Let’s begin with commutative rings. Commutative rings are algebra structures which have good properties and close to some sets we have learned well, such as $\\mathbb {R}$ and $\\mathbb{Z}$.<\/p>\n
Definition 1.0.1<\/strong> $f:A\\to B,g:B\\to C$ and $h:A\\to C$ are functions. And the diagram<\/strong> of them is: Definition 1.0.2<\/strong> A binary operation<\/strong> on a set $R$ is a function $\\cdot:R\\times R\\to R,(a,b)\\mapsto a\\cdot b$. If the binary operation is multiplication, we always write it as $ab$ instead of $a\\cdot b$.<\/p>\n Definition 1.0.3<\/strong> $R$ is a set, and there are two binary operations, addition $+$ and multiplication $\\cdot$, on it, which satisfy:<\/p>\n And then we call $(R,+,\\cdot)$ a ring<\/strong>, usually denote it as $R$.<\/p>\n If $\\forall a\\in R^\\times:=R\\textbackslash\\{0\\}$, $a$ have an (multiplicative) inverse<\/strong> $b\\in R^\\times$, which means $ab=ba=1$, then we call $R$ is an division ring (we always denote $b$ as $a^{-1}$).<\/p>\n We should know that not every binary operation have associativity, such as the subtraction on $\\mathbb R$. $(a-b)-c=a-(b-c)$ is not always true.<\/p>\n Here is a diagram of associativity: Example 1.0.1<\/strong> Here are some examples of rings.<\/p>\n Proposition 1.0.1<\/strong> Let $R$ be a ring. Here are some trivial propositions about rings which I won’t prove in my note.<\/p>\n Definition 1.0.4<\/strong> Let $R$ be a ring. A subset $S$ is a subring<\/strong> of $R$ if<\/p>\n If $S$ is a proper subset of $R$ ($S\\ne R$), $S$ is a proper subring<\/strong> of $R$. It’s easy to show that $S$ is a ring.<\/p>\n Definition 1.0.5<\/strong> A ring $R$ is commutative<\/strong> if $ab=ba,\\forall a,b\\in R$.<\/p>\n The sets $\\mathbb {Z},\\mathbb Q,\\mathbb R$ and $\\mathbb C$ are commutative rings with the usual addition and multiplication. And then all the rings in the rest of this chapter are commutative<\/strong> unless we say otherwise.<\/p>\n Proposition 1.0.2 (Binomial Theorem)<\/strong> Let $R$ be a commutative ring. If $a,b\\in R$, then Example 1.0.2<\/strong> Here is an example of a commutative ring from set theory. <\/p>\n If $A,B\\subset X$, then their symmetric difference<\/strong> is $A+B=(A\\cup B)\\textbackslash(A\\cap B)$. And define $A\\cdot B=A\\cap B$. Then it’s not difficult to show $(P(X),+,\\cdot)$ is a commutative ring. And we tell it a Boolean ring<\/strong>.<\/p>\n We can find that $0=\\emptyset$ and $1=X$. Using Boolean ring we can prove the de Morgan law The familiar examples of commutative rings, $\\mathbb Z,\\mathbb Q,\\mathbb R,$ and $\\mathbb C$ are domains; the zero ring is not a domain.<\/p>\n Proposition 1.0.3<\/strong> A nonzero commutative ring $R$ is a domain if and only if $\\forall a,b\\in R,ab=0\\Rightarrow(a=0\\lor b=0)$.<\/p>\n Proof.<\/p>\n ($\\Rightarrow$).<\/p>\n Let’s assume that $a\\ne 0$.<\/p>\n Then $ab=0=0b\\Rightarrow b=0$.<\/p>\n<\/li>\n ($\\Leftarrow$).<\/p>\n $ab=bc\\Rightarrow (a-b)c=0$.<\/p>\n Because $c\\ne 0$, $a-b=0$. <\/p>\n So $a=b$. $\\square$<\/p>\n<\/li>\n<\/ul>\n Proposition 1.0.4<\/strong> If a nonzero commutative ring $R$ is a division ring, it is a domain.<\/p>\n Corollary 1.0.1<\/strong> The commutative ring $\\mathbb Z_p$ is a domain if and only if $p$ is a prime.<\/p>\n Definition 1.0.7<\/strong> Let $a$ and $b$ be elements of a commutative ring $R$. Then $a$ divides<\/strong> $b$ in $R$ (or $a$ is a divisor<\/strong> of $b$ or $b$ is a multiple<\/strong> of $a$), denoted by Definition 1.0.8<\/strong> An element $u$ in a commutative ring $R$ is called a unit<\/strong> or invertible element<\/strong> if $u\\mid 1$. In other words, $u$ has an inverse $u^{-1}$ in $R$.<\/p>\n Definition 1.0.9<\/strong> If $R$ is a nonzero commutative ring, then the group of units<\/strong> of $R$ is $U(R):=\\{\\text{all units in } R\\}$.<\/p>\n It’s easy to check that $U(R)$ is a multiplicative group.<\/p>\n Definition 1.0.10<\/strong> A nonzero commutative ring $R$ is a field, if $U(R)=R^\\times$.<\/p>\n Proposition 1.0.5<\/strong> Every field $F$ is a domain.<\/p>\n It’s easy to prove by using Proposition-1.0.4.<\/p>\n Theorem 1.0.1<\/strong> If $R$ is a domain, then there is a field containing $R$ as a subring.<\/p>\n Moreover, such a field $F$ can be chosen so that, for each $f\\in F$. there are $a,b\\in R$ with $b\\ne 0$ and $f=ab^{-1}$.