1. Factor the ideal 𝔞 into prime ideals in $𝒪_K$ in the following cases
2. Let $K=𝐐(\sqrt{10})$. Show that $𝒪_K$ is not a principal ideal domain.

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   Since $10 ≡ 2 \pmod 4$, we have $𝒪_K = 𝐙[\sqrt{10}]$.
   Consider the ideals $I = (2, \sqrt{10})$.
   Assume that $I$ is a principal ideal, i.e., $I^2=(2)⇒N(I)=±2$.
   Suppose $I=a+b\sqrt{10}$, then $a^2-10b^2=±2$. Quadratic residues mod 10 are 0,1,4,5,6,9, so no solution. so $I$ cannot be principal.
   Therefore, $𝒪_K$ is not a PID.
3. Suppose that $K=𝐐(\sqrt{d})$, $d$ squarefree and $d ≠ 0,1$. Show that whether (2) ramifies, is inert, or splits completely on $𝒪_K$ depends only on the value of $d\pmod8$, and classify the different possibilities.

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   Dedekind’s criterion applies, since $𝒪_K=𝐙[α],α=\begin{cases}\sqrt{d}&d≡2,3\pmod4\\\frac{1+\sqrt{d}}2&d≡1\pmod4\end{cases}$ by 2.17
   When $d≡3\pmod4$, minimal polynomial of α is $m=X^2-d,\bar m=(X+1)^2$ in $𝔽_2$, so (2) factors to $(2,α+1)^2$, so (2) ramifies.
   When $d≡2\pmod4$, $m=X^2-d,\bar m=X^2$ in $𝔽_2$, so (2) factors to $(2,α)^2$, so (2) ramifies.
   When $d≡1\pmod4$, $m=X^2-X+\frac{1-d}4$, separate 2 cases:
   • $d≡1\pmod8$, $\bar m=X(X+1)$, so (2) factors to $(2,α)(2,α+1)$, so (2) splits completely.
   • $d≡5\pmod8$, $\bar m=X^2-X+1$ is irreducible, so (2) is inert.
4. Let $K=𝐐(\sqrt{15})$. Show that there is a unique ideal in $𝒪_K$ of norm 12. Is it principal? Find all ideals containing 8.

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   $𝒪_K=𝐙[\sqrt{15}]$. By Dedekind $(2,α+1)^2=(2),(3,α)^2=(3)$
   $(2,α+1)^2=(2)⇒N((2,α+1))=2$
   $(3,α)^2=(3)⇒N((3,α))=2$
   If an ideal $I$ is of norm 12, then $I$ divides $(12)=(2,α+1)^4(3,α)^2$, so $I=(2,α+1)^i(3,α)^j,0≤i≤4,0≤j≤2$, so $N(I)=2^i3^j=12$, so $i=2,j=1$, so $I=2(3,α)$. To prove $I$ is not principal, suppose $I=(x_1+x_2\sqrt{15})$ for integers $x_1,x_2$, $N(x_1+x_2\sqrt{15})=±N(I)=±12⇔x_1^2-15x_2=±12$, both sides mod 5, $x_1^2≡±2\pmod5$, no solution. $$(8)=(2)^3=(2,α+1)^6$$ so there are 7 ideals $(2,α+1)^i,0≤i≤6$ containing 8.
   Why $(2,α+1)^3$ is not principal? $(2,α+1)^2=(2)$ so if $(y)=(2,α+1)^3$ is principal, then $(2)(2,α+1)=(y)$ implies $(2,α+1)$ is principal.
5. Let $𝒪_K$ be the ring of integers of a number field, and let $p$ be a rational prime. Show that $p$ ramifies in $𝒪_K$ if and only if the ring $𝒪_K/(p)$ has a nilpotent element, that is to say a nonzero element $x$ for which $x^n=0$ for some $n$.

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   Let \((p) = p𝒪_K = \prod_i 𝔭_i^{e_i}\) for prime ideals \(𝔭_i ⊴ 𝒪_K\) and \(e_i∈ℕ\).
   (⟹). Let \(p\) ramify in \(𝒪_K\), then \(𝒪_K/p𝒪_K ≅ \prod_i 𝒪_K/𝔭_i^{e_i}\), where at least one \(e_i > 1\), let us say \(e_1\). Then, \(𝒪_K/𝔭_1^{e_1}\) has a nilpotent element since, for \(x ∈ 𝔭_1 ∖ 𝔭_1^{e_1}\), 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 \(𝒪_K\), then \(𝒪_K/p𝒪_K ≅ \prod_i 𝒪_K/𝔭_i\), each of which is a field since \(𝔭_i\) is maximal in \(𝒪_K\). Each of these fields is finite by 4.6. If \(𝒪_K/p𝒪_K\) has a nilpotent element, then has a nonzero entry in $i$-th position, its projection onto $𝒪_K/𝔭_i$ is nilpotent but $𝒪_K/𝔭_i$ is a field so doesn't have nilpotent elements.
6. Let $K=𝐐(\sqrt{-210})$. Show that $8∣h_K$.

