一点代数学(可能长期更新)

Aries

Rings, ideals, modules

In this note, we will not recall basic concepts. The definition of ring, ideal,module, homomorphism of rings are not reproduced. In this course, any ring A always be:
$$\begin{gathered} 1.commutative:x\cdot y=y\cdot x\text{ for any }x,y\in A \\ 2.associative:x(yz)=(xy)z\text{ for any }x,y,z\in A \\ 3.unotary:\text{there exists }1\in A\text{ such that }1\cdot x=x\text{ for any }x\in A. \end{gathered}$$
So,for any ring homomorphism $f:A\to B$ preserves the units:$f(1)=1$.We denote by $A^{\times }$ the multiplicative group of invertible elements of A.

We will say that an element $a\in A$ is
nilpotent:if there exist n∈Nsuchthatan=0divisorofzero:if there is b∈A\{0}such that ab=0regular:if it is not a divider of zero. nilpotent:\text{if there exist n}\in \mathbb{N} such that a^{n}=0\\ divisor of zero:\text{if there is b}\in A \backslash\{0\} \text{such that ab=0}\\ regular:\text{if it is not a divider of zero}. nilpotent:if there exist n∈Nsuchthatan=0divisorofzero:if there is b∈A\{0}such that ab=0regular:if it is not a divider of zero.
We say that A is integral (resp. Is a field) if A≠{0}A \neq\{0\}A​={0} and if any non-zero element of A is regular (resp. Invertible).(The latter conditions are more important)

An ideal $I\subset A$ is principal if there exists $a\in A$ such that I=Aa:{ab:b∈A}I=Aa:\{ab:b\in A\}I=Aa:{ab:b∈A}.We say that $I$ is of finite type, if it exits $a_1,a_2,\dots ,a_n\in A$(for some $n\in \mathbb{N}$) such that $I=A{a}_1+A{a}_2+\cdots +A{a}_n$;in this case, the quotient ring $A/I$ is often denoted :
$$A/\left(a_1,\dots ,a_n\right).$$
We say that A is principal if all its ideals are principal.

Example 1.1 (1) Ring $\mathbb{Z}$ is principal;to show this we must show that any ideal $I\subset \mathbb{Z}$ is principal. if $I=0$,this is trivial, and if not, let $a\in I$ be the smallest element which >0.For all $b\in I$ there exist $q,r\in \mathbb{Z}$ such that $b=aq+r$ and $0\le r\le a$; we can see $r\in I$,therefore $r=0$, by the minimality of $a$, hence $I=a\mathbb{Z}$.

(2) if $K$ is a field, the same argument applies to the ring of polynomials $K[X]$:for any non-zero ideal $I\subset K[X]$ we choose $p(X)\in I$ which is non zero of minimal degree polynomial. If $b(X)\in I$,by Euclidean division, we can give $q(X),r(X)\in K[X]$ such that $b(X)=p(X)\cdot q(X)+r(X)$ with either $r(X)=0$ or $de{g}_{X}r(X)<de{g}_{X}p(X)$. But $r(X)\in I$,so finally $r(X)=0$ by the minimality of $de{g}_{X}p(X)$, hence $I=p(X)\cdot K[X]$.

Algebras

An $A-algebra$ is dual $(B,f)$ consisting of a ring $B$ and a homomorphism of rings $f:A\to B$, called the structural morphism of $B$. A homomorphism of $A-algebra$
$$g:(B,f)\to \left(B^{\prime },f^{\prime }\right)$$
is a homomorphism of rings $g:B\to B^{\prime }$ switching the diagram

Obviously, the composition of two homomorphism of $A-algebra$ $g:B\to B^{\prime }$ and $g^{\prime }:B^{\prime }\to B^{\prime \prime }$ is homomorphism of $A-algebra$ $g^{\prime }\circ g:B\to B^{\prime \prime }$. We denote by
$$A-Alg(B,B^{\prime })$$
the set of homomorphism of $A-algebra$ $B\to B^{\prime }$.

