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Algebraic number theory problem sheet 3

 
  • Jan 11, 2024
  1. Factor the ideal π”ž into prime ideals in $π’ͺ_K$ in the following cases
  2. Let $K=𝐐(\sqrt{10})$. Show that $π’ͺ_K$ is not a principal ideal domain.
    Since $10 ≑ 2 \pmod 4$, we have $π’ͺ_K = 𝐙[\sqrt{10}]$.
    Consider the ideals $I = (2, \sqrt{10})$.
    Assume that $I$ is a principal ideal, i.e., $I^2=(2)β‡’N(I)=Β±2$.
    Suppose $I=a+b\sqrt{10}$, then $a^2-10b^2=Β±2$. Quadratic residues mod 10 are 0,1,4,5,6,9, so no solution. so $I$ cannot be principal.
    Therefore, $π’ͺ_K$ is not a PID.
  3. Suppose that $K=𝐐(\sqrt{d})$, $d$ squarefree and $d β‰  0,1$. Show that whether (2) ramifies, is inert, or splits completely on $π’ͺ_K$ depends only on the value of $d\pmod8$, and classify the different possibilities.
    Dedekind’s criterion applies, since $π’ͺ_K=𝐙[Ξ±],Ξ±=\begin{cases}\sqrt{d}&d≑2,3\pmod4\\\frac{1+\sqrt{d}}2&d≑1\pmod4\end{cases}$ by 2.17
    When $d≑3\pmod4$, minimal polynomial of Ξ± is $m=X^2-d,\bar m=(X+1)^2$ in $𝔽_2$, so (2) factors to $(2,Ξ±+1)^2$, so (2) ramifies.
    When $d≑2\pmod4$, $m=X^2-d,\bar m=X^2$ in $𝔽_2$, so (2) factors to $(2,Ξ±)^2$, so (2) ramifies.
    When $d≑1\pmod4$, $m=X^2-X+\frac{1-d}4$, separate 2 cases:
    β€’ $d≑1\pmod8$, $\bar m=X(X+1)$, so (2) factors to $(2,Ξ±)(2,Ξ±+1)$, so (2) splits completely.
    β€’ $d≑5\pmod8$, $\bar m=X^2-X+1$ is irreducible, so (2) is inert.
  4. Let $K=𝐐(\sqrt{15})$. Show that there is a unique ideal in $π’ͺ_K$ of norm 12. Is it principal? Find all ideals containing 8.
    $π’ͺ_K=𝐙[\sqrt{15}]$. By Dedekind $(2,Ξ±+1)^2=(2),(3,Ξ±)^2=(3)$
    $(2,Ξ±+1)^2=(2)β‡’N((2,Ξ±+1))=2$
    $(3,Ξ±)^2=(3)β‡’N((3,Ξ±))=2$
    If an ideal $I$ is of norm 12, then $I$ divides $(12)=(2,Ξ±+1)^4(3,Ξ±)^2$, so $I=(2,Ξ±+1)^i(3,Ξ±)^j,0≀i≀4,0≀j≀2$, so $N(I)=2^i3^j=12$, so $i=2,j=1$, so $I=2(3,Ξ±)$. To prove $I$ is not principal, suppose $I=(x_1+x_2\sqrt{15})$ for integers $x_1,x_2$, $N(x_1+x_2\sqrt{15})=Β±N(I)=Β±12⇔x_1^2-15x_2=Β±12$, both sides mod 5, $x_1^2≑±2\pmod5$, no solution. $$(8)=(2)^3=(2,Ξ±+1)^6$$ so there are 7 ideals $(2,Ξ±+1)^i,0≀i≀6$ containing 8.
    Why $(2,Ξ±+1)^3$ is not principal? $(2,Ξ±+1)^2=(2)$ so if $(y)=(2,Ξ±+1)^3$ is principal, then $(2)(2,Ξ±+1)=(y)$ implies $(2,Ξ±+1)$ is principal.
  5. Let $π’ͺ_K$ be the ring of integers of a number field, and let $p$ be a rational prime. Show that $p$ ramifies in $π’ͺ_K$ if and only if the ring $π’ͺ_K/(p)$ has a nilpotent element, that is to say a nonzero element $x$ for which $x^n=0$ for some $n$.
    Let \((p) = pπ’ͺ_K = \prod_i 𝔭_i^{e_i}\) for prime ideals \(𝔭_i ⊴ π’ͺ_K\) and \(e_iβˆˆβ„•\).