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   We have $-210≡2\pmod4$, so the Minkowski bound is $\frac4π\sqrt{210}<19$. 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 &≅ x^2 \pmod2 \\ x^2 + 210 &≅ x^2 \pmod3 \\ x^2 + 210 &≅ x^2 \pmod5 \\ x^2 + 210 &≅ x^2 \pmod7 \end{align*} Then let $𝔭_2 = (2,α),𝔭_3 = (3,α),𝔭_5 = (5,α),𝔭_7 = (7,α)$.
   Thus $[𝔭_2],[𝔭_3],[𝔭_5],[𝔭_7]$ generate $\operatorname{Cl}(K)$. \[𝔭_2𝔭_3𝔭_5𝔭_7=(210,α)=(α)\] 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.
   $𝔭_2𝔭_3=(6,α)$ has norm 6 so not principal (no element of $𝒪_K$ has norm 6)
   Similarly $𝔭_2^i𝔭_3^j𝔭_5^k$ is principal then $2|i,2|j,2|k$.
   Thus $\operatorname{Cl}(K)≅C_2×C_2×C_2$, 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 $α$ is a root of $f$ and $K=𝐐(α)$ then $e_1, e_2, e_3$ is an integral basis for $𝒪_K$, where $e_1=1$, $e_2=α$ and $e_3=\frac{1}{2} α(α+1)$.
   • Show that the linear maps $ψ_v: 𝒪_K → 𝐅_2$ 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 $𝒪_K/(2) ≅ 𝐅_2^3$.

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     Define $ψ:𝒪_K→𝐅_2^3$ 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 $(2)⊂\kerψ$. On the other hand $N((2))=2^3=8$, $N(\kerψ)={|𝒪_K/\kerψ|}={|𝐅_2^3|}=8$, so $N((2))=N(\kerψ)$, so $\kerψ=(2)$.
   • Explain why it follows from (i) that (2) splits completely in $𝒪_K$.

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     From (i) $𝒪_K/(2) ≅ 𝐅_2^3$, 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 $𝒪_K$. \begin{array}{lll} \kerψ_{(1,0,0)}=(e_2,e_3)=(α,α(α+1)/2)\\ \kerψ_{(1,1,0)}=(e_1-e_2,e_3)=(1-α,α(α+1)/2)\\ \kerψ_{(1,0,1)}=(e_1-e_3,e_2)=(1-α(α+1)/2,α) \end{array}
   • Show that $𝒪_K$ does not have a power integral basis.

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     If $𝒪_K$ 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 $𝒪_K/(2) ≅ 𝐅_2^3$ has no nilpotent elements, contradicting Q5.
8. Let $𝒪_K$ be the ring of integers of a number field.
   • Let $𝔭$ be a prime ideal in $𝒪_K$ and $n$ a positive integer. Explain why every proper ideal of the quotient ring $𝒪_K/𝔭^n$ is of the form $𝔭^i/𝔭^n$ for some $i$.

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     Any proper ideal of $𝒪_K/𝔭^n$ is of the form $𝔞/𝔭^n$ for some ideal $𝔞⊃𝔭^n$ by correspondence theorem, so $𝔞|𝔭^n$, so $𝔞=𝔭^i$ for some $i≤n$.
   • Show that $𝔭^i/𝔭^n$ is principal. (Hint: first try the case $n=i+1$)

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     Pick $α∈𝔭^i∖𝔭^{i+1}$, then $𝔭^{i+1}⊊α+𝔭^{i+1}⊂𝔭^i$, by unique factorization $α+𝔭^{i+1}=𝔭^i$, so $α/𝔭^{i+1}$ generates $𝔭^i/𝔭^{i+1}$.
     In general, assume $n>i$ (otherwise it's the zero ideal). $∃α∈𝔭^i∖𝔭^{i+1}$ \[𝔭^i/𝔭^n⊃α+𝔭^i/𝔭^n⊋𝔭^{i+1}/𝔭^n\] $α+𝔭^i/𝔭^n$ generates $𝔭^i/𝔭^n$.
   • Let $𝔞$ be an arbitrary ideal in $𝒪_K$. Show that every ideal in $𝒪_K/𝔞$ is principal.

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     Factor $𝔞=\prod_i𝔭_i^{e_i}$ for $𝔭_i$ distinct. By CRT $𝒪_K/𝔞≅\prod_i𝒪_K/𝔭_i^{e_i}$ each factor is principal.
   • Conclude that every ideal $𝔞$ in $𝒪_K$ is generated by at most two elements.

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     Take $0≠a∈𝔞$, by (iii) $𝔞/(a)$ is principal so ∃$b∈𝒪_K,𝔞/(a)=(b)/(a)$ so $(a,b) =𝔞$.