For example, for all $n\in \mathbb{N}$, the ring of polynomials of $n$ variables $A[X_1,\dots ,X_n]$ with coefficients in $A$ is provided with a canonical structure of $A-algebra$, whose structural morphism is the natural inclusion $A\to A[X_1,\dots ,X_n]$ which identifies A with the sub-ring of polynomials of total degree 0. Also, for all ideal $I\subset A$, the canonical projection $A\to A/I$ provides the quotient ring $A/I$ with a natural structure of $A-algebra$.

Remark 1.2 (1) Every ring A admits a unique homomorphism $\mathbb{Z}\to A$, therefore every ring is canonically a $\mathbb{Z}-algebra$.
(2) Let $A$ be a ring, $B$ is an $A-algebra$, $n\in \mathbb{N}$, and $(b_1,\dots ,b_n)\in B^n$. Note that there is a unique homomorphism of $A-algebra$ $f:A[X_1,\dots ,X_n]\to B$ such that $f(X_i)=b_i$ for $i=1,\dots ,n$:this is the homomorphism defined by
$$f(P):=P(b_1,\dots ,b_n)\forall P\in A[X_1,\dots ,X_n].$$
In other words, for all $A-algebra$ $B$ and all $n\in \mathbb{N}$ there is a natural bijection
$$B^n\stackrel{\sim }{\to }A-\mathrm{Alg}\left(A\left[X_1,\dots ,X_n\right],B\right).$$
We will see in section 2.2 how this property characterizes $A[X_1,\dots ,X_n]$ except for canonical isomorphism.

We say that an $A-algebra$ $B$ is ring of finite type, if there exists a surjective homomorphism of $A-algebra$ $\pi :A[X_1,\dots ,X_n]\to B$, for some $n\in \mathbb{N}$. In view of remark 1.2(2), this amounts to saying that there exists a finite system $b_{\bullet }:=\left(b_1,\dots ,b_n\right)$ of elements of $B$ such that all $b\in B$ is written under the form $b=P(b_1,\dots ,b_n)$ for some polynomial $P\in A[X_1,\dots ,X_n]$; We say that $b_{\bullet }$ is a finite system of generators of the $A-algebra$ $B$, and we also write $B=A[b_1,\dots ,b_n]$. We say that $B$ is an Algebra of finite presentation, if one can find a $\pi$ surjection as above, whose kernel ${\pi }^{-1}(0)$ is an ideal of finite type; In this case $B$ is isomorphic to a quotient $A[X_1,\dots ,X_n]/I$, with $I\subset A[X_1,\dots ,X_n]$ an ideal of finite type. if $(B,f)$ is an $A-algebra$ of finite type.(resp. finite presentation), we also say that $f$ is a homomorphism of rings of finite type (resp. finite presentation).

Exercise 1.3 If $A\stackrel{f}{\to }B\stackrel{g}{\to }C$ are two homomorphisms of rings of finite type(resp. finite presentation), show that it is the same for $g\circ f$.

Field of fractions

For any integral ring A we denote by
$$FracA$$
the field of fractions of A. Recall that this is the set of fractions $a/b$ (also noted $b^{-1}a$), i.e. the equivalence classes of couples $(a,b)$ with $a,b\in A$ and $b\ne 0$; two such couples $(a,b),(a^{\prime },b^{\prime })$ are equivalent $\iff a{b}^{\prime }=a^{\prime }b$. We define the laws of addition and multiplication of $FracA$ by the obvious formulas:
$$a/b+c/d:=(ad+cb)/bda/b\cdot c/d:=ac/bd.$$
We can easily verify that these laws do not depend on the choice of the representatives $(a,b),(c,d)$ for $a/b,c/d$; the identity elements of addition and multiplication are $0/1$ and $1/1$, so that $-(a/b)=(-a)/b$ and $(a/b{)}^{-1}=b/a$ if $a\ne 0$. Furthermore, we have the injective homomorphism of $A\to FracA:a\to a/1$;thus, $FracA$ is the smallest field containing $A$, which is unique except for isomorphism. We leave the details to the reader, as we will see a more general construction in section 2.3.

