{"id":405,"date":"2023-06-20T16:40:03","date_gmt":"2023-06-20T08:40:03","guid":{"rendered":"https:\/\/qwq.cafe\/?p=405"},"modified":"2023-06-20T16:40:03","modified_gmt":"2023-06-20T08:40:03","slug":"abstract-algebra-1-5-maximal-ideals-and-prime-ideals","status":"publish","type":"post","link":"https:\/\/qwq.cafe\/?p=405","title":{"rendered":"Abstract Algebra \u2013 1.5 Maximal Ideals and Prime Ideals"},"content":{"rendered":"<p><strong>Definition 1.5.1<\/strong> An ideal $I$ in a commutative ring $R$ is called a <strong>maximal ideal<\/strong> if $I$ is a proper ideal for which there is no proper ideal $J$ with $I\\subsetneq J$.<\/p>\n<p><strong>Proposition 1.5.1<\/strong> A proper ideal $I$ in a commutative ring $R$ is a maximal ideal if and only if $R\/I$ is a field.<\/p>\n<p>Proof.<\/p>\n<ul>\n<li>\n<p>($\\Rightarrow$).<\/p>\n<p>Since $I$ is a maximal ideal, then $I\/I=(0)$ is a maximal ideal in $R\/I$ for Proposition1.3.5.<\/p>\n<p>Then because a commutative ring only having trivial ideals is a field, $R\/I$ is a field.<\/p>\n<\/li>\n<li>\n<p>($\\Leftarrow$).<\/p>\n<p>Since $R\/I$ is a field, $I\/I$ is a maximal ideal in $R\/I$.<\/p>\n<p>Because of Proposition1.3.5, $I$ is a maximal ideal in $R$. $\\square$<\/p>\n<\/li>\n<\/ul>\n<p><strong>Example 1.5.1<\/strong><\/p>\n<ul>\n<li>If $p$ is a prime number, then $(p)$ is a maximal ideal in $\\mathbb Z$, for $\\mathbb Z_p$ is a field.<\/li>\n<li>If $K$ is a field, then $(x)$ is a maximal ideal in $K[x]$, for $K[x]\/(x)\\cong K$.<\/li>\n<li>$(x^2+1)$ is a maximal ideal in $\\mathbb R[x]$, for $\\mathbb R[x]\/(x^2+1)\\cong \\mathbb C$.<\/li>\n<\/ul>\n<p><strong>Proposition 1.5.2<\/strong> If $K$ is a field, then $I=(x_1-a_1,\\dots,x_n-a_n)$ is a maximal ideal in $K[x_1,\\dots,x_n]$ whenever $a_1,\\dots,a_n\\in K$.<\/p>\n<p><strong>Definition 1.5.2<\/strong> An ideal $I$ in a commutative ring $R$ is called a <strong>prime ideal<\/strong> if $I$ is a proper ideal such that $ab\\in I\\Rightarrow (a\\in I\\lor b\\in I)$.<\/p>\n<p><strong>Proposition 1.5.3<\/strong> If $I$ is a proper ideal in a commutative ring $R$, then $I$ is a prime ideal if and only if $R\/I$ is a domain.<\/p>\n<p>Proof.<\/p>\n<ul>\n<li>\n<p>($\\Rightarrow$).<\/p>\n<p>If $(a+I)(b+I)=0+I$, then $ab\\in I\\Rightarrow (a\\in I\\lor b\\in I)\\Rightarrow (a+I=0+I\\lor b+I=0+I)$.<\/p>\n<\/li>\n<li>\n<p>($\\Leftarrow$).<\/p>\n<p>If $ab\\in I$, then $(a+I)(b+I)=0+I\\Rightarrow (a+I=0+I\\lor b+I=0+I)\\Rightarrow (a\\in I\\lor b\\in I)$. $\\square$<\/p>\n<\/li>\n<\/ul>\n<p><strong>Corollary 1.5.1<\/strong> Every maximal ideal is a prime ideal.