<\/p>\n Proof.<\/p>\n Define a relation $\\equiv$ on $R\\times R^\\times$ by $(a,b)\\equiv (c,d)$ if $ad=bc$.<\/p>\n Then, let’s prove $\\equiv$ is an equivalence relation.<\/p>\n Denote the equivalence class of $(a,b)$ by $[a,b]$, define $F=R\/\\equiv$ and equip $F$ with the following addition and multiplication (if we pretend that $[a,b]$ is the fraction $a\/b$, then these are just the familiar formulas): Finally if $b\\ne 0$, then $[1,b]=[b,1]^{-1}$, and so $[a,b]=[a,1][b,1]^{-1}$. $\\square$<\/p>\n Definition 1.0.11<\/strong> The field $F$ constructed from $R$ in Theorem-1.0.1 is called the fraction field<\/strong> of $R$; we denote it by We can find that the fraction field of $\\mathbb Z$ is $\\mathbb Q$.<\/p>\n Definition 1.0.12<\/strong> A subfield<\/strong> of a field $K$ is a subring $k$ of $K$ that is also a field.<\/p>\n It’s easy to see that a subset $k$ of a field $K$ is a subfield if and only $k$ is a subring that is closed under inverses; that is, if $a\\in k^\\times$, then $a^{-1}\\in k^\\times$. It is also routine to see that any intersection of subfields of $K$ is itself a subfield of $K$.<\/p>\n","protected":false},"excerpt":{"rendered":" Let’s begin with commutative rings. Commutative r […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[12,6],"tags":[11,8],"class_list":["post-393","post","type-post","status-publish","format-standard","hentry","category-abstract-algebra","category-6","tag-11","tag-8"],"_links":{"self":[{"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts\/393","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=393"}],"version-history":[{"count":1,"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts\/393\/revisions"}],"predecessor-version":[{"id":394,"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts\/393\/revisions\/394"}],"wp:attachment":[{"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=393"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=393"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=393"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n$$\\xymatrix{
\nA \\ar@{->}[r]^{f} \\ar@{->}[rd]_{h} & B \\ar@{->}[d]^{g} \\\\
\n & C
\n}
\n$$
\nAnd a commutative diagram<\/strong> is a diagram such that all directed paths in the diagram with the same start and endpoints lead to the same result. If we say this diagram is commutative, it means $h=g\\circ f$.<\/p>\n\n
\n
\n$$\\xymatrix{
\nR\\times R\\times R \\ar@{->}[d]_{1\\times\\cdot} \\ar@{->}[rr] \\ar@{->}[rr] \\ar@{->}[rr]^{\\cdot\\times 1} & & R\\times R \\ar@{->}[d]^{\\cdot} \\\\
\nR\\times R \\ar@{->}[rr]^{\\cdot} & & R
\n}
\n$$
\nThe function $\\cdot\\times 1:R\\times R\\times R\\to R\\times R$ is defined by $(a,b,c)\\mapsto (ab,c)$ while $1\\times\\cdot:R\\times R\\times R\\to R\\times R$ is defined by $(a,b,c)\\mapsto(a,bc)$. Associativity says that this diagram is commutative, and in other words the two composite functions $R\\times R\\times R$ are equal.<\/p>\n\n
\n
\n
\n$$(a+b)^n=\\sum_{r=0}^n C_n^ra^rb^{n-r}
\n$$
\nAnd by using mathematical induction, it’s easy to prove.<\/p>\n
\n$$(A\\cup B)^c=A^c\\cap B^c
\n$$
\nTo prove it using set-theoretic methods needs too much on the meaning of the words and<\/em>, or<\/em>, and not<\/em>. But using algebraic method is brief and clear. We can see $A\\cup B=A+B+AB$ and $A^c=1+A$, and then proves
\n$$(A\\cup B)^c=1+(A+B+AB)=(1+A)(1+B)=A^c\\cap B^c\\ \\square
\n$$
\nDefinition 1.0.6<\/strong> A domain<\/strong> (often called an integral domain<\/strong>) is a commutative ring $R$ that satisfies two extra axioms:<\/p>\n\n
\n
\n$$a\\mid b,
\n$$
\nif $\\exists c\\in R$ such that $b=ca$.<\/p>\n\n
\n$$[a,b]+[c,d]=[ad+bc,bd] \\text{ and } [a,b][c,d]=[ac,bd].
\n$$
\nIt’s easy to show that addition and multiplication are well-defined and $F$ is a commutative ring. We can find that the family $R'=\\{[a,1];a\\in R\\}$ is a subring of $F$ and we identify $a\\in R$ with $[a,1]\\in R'$. To see that $F$ is a field, observe that if $[a,b]\\ne [0,1]$, then $a\\ne 0$, and the inverse of $[a,b]$ is $[b,a]$.<\/p>\n
\n$$\\operatorname{Frac}(R),
\n$$
\nand we denote $[a,b]\\in \\operatorname{Frac}(R)$ by $a\/b$; in particular, the elements $[a,1]$ of $F$ are denoted by $a\/1$ or, more simply, by $a$.<\/p>\n