    (⟹). Let \(p\) ramify in \(π’ͺ_K\), then \(π’ͺ_K/pπ’ͺ_K β‰… \prod_i π’ͺ_K/𝔭_i^{e_i}\), where at least one \(e_i > 1\), let us say \(e_1\). Then, \(π’ͺ_K/𝔭_1^{e_1}\) has a nilpotent element since, for \(x ∈ 𝔭_1 βˆ– 𝔭_1^{e_1}\), we get \((x+𝔭_1^{e_1})^{e_1}=x^{e_1} + 𝔭_1^{e_1} = 𝔭_1^{e_1}\).
    (⟸). If \(p\) does not ramify in \(π’ͺ_K\), then \(π’ͺ_K/pπ’ͺ_K β‰… \prod_i π’ͺ_K/𝔭_i\), each of which is a field since \(𝔭_i\) is maximal in \(π’ͺ_K\). Each of these fields is finite by 4.6. If \(π’ͺ_K/pπ’ͺ_K\) has a nilpotent element, then has a nonzero entry in $i$-th position, its projection onto $π’ͺ_K/𝔭_i$ is nilpotent but $π’ͺ_K/𝔭_i$ is a field so doesn't have nilpotent elements.
  6. Let $K=𝐐(\sqrt{-210})$. Show that $8∣h_K$.
    We have $-210≑2\pmod4$, so the Minkowski bound is $\frac4Ο€\sqrt{210}<19$. So we only need to look at $p=2,3,5,7,11,13,17$.
    $11,13,17$ are inert, so the class group generated by the prime ideals dividing $(2),(3),(5),(7)$. Then $x^2 + 210$ reduces modulo $p$ as \begin{align*} x^2 + 210 &β‰… x^2 \pmod2 \\ x^2 + 210 &β‰… x^2 \pmod3 \\ x^2 + 210 &β‰… x^2 \pmod5 \\ x^2 + 210 &β‰… x^2 \pmod7 \end{align*} Then let $𝔭_2 = (2,Ξ±),𝔭_3 = (3,Ξ±),𝔭_5 = (5,Ξ±),𝔭_7 = (7,Ξ±)$.
    Thus $[𝔭_2],[𝔭_3],[𝔭_5],[𝔭_7]$ generate $\operatorname{Cl}(K)$. \[𝔭_2𝔭_3𝔭_5𝔭_7=(210,Ξ±)=(Ξ±)\] Thus $[𝔭_2],[𝔭_3],[𝔭_5]$ generate $\operatorname{Cl}(K)$. \begin{align*} 𝔭_2^2=&(2)\\ 𝔭_3^2=&(3)\\ 𝔭_5^2=&(5)\\ 𝔭_7^2=&(7) \end{align*} Thus $[𝔭_2],[𝔭_3],[𝔭_5]$ all have order 2.
    $𝔭_2𝔭_3=(6,Ξ±)$ has norm 6 so not principal (no element of $π’ͺ_K$ has norm 6)
    Similarly $𝔭_2^i𝔭_3^j𝔭_5^k$ is principal then $2|i,2|j,2|k$.
    Thus $\operatorname{Cl}(K)β‰…C_2Γ—C_2Γ—C_2$, so $h_K=8$.
  7. Let $f(X)=X^3-X^2-2 X-8$. Recall from Sheet 1 that $f$ is irreducible, and that if $Ξ±$ is a root of $f$ and $K=𝐐(Ξ±)$ then $e_1, e_2, e_3$ is an integral basis for $π’ͺ_K$, where $e_1=1$, $e_2=Ξ±$ and $e_3=\frac{1}{2} Ξ±(Ξ±+1)$.
    1. Show that the linear maps $ψ_v: π’ͺ_K β†’ 𝐅_2$ defined by $ψ_v(e_i)=v_i$ are ring homomorphisms, for the following values of $v$:
      • $v=(1,0,0)$;
      • $v=(1,1,0)$;
      • $v=(1,0,1)$.
      Conclude that $π’ͺ_K/(2) β‰… 𝐅_2^3$.
      Define $ψ:π’ͺ_K→𝐅_2^3$ by $ψ=(ψ_{(1,0,0)},ψ_{(1,1,0)},ψ_{(1,0,1)})$, then \begin{array}l ψ(e_1)=(1,1,1)\\ ψ(e_2)=(0,1,0)\\ ψ(e_3)=(0,0,1) \end{array} To prove ψ is a ring homomorphism.