<\/p>\n<p><strong>Definition 1.5.3<\/strong> If $I$ and $J$ are ideals in a commutative ring, then<br \/>\n$$IJ:=\\{\\text{all finite sums}\\sum_{\\mathscr l}a_{\\mathscr l}b_\\mathscr l;a_{\\mathscr l}\\in I,b_\\mathscr l\\in J\\}<br \/>\n$$<br \/>\nIt&#8217;s easy to see that $IJ$ is an ideal in $R$.<\/p>\n<p><strong>Proposition 1.5.4<\/strong> Let $P$ be a prime ideal in a commutative ring $R$. If $I$ and $J$ are ideals with $IJ\\subset P$, then $I\\subset P$ or $J\\subset P$.<\/p>\n<p>Proof.<\/p>\n<p>If $I\\not\\subset P$ and $J\\not\\subset P$, then there are $a\\in I$ and $b\\in J$ with $a,b\\notin P$. <\/p>\n<p>But $ab\\in IJ\\subset P$, contradicting $P$ being prime. $\\square$<\/p>\n<p><strong>Proposition 1.5.5<\/strong> If $K$ is a field and $I=(f)$, where $f(x)$ is a nonzero polynomial in $K[x]$, then the following are equivalent:<\/p>\n<ol>\n<li>\n<p>$f$ is irreducible;<\/p>\n<\/li>\n<li>\n<p>$K[x]\/I$ is a field;<\/p>\n<\/li>\n<li>\n<p>$K[x]\/I$ is a domain.<\/p>\n<\/li>\n<\/ol>\n<p>Proof.<\/p>\n<ul>\n<li>\n<p>(1) $\\Rightarrow$ (2).<\/p>\n<p>If there is an ideal $J$ on $K[x]$, with $I\\subset J$.<\/p>\n<p>Since every ideal in $K[x]$ is principle, there is $d(x)\\in K[x]$ such that $J=(d)$.<\/p>\n<p>Then because $f\\in(f)\\subset (d)$, there is $g(x)\\in K[x]$ with $f=gd$.<\/p>\n<p>Then $d$ is a unit or $g$ is a unit for $f$ is irreducible.<\/p>\n<p>Therefore, $(d)=K[x]$ or $(d)=(f)$. $I$ is a maximal ideal.<\/p>\n<\/li>\n<li>\n<p>(2) $\\Rightarrow$ (3).<\/p>\n<p>Every field is a domain.<\/p>\n<\/li>\n<li>\n<p>(3) $\\Rightarrow$ (1).<\/p>\n<p>Since $K[x]\/I$ is a domain, $I$ is a prime ideal in $K[x]$.<\/p>\n<p>If some polynomials $g(x),h(x)\\in K[x]$ such that $f=gh$, then $gh=f\\in I\\Rightarrow (g\\in I\\lor h\\in I)$.<\/p>\n<p>Since $\\deg(g)\\le \\deg(f)$ and $\\deg(h)\\le \\deg(f)$, $g$ is a unit or $h$ is a unit. $\\square$<\/p>\n<\/li>\n<\/ul>\n<p><strong>Proposition 1.5.6<\/strong> Let $k$ be a field, let $p(x)$ be a monic irreducible polynomial in $k[x]$ of degree $d$, let $K=k[x]\/I$, where $I=(p)$, and let $\\beta=x+I\\in K$. Then:<\/p>\n<ol>\n<li>$K$ is a field and $k&#039;=\\{a+I;a\\in k\\}$ is a subfield of $K$ isomorphic to $k$.<\/li>\n<li>$\\beta$ is a root of $p$ in $K$.<\/li>\n<li>If $g(x)\\in k[x]$ and $\\beta$ is a root of $g$ in $K$, then $p\\mid g$ in $k[x]$.<\/li>\n<li>$p$ is the unique monic irreducible polynomial in $k[x]$ having $\\beta$ as a root.<\/li>\n<li>The list $1,\\beta,\\beta^2,\\dots,\\beta^{d-1}$ is a basis of $K$ as a vector space over $k$, and so $\\dim_{k}(K)=d$.<\/li>\n<\/ol>\n<p>Proof.<\/p>\n<ol>\n<li>\n<p>Trivial.<\/p>\n<\/li>\n<li>\n<p>Let $p(x)=a_0+a_1x+\\cdots+a_{d-1}x^{d-1}+x^d$, where $a_i\\in k$ for all $i$. In $K=k[x]\/I$, we have<br \/>\n$$<br \/>\n\\begin{align}<br \/>\np(\\beta)&amp;=(a_0+I)+(a_1+I)\\beta+\\cdots+(1+I)\\beta^d\\<br \/>\n&amp;=(a_0+I)+(a_1+I)(x+I)+\\cdots+(1+I)(x+I)^d\\<br \/>\n&amp;=(a_0+I)+(a_1x+I)+\\cdots+(1x^d+I)\\<br \/>\n&amp;=a_0+a_1x+\\cdots+x^d+I\\<br \/>\n&amp;=p(x)+I=0+I.