Prime and maximum ideals

We recall that an ideal $I\subset A$ is said:
$$\begin{gathered} \text{prime}:\text{if}1\notin I\text{ and }x,y\notin I\Rightarrow xy\notin I\text{ for all }x,y\in A \\ \text{maximal}:if1\notin I\text{and the only ideals of A containing I are I and A.} \end{gathered}$$
Proposition 1.4. Let $A$ be a ring, $I\subset A$ an ideal. We have:
(i) I is prime if and only if $A/I$ is an integral ring.
(ii) I is maximal if and only if $A/I$ is a field.
(iii) Any maximum ideal is prime.

Demonstration: (i) Let $x,y\in A$, and denote by $\bar{x},\bar{y}\in A/I$ the classes of $x$ and $y$.if $\bar{x},\bar{y}\ne 0$, we have $x,y\notin I$;if now $I$ is prime, we deduce $xy\notin I$,and therefore $\bar{x}\cdot \bar{y}\ne 0$, which shows that $A/I$ is integral. Conversely, if $A/I$ is integral, we have $\bar{x}\cdot \bar{y}\ne 0$, i.e. $xy\notin I$, and then $I$ is prime.
(ii) Let $x\in A$ such that $x\notin I$, therefore $\bar{x}\ne 0$. if $A/I$ is a field, there exists $y\in A$ such that $\bar{x}\cdot \bar{y}=1$ in $A/I$, therefore $xy-1\in I$, hence $I+Ax=A$; (Note that the sum of any two ideals is an ideal. )Since $x$ is arbitrary, we deduce that the only ideals that contain $I$ are $I$ and $A$, i.e. $I$ is maximal. On the other hand, if $I$ is maximal, the hypothesis $x\notin I$ Implies that we have $I+Ax=A$, so there exists $a\in I,y\in A$ such that $xy+a=1$, hence $\bar{x}\cdot \bar{y}=1$, which shows that $A/I$ is a field. Assertion (iii) follows Immediately from (i) and (ii).

We note:
Max$A$: the set of maximum ideals of $A$(maximum spectrum of A);
Spec$A$: the set of prime ideals of $A$(prime spectrum of $A$).
One of the objectives of this course is to explain why Max$A$ and Spec$A$ are ‘geometric objects’. According to Proposition 1.4(iii), we have: Max$A\subset$Spec$A$.

Exercise 1.5. Let $A$ be an integral principal ring. Show that Spec$A$={0}$\bigcup$Max$A$.

This follow prime ideals in integral principal ring is maximal.

Lemma 1.6 If $I\subset A$ is an ideal, we have a canonical bijection:
{ideals $J$ of $A$ such that $I\subset J$}$⟷${ideals of $A/I$} ($J\subset A$)$\to$($J/I\subset A/I$).
This bijection induced by restriction of bijections:
{p∈Spec⁡A∣I⊂p}⟷Spec⁡A/I{m∈Max⁡A∣I⊂m}⟷Max⁡A/I. \{\mathfrak{p} \in \operatorname{Spec} A | I \subset \mathfrak{p}\} \longleftrightarrow \operatorname{Spec} A / I \quad\{\mathfrak{m} \in \operatorname{Max} A | I \subset \mathfrak{m}\} \longleftrightarrow \operatorname{Max} A / I. {p∈SpecA∣I⊂p}⟷SpecA/I{m∈MaxA∣I⊂m}⟷MaxA/I.
Demonstration: Let $\pi :A\to A/I$ be the canonical projection; the reciprocal bijection associates with any ideal $\bar J $of$ A /I$, the ideal ${\pi }^{-1}(\bar{J})\subset A$. If $\mathfrak{p}$ is an ideal of $A$ and $I\subset \mathfrak{p}$, we have $A/\mathfrak{p}=(A/I)/(\mathfrak{p}/I)$, therefore $A/\mathfrak{p}$ is integral (resp. a field) if and only if $(A/I)/(\mathfrak{p}/I)$ is integral (resp. a filed), and with proposition 1.4 we deduce that $\mathfrak{p}$ is prime (resp. maximal) in $A$ if and only if $\mathfrak{p}/I$ is prime (resp. maximum) in $A/I$.