      $\begin{array}{c|ccc} &e_1&e_2&e_3\\\hline e_1&e_1&e_2&e_3\\ e_2&e_2&2e_3-e_2&2e_3+4e_1\\ e_3&e_3&2e_3+4e_1&3e_3-2(e_2+3e_1) \end{array}$ this table mod 2 is the table $\begin{array}{c|ccc} &v_1&v_2&v_3\\\hline v_1&v_1&v_2&v_3\\ v_2&v_2&v_2&0\\ v_3&v_3&0&v_3 \end{array}$
      Three values of $v$ all satisfy the table, so $ψ(e_i)ψ(e_j)=ψ(e_ie_j)$ for $1≀i≀j≀3$.
      To prove ψ is surjective.
      $(1,1,1),(0,1,0),(0,0,1)$ is a basis for $𝐅_2^3$.
      To prove ker ψ=(2).
      Clearly $(2)βŠ‚\kerψ$. On the other hand $N((2))=2^3=8$, $N(\kerψ)={|π’ͺ_K/\kerψ|}={|𝐅_2^3|}=8$, so $N((2))=N(\kerψ)$, so $\kerψ=(2)$.
    2. Explain why it follows from (i) that (2) splits completely in $π’ͺ_K$.
      From (i) $π’ͺ_K/(2) β‰… 𝐅_2^3$, so $(2)=\kerψ_{(1,0,0)}∩\kerψ_{(1,1,0)}∩\kerψ_{(1,0,1)}$ and they are coprime so $(2)=\kerψ_{(1,0,0)}\kerψ_{(1,1,0)}\kerψ_{(1,0,1)}$ in $π’ͺ_K$. \begin{array}{lll} \kerψ_{(1,0,0)}=(e_2,e_3)=(Ξ±,Ξ±(Ξ±+1)/2)\\ \kerψ_{(1,1,0)}=(e_1-e_2,e_3)=(1-Ξ±,Ξ±(Ξ±+1)/2)\\ \kerψ_{(1,0,1)}=(e_1-e_3,e_2)=(1-Ξ±(Ξ±+1)/2,Ξ±) \end{array}
    3. Show that $π’ͺ_K$ does not have a power integral basis.
      If $π’ͺ_K$ has a power integral basis, then Dedekind’s criterion applies, (ii) says (2) splits completely into 3 factors, so $\bar{m}$ splits into 3 factors, in $𝐅_2[x]$ there are only two monic linear irreducible $x,x+1$, so $\bar{m}$ has a repeated factor, so 2 ramifies, but $π’ͺ_K/(2) β‰… 𝐅_2^3$ has no nilpotent elements, contradicting Q5.
  8. Let $π’ͺ_K$ be the ring of integers of a number field.
    1. Let $𝔭$ be a prime ideal in $π’ͺ_K$ and $n$ a positive integer. Explain why every proper ideal of the quotient ring $π’ͺ_K/𝔭^n$ is of the form $𝔭^i/𝔭^n$ for some $i$.
      Any proper ideal of $π’ͺ_K/𝔭^n$ is of the form $π”ž/𝔭^n$ for some ideal $π”žβŠƒπ”­^n$ by correspondence theorem, so $π”ž|𝔭^n$, so $π”ž=𝔭^i$ for some $i≀n$.
    2. Show that $𝔭^i/𝔭^n$ is principal. (Hint: first try the case $n=i+1$)
      Pick $Ξ±βˆˆπ”­^iβˆ–π”­^{i+1}$, then $𝔭^{i+1}⊊α+𝔭^{i+1}βŠ‚π”­^i$, by unique factorization $Ξ±+𝔭^{i+1}=𝔭^i$, so $Ξ±/𝔭^{i+1}$ generates $𝔭^i/𝔭^{i+1}$.
      In general, assume $n>i$ (otherwise it's the zero ideal). $βˆƒΞ±βˆˆπ”­^iβˆ–π”­^{i+1}$ \[𝔭^i/𝔭^nβŠƒΞ±+𝔭^i/𝔭^nβŠ‹π”­^{i+1}/𝔭^n\] $Ξ±+𝔭^i/𝔭^n$ generates $𝔭^i/𝔭^n$.
    3. Let $π”ž$ be an arbitrary ideal in $π’ͺ_K$. Show that every ideal in $π’ͺ_K/π”ž$ is principal.
      Factor $π”ž=\prod_i𝔭_i^{e_i}$ for $𝔭_i$ distinct. By CRT $π’ͺ_K/π”žβ‰…\prod_iπ’ͺ_K/𝔭_i^{e_i}$ each factor is principal.
    4. Conclude that every ideal $π”ž$ in $π’ͺ_K$ is generated by at most two elements.
      Take $0β‰ aβˆˆπ”ž$, by (iii) $π”ž/(a)$ is principal so βˆƒ$b∈π’ͺ_K,π”ž/(a)=(b)/(a)$ so $(a,b) =π”ž$.