<br \/>\n\\end{align}<br \/>\n$$<\/p>\n<\/li>\n<li>\n<p>We have $g(x)=h(x)p(x)+r(x)$ in $k[x]$ with $r=0\\lor\\deg(r)&lt;\\deg(p)$ for the Division Algorithm.<\/p>\n<p>Then $0+I=g(\\beta)=h(\\beta)p(\\beta)+r(\\beta)=r(\\beta)$; that is $r(x)\\in I\\Rightarrow r=0$. <\/p>\n<p>Therefore $p\\mid g$.<\/p>\n<\/li>\n<li>\n<p>(3) $\\Rightarrow$ (4) is trivial.<\/p>\n<\/li>\n<li>\n<p>If $a_0+a_1\\beta+\\cdots+a_{d-1}\\beta^{d-1}=0$, letting $g(x)=a_0+a_1x+\\cdots+a_{d-1}x^{d-1}$, then $g=0$ for (3). $\\square$<\/p>\n<\/li>\n<\/ol>\n<p><strong>Definition 1.5.4<\/strong> If $K$ is a field containing $k$ as a subfield, then $K$ is called an <strong>extension field<\/strong> of $k$, and we denote an extension field by<br \/>\n$$K\/k.<br \/>\n$$<br \/>\nAn extension field $K\/k$ is a <strong>finite extension<\/strong> if $K$ is a finite-dimensional vector space over $k$. The dimension of $K$, denoted by<br \/>\n$$[K:k],<br \/>\n$$<br \/>\nis called the <strong>degree<\/strong> of $K\/k$.<\/p>\n<p><strong>Definition 1.5.5<\/strong> Let $K\/k$ be an extension field. An element $a\\in K$ is <strong>algebraic<\/strong> over $k$, if there is some nonzero polynomial $f(x)\\in k[x]$ having $\\alpha$ as a root; otherwise, $\\alpha$ is <strong>transcendental<\/strong> over $k$. An extension field $K\/k$ is <strong>algebraic<\/strong> if every $\\alpha\\in K$ is algebraic over $k$.<\/p>\n<p><strong>Proposition 1.5.7<\/strong> If $K\/k$ is a finite extension field, then $K\/k$ is an algebraic extension.<\/p>\n<p>Proof.<\/p>\n<p>Let $n=\\dim_k(K)$, for all $\\beta\\in K$, then $1,\\beta,\\dots,\\beta^n$ are dependent.<\/p>\n<p>Therefore, there are $a_0,a_1,\\dots,a_n\\in k$ such that $a_0+a_1\\beta+\\dots+a_n\\beta^n=0$. $\\square$<\/p>\n<p><strong>Definition 1.5.6<\/strong> If $K\/k$ is an extension field and $\\alpha\\in K$, then<br \/>\n$$k(\\alpha)<br \/>\n$$<br \/>\nis the intersection of all those subfields of $K$ containing $k$ and $\\alpha$; we call $k(\\alpha)$ the subfield of $K$ obtained by <strong>adjoining<\/strong> $\\alpha$ to $k$.<\/p>\n<p>More generally, if $A$ is a (possibly infinite) subset of $K$, define $k(A)$ to be the intersection of all the subfields of $K$ containing $k\\cup A$; we call $k(A)$ the subfield of $K$ obtained by <strong>adjoining<\/strong> $A$ to $k$. In particular, if $A=\\{z_1,\\dots,z_n\\}$ is a finite subset, then we may denote $k(A)$ by $k(z_1,\\dots,z_n)$.