Definition 1.7. Let $A$ be a ring, $a\in A$ non-zero element.
(i)We say that $a$ is prime if the ideal $Aa$ is prime.
(ii)We say that $a$ is reducible, if it is the product of two non-reducible and non-invertible.
(iii)We say that $A$ is factorial if it is integral and if any non-zero and non-invertible element of $A$ is written as a product of prime elements.

Remark 1.8 (i) Let $A$ be a ring. In the following, we will use the notation $\cdot |\cdot$ for the divisiblity relation in $A$: therefore, $a|b$ means that $a,b\in A$ and $b\in Aa$.
(ii) if A is factorial and $a\in A\backslash 0$, the factorization $a=u{p}_1\cdots p_t$ with $u\in A^{\times }$ and $p_1,\cdots ,p_t$ prime elements is essentially unique:if $a=v{q}_1\cdots q_s$ is a other factorization, we have $s=t$ and there exists a premutation σ:{1,…,t}→∼{1,…,t}\sigma:\{1, \ldots, t\} \stackrel{\sim}{\rightarrow}\{1, \ldots, t\}σ:{1,…,t}→∼{1,…,t} such that $p_i^{-1}{q}_{\sigma (i)}\in A^{\times }$ for $i=1,\dots ,t$. For the proof, we reason by induction on $t$: we have $t=0$ if and only if $a\in A^{\times }$, and in this case it is clear that $s=0$. Suppose that $t\ge 1$ and that the uniqueness of the factors is already known for the products of $t-1$ prime elements; since the ideal $A{p}_1$ is prime, we have $p_1|q_i$ for some $i\le s$, and leaves to permute the factors we can assume that $i=1$. There exists $u_1\in A$ such that $q_1=p_1{u}_1$, hence $u_1\in A^{\times }$ and $a^{\prime }:=u{p}_2\cdots p_t=u_{1}v\cdot q_2\cdots q_s$; by induction hypothesis, we have the uniqueness of the factorization of $a^{\prime }$ with premutation of the factors close, hence the same foa $a$.
(iii) Let a,b,d∈A\{0}a,b,d\in A\backslash \{0\}a,b,d∈A\{0}; we say that $d$ is a greatest common divisor (abbreviated gcd) of $a$ and $b$, if $a,b\in dA$ and if for all $c\in A$ with $a,b\in cA$, we have $d\in cA$. Symmetrically, $e\in A$ is a smaller common multiple (abbreviated lcm) of $a$ and $b$ if $e\in Aa\bigcap Ab$ and if for all $x\in Aa\bigcap Ab$ we have $x\in Ae$;i.e. $Ae=Aa\bigcap Ab$. Note that if $A$ is integral, the gcd and the lcm of $a$ and $b$, when they exist, are determined to the nearest multiplication of invertible elements: because if $d$ and $d^{\prime }$ are gcd of $a$ and $b$, we have $d^{\prime }|d$ and $d|d^{\prime }$, i.e. $d=d^{\prime }u,d^{\prime }=dv$ for some $u,v\in A$ hence $d=duv$, and therefore $uv=1$, because $A$ is integral; we reason the same for the lcm.
(iv) if $ A$ is integral and if lcm $(a,b)$ exists, then lcm $(a,b)$ exists, and we have:
$$gcd(a,b)\cdot lcm(a,b)=ab$$
with multiplication of invertible elements near, i.e. if $e$ is a lcm of $a$ and $b$, then $e^{-1}ab\in A$ is gcd of $a$ and $b$: indeed, there exists $d\in A$ such that $ab=ed,$ and since on the one hand $a|e$, it follows that $d|b$, and on the other hand, $b|e$, therefore $d|a$ too. Now, if $c$ divides $a$ and $b$, say $a=cx,b=cy,$ we see that $cxy$ is a common multiple of $a$ and $b$. therefore $e|cxy$, and of $ed=c\cdot cxy$ it comes that $c|d$, as wished. (On the other hand, the existence of the gcd does not entail that of the lcm: see Remark 1.111 ).
(v) if $A$ is factorial, any pair $(a,b)$ of non-zero elements of $A$ admits a gcd and lcm. Indeed, thanks to (iv) it suffices to show the existence of the lcm; so let $a=u{p}_1^{{\nu }_1}\cdots p_k^{{\nu }_k}$ and $b=v{p}_1^{{\mu }_1}\cdots p_k^{{\mu }_k}$ factorization with $u,v\in A^{\times }$, $v_i,u_i\in \mathbb{N}$ and $p_i$ for all $i=1,\dots ,k$, with $A{p}_i\ne A{p}_j$ for $i\ne j$. Given (iii), we can easily see that ${\prod }_{i=1}^k{p}_i^{\max \left({\nu }_i,{\mu }_i\right)}$ is a lcm of $a$ and $b$; it is also deduced that ${\prod }_{i=1}^k{p}_i^{\min \left({\nu }_i,{\mu }_i\right)}$ is a gcd of $a$ and $b$: the details are left to the reader.

