $f=\sum_{j=0}^{m} f_{j}\left(X_{1}, \ldots, X_{n-1}\right) X_{n}^{j}$. One can assume that $m>0$ and $f_{m}\left(X_{1}, \ldots, X_{n-1}\right) \neq 0$ (how?). Let $x_{i}$ denote the class of $X_{i}$ modulo $(f)$. Then $\sum_{j=0}^{m} f_{j}\left(x_{1}, \ldots, x_{n-1}\right) x_{n}^{j}=0$, showing that $\operatorname{trdeg}{k}(B /(f)) \leq n-1$. On the other hand, $\left{x{1}, \ldots, x_{n-1}\right}$ is algebraically independent over $k$. Indeed, otherwise an equation of algebraic dependence would yield a nonzero polynomial $g \in k\left[X_{1}, \ldots, X_{n-1}\right]$ such that $g \in(f)$, which is absurd since $f$ has a nonzero term involving $X_{n}$.

The preceding proposition has no obvious generalization to arbitrary prime ideals. However, one can state the following weak version.

Proposition 1.2.9 (Going modulo a prime ideal). Let $B$ be a finitely generated domain over a field $k$ and let $P \subset B$ be a prime ideal. Then $\operatorname{trdeg}{k}(B / P) \leq \operatorname{trdeg}{k}(B)$, with equality (if and) only if $P={0}$.

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