<\/p>\n<p><strong>Theorem 1.5.1<\/strong><\/p>\n<ol>\n<li>\n<p>If $K\/k$ is an extension field and $\\alpha\\in K$ is algebraic over $k$, then there is a unique monic irreducible polynomial $p(x)\\in k[x]$ having $\\alpha$ as a root. Moreover, if $I=(p)$, then $k[x]\/I\\cong k(\\alpha)$; indeed, there exists an isomorphism<br \/>\n$$<br \/>\n\\varphi:k[x]\/I\\to k(\\alpha)<br \/>\n$$<br \/>\nwith $\\varphi(x+I)=\\alpha$ and $\\varphi(c+I)=c,\\forall c\\in k$.<\/p>\n<\/li>\n<li>\n<p>If $\\alpha&#039;\\in K$ is another root of $p(x)$, then there is an isomorphism<br \/>\n$$<br \/>\n\\theta:k(\\alpha)\\to k(\\alpha&#8217;)<br \/>\n$$<br \/>\nwith $\\theta(\\alpha)=\\alpha&#039;$ and $\\theta(c)=c,\\forall c\\in k$.<\/p>\n<\/li>\n<\/ol>\n<p><strong>Definition 1.5.7<\/strong> If $K\/k$ is an extension field and $\\alpha\\in K$ is algebraic over $k$, then the unique monic irreducible polynomial $p(x)\\in k[x]$ having $\\alpha$ as a root is called the <strong>minimal polynomial<\/strong> of $\\alpha$ over $k$; it is denoted by<br \/>\n$$\\operatorname{irr}(\\alpha,k)=p(x).<br \/>\n$$<br \/>\n<strong>Theorem 1.5.2<\/strong> Let $k\\subset E\\subset K$ be fields, with $E$ a finite extension field of $k$ and $K$ a finite extension field of $E$. Then $K$ is a finite extension field of $k$ and<br \/>\n$$[K:k]=[K:E][E:k].<br \/>\n$$<br \/>\nProof.<\/p>\n<p>Let $n=[K:E],m=[E:k]$.<\/p>\n<p>Then let $\\alpha_1,\\dots,\\alpha_n$ be a base of vector space $K$ over $E$, and $\\beta_1,\\dots,\\beta_m$ be a base of vector space $E$ over $k$.<\/p>\n<p>For all $\\alpha\\in K$, there are $b_1,\\dots,b_n\\in E$ with $b_1\\alpha_1+\\cdots+b_n\\alpha_n=\\alpha$.<\/p>\n<p>And for all $i(1\\le i\\le n)$, there are $a_{i1},\\dots,a_{im}$ such that $b_i=a_{i1}\\beta_1+\\cdots +a_{im}\\beta_m$.<\/p>\n<p>Therefore $\\alpha=\\sum_{i,j} a_{ij}\\alpha_i\\beta_j$, which means $\\{\\alpha_i\\beta_j;1\\le i\\le n,1\\le j\\le m\\}$ is a base of vector space $K$ over $k$.<\/p>\n<p>Then let&#8217;s prove $\\{\\alpha_i\\beta_j;1\\le i\\le n,1\\le j\\le m\\}$ is independent.<\/p>\n<p>Suppose that $\\sum_{i,j}a_{ij}\\alpha_i\\beta_j=0$, and then $\\sum_{i} (\\sum_j a_{ij}\\beta_j)\\alpha_i=0$.<\/p>\n<p>Since $\\alpha_1,\\dots,\\alpha_n$ is a base, $\\sum_ja_{ij}\\beta_j=0$ for all $i$.<\/p>\n<p>Because $\\beta_1,\\dots,\\beta_m$ is also a base, $a_{ij}=0$ for all $i,j$. $\\square$<\/p>\n<p><strong>Example 1.5.2<\/strong> Let $f(x)=x^4-10x^2+1\\in \\mathbb Q[x]$. If $\\beta$ is a root of $f$, then the quadratic formula gives $\\beta^2=5\\pm 2\\sqrt 6$. But the identity $a+2\\sqrt{ab}+b=(\\sqrt a+\\sqrt b)^2$ gives $\\beta=\\pm (\\sqrt 2+\\sqrt 3)$. Similarly, $5-2\\sqrt{6}=(\\sqrt 2-\\sqrt 3)^2$, so that the roots of $f$ are<br \/>\n$$\\sqrt 2+\\sqrt 3,-\\sqrt{2}-\\sqrt 3,\\sqrt 2-\\sqrt 3,-\\sqrt 2+\\sqrt 3.