Exercise 1.9. (i) Show that every integral and main ring is factorial.
(ii) Show that the ring $\mathbb{Z}[\sqrt{-5}]$ is not factorial.
(iii) Let $A$ be an integral ring; show that any pair of elements of A\{0}A\backslash \{0\}A\{0} admits a gcd $\iff$ any pair of elements of A\{0}A\backslash \{0\}A\{0} admits a lcm.

With exercise 1.9 (i) and example 1.1 we see that $\mathbb{Z}$ is factorial, and similarly for $K[X]$, if $K$ is an arbitrary field. The prime elements of $\mathbb{Z}$ are obviously the usual integers; the first $K[X]$ are the irreducible polynomials, i.e. the $P\in K[X]$ with $d:=de{g}_{X}P>0$, which are not products of polynomials of $degrees<d$. In both cases, the key point of the proof is the existence of a Euclidean division for any pair of non-zero elements; the following problem axiomatizes the properties required to reproduce this argument:

Problem 1.10. Ring $A$ is said to be Euclidean if it is integral and there is an application
∣⋅∣:A\{0}→N |\cdot|: A \backslash\{0\} \rightarrow \mathbb{N} ∣⋅∣:A\{0}→N
satisfying the following condition:

–for all $a, b \in A \backslash {0} $ there exists $q,r\in A $ such that $a = bq + r $ and either $r=0$, or $|r|<|b|$ (Euclidean division of $a$ by $b$).
(i) Show that any Euclidean ring is principal (and therefore, factorial).
(ii) Show that $\mathbb{Z}[i\sqrt{n}]$ is a Euclidean ring for $n=1,2$.
(iii) The rest of the problem is devoted to a classical arithmetic application of the ring $\mathbb{Z}[i]$ of Gauss integers. First, let $p\in \mathbb{N}$ be a prime number with $p\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}4)$. Show that there exists $x\in \mathbb{Z}$ such that $x^2\equiv -1(\mathrm{m}\mathrm{o}\mathrm{d}p)$.
(iv) Deduce from (ii) and (iii) the following Fermat Theorem: Let $p\in \mathbb{N}$ be an odd prime number; then p is the sum of two squares $p=a^2+b^2$ of integers $a,b\in \mathbb{N}$ if and only if $p\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}4)$.