<br \/>\n$$<br \/>\nWe claim that $f$ is irreducible in $\\mathbb Q[x]$. If $g$ is a quadratic factor of $f$ in $\\mathbb Q[x]$, then<br \/>\n$$g(x)=(x-a\\sqrt 2-b\\sqrt 3)(x-c\\sqrt 2-d\\sqrt 3),<br \/>\n$$<br \/>\nthere $a,b,c,d\\in \\{1,-1\\}$. Multiplying,<br \/>\n$$g(x)=x^2-\\left((a+c)\\sqrt 2+(b+d)\\sqrt 3\\right)x+2ac+3bd+(ad+bc)\\sqrt 6.<br \/>\n$$<br \/>\nWe check easily that $(a+c)\\sqrt 2+(b+d)\\sqrt 3$ is rational if and only if $a+c=0=b+d$; but these equations force $ad+bc\\ne 0$, and so the constant term of $g$ is not rational. Therefore, $g\\notin \\mathbb Q[x]$, and so $f$ is irreducible in $\\mathbb Q[x]$. If $\\beta = \\sqrt 2+\\sqrt 3$, then $f(x)=\\operatorname{irr}(\\beta,\\mathbb Q)$.<\/p>\n<p>Consider the field $E=\\mathbb Q(\\beta)=\\mathbb Q(\\sqrt 2+\\sqrt 3).$ There is a tower of fields $\\mathbb Q\\subset E\\subset F$, where $F=\\mathbb Q(\\sqrt 2,\\sqrt 3)$, and so<br \/>\n$$[F:\\mathbb Q]=[F:E][E:\\mathbb Q].<br \/>\n$$<br \/>\nSince $E=\\mathbb Q$ and $\\beta$ is a root of an irreducible polynomial of degree $4$, namely, $f$, we have $[E:\\mathbb Q]=4$.<\/p>\n<p>On the other hand,<br \/>\n$$[F:\\mathbb Q]=[F:\\mathbb Q(\\sqrt 2)][\\mathbb Q(\\sqrt 2):\\mathbb Q].<br \/>\n$$<br \/>\nNow $[\\mathbb Q(\\sqrt2):\\mathbb Q]=2$, because $\\sqrt 2$ is a root of the irreducible quadratic $x^2-2$ in $\\mathbb Q[x]$. We claim that $[F:\\mathbb Q(\\sqrt 2)]\\le 2$. The field $F$ arises by adjoining $\\sqrt 3$ to $\\mathbb Q(\\sqrt 2)$; either $\\sqrt 3\\in \\mathbb Q(\\sqrt 2)$, in which case the degree is $1$, or $x^2-3$ is irreducible in $\\mathbb Q(\\sqrt 2)[x]$, in which case the degree is $2$ (in fact, the degree is $2$). It follows that $[F:\\mathbb Q]\\le 4$, and so the equation $[F:\\mathbb Q]=[F:E][E:\\mathbb Q]$ gives $[F:E]=1$; that is $F=E$.<\/p>\n<p>Let us note that $F$ arises from $\\mathbb Q$ by adjoining all the roots of $f$, but it also arises from $\\mathbb Q$ by adjoining all the roots of the reducible polynomial $g(x)=(x^2-2)(x^2-3)$.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Definition 1.5.1 An ideal $I$ in a commutative ring $R$ [&hellip;]<\/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-405","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\/405","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=405"}],"version-history":[{"count":1,"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts\/405\/revisions"}],"predecessor-version":[{"id":406,"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts\/405\/revisions\/406"}],"wp:attachment":[{"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=405"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=405"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=405"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}