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Algebraic number theory problem sheet 1

 
  • Jan 11, 2024
$\DeclareMathOperator{\tr}{tr}\DeclareMathOperator{\disc}{disc}$
  1. Let $K=𝐐(α)$, where $α=2^{1/3}$. Evaluate $\disc_{K/𝐐}(1, α, α^2)$.
    There are three embeddings of $K$ defined by $Οƒ_i(Ξ±)=Ο‰^{i-1}Ξ±\,(i=1,2,3)$ \begin{align*}\det M&= \begin{vmatrix} Οƒ_1(1) & Οƒ_1(Ξ±) & Οƒ_1(Ξ±^2) \\ Οƒ_2(1) & Οƒ_2(Ξ±) & Οƒ_2(Ξ±^2) \\ Οƒ_3(1) & Οƒ_3(Ξ±) & Οƒ_3(Ξ±^2) \end{vmatrix}= \begin{vmatrix} 1 & Ξ± & Ξ±^2 \\ 1 & ωα & (ωα)^2 \\ 1 & Ο‰^2Ξ± & (Ο‰^2Ξ±)^2 \end{vmatrix}\\ \disc_{K/𝐐}(1, Ξ±, Ξ±^2)&=(\det M)^2\text{ by the Vandermonde determinant} \\&=(1-Ξ±)^2(1-ωα)^2(ωα-Ο‰^2Ξ±)^2= -108.\end{align*} $-4p^3-27q^2=-4β‹…0^3-27β‹…(-2)^2=-108$ agrees with Q5
  2. Let $K=𝐐(\sqrt{d}), d<0$ squarefree, be an imaginary quadratic field. Write $u(d)$ for the number of units in $π’ͺ_K$. What are the possible values of $u(d)$, and for which fields are these values attained?
    By 2.17 $π’ͺ_K=π™βŠ•\sqrt{d}𝐙$ if $d≑2,3\pmod 4$; $π’ͺ_K=π™βŠ•\frac{1+\sqrt{d}}2𝐙$ if $d≑1\pmod 4$.
    Case 1. If $d≑2,3\pmod 4$, elements of $π’ͺ_K$ are $a+b\sqrt{d}$ for $a,b ∈ 𝐙$. By 2.10 units in $π’ͺ_K$ have norm $Β±1$, but $d<0$, the norms are positive, so units in $π’ͺ_K$ have norm $1$, \[N(a+b\sqrt{d})=a^2-b^2d=1\] If $d=-1$ then $a^2+b^2=1$, so $(a,b)=(Β±1,0)$ or $(0,Β±1)$.
    If $d<-1$ then $b=0$, so $(a,b)=(Β±1,0)$.
    Case 2. If $d≑1\pmod 4$, elements of $π’ͺ_K$ are $\frac{a+b\sqrt{d}}{2}$ for $a,b ∈ 𝐙$ with $2|a-b$. Similarly \[N\left(\frac{a + b \sqrt{d}}{2}\right)=\frac{a^2-db^2}{4}=1\] noting that the condition $2|a-b$ is unnecessary because if $a-b$ is odd, $a^2-db^2$ would be odd and hence indivisible by 4.
    If $-d>4$, then $a^2-db^2>4$ for $b≠0$, so only units are $±1$.
    For the remaining case $d=-3$, the solutions of $a^2+3b^2=4$ are $(a,b)=(Β±1,Β±1)(Β±2,0)$ \[U(π’ͺ_K)= \begin{cases} ⟨ \frac{1+\sqrt{-3}}{2} ⟩ &d=-3\\ ⟨ \sqrt{-1} ⟩ &d=-1\\ ⟨ -1 ⟩ &d<-4\end{cases}\] so \[u(d)=\begin{cases}6&d=-3\\4&d=-1\\2&d<-4\end{cases}\]
  3. Suppose that $Ξ²$ is a root of $X^3+p X+q=0$, where $X^3+p X+q$ is an irreducible polynomial in $𝐙[X]$, and let $K=𝐐(Ξ²)$. Compute $\tr_{K/𝐐}(Ξ²^i)$ for $i=0,1, …, 4$. Deduce that $\disc_{K/𝐐}(1, Ξ², Ξ²^2)=-4 p^3-27 q^2$. Hence, give an example of a cubic number field $K$ such that $π’ͺ_K$ has a power integral basis.
    $\{1,β,β^2\}$ is a 𝐐-basis of $K$. Since $β^3=-q-pβ$
    $\tr_{K/𝐐}(1)=\tr\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=3$
    $\tr_{K/𝐐}(β)=\tr\begin{pmatrix}0&1&0\\0&0&1\\-q&-p&0\end{pmatrix}=0$
    $\tr_{K/𝐐}(β^2)=\tr\begin{pmatrix}0&0&1\\-q&-p&0\\0&-q&-p\end{pmatrix}=-2p$
    $\tr_{K/𝐐}(β^3)=\tr\begin{pmatrix}-q&-p&0\\0&-q&-p\\pq&p^2&-q\end{pmatrix}=-3q$
    $\tr_{K/𝐐}(Ξ²^4)=\tr\begin{pmatrix}0&-q&-p\\pq&p^2&-q\\q^2&2pq&p^2\end{pmatrix}=2p^2$ \begin{align*}\disc_{K/𝐐}(1, Ξ², Ξ²^2)&=\begin{vmatrix}\tr_{K/𝐐}(1)&\tr_{K/𝐐}(Ξ²)&\tr_{K/𝐐}(Ξ²^2)\\\tr_{K/𝐐}(Ξ²)&\tr_{K/𝐐}(Ξ²^2)&\tr_{K/𝐐}(Ξ²^3)\\\tr_{K/𝐐}(Ξ²^2)&\tr_{K/𝐐}(Ξ²^3)&\tr_{K/𝐐}(Ξ²^4)\\\end{vmatrix}\\&=\begin{vmatrix}3&0&-2p\\ 0&-2p&-3q\\-2p&-3q&2p^2\end{vmatrix} \\&=-4 p^3 - 27 q^2\end{align*} When $p=q=1$, we check $x^3+x+1$ is irreducible and $\disc_{K/𝐐}(1, Ξ², Ξ²^2)=-31$ is squarefree. Since $m_{𝐐,Ξ²}=x^3+x+1$ we have $β∈π’ͺ_K$, by 2.4 $Ξ²^2∈π’ͺ_K$, by 2.18 $K=𝐐(Ξ²)$ is a cubic number field such that $π’ͺ_K$ has a power integral basis $\{1, Ξ², Ξ²^2\}$.
  4. Let $f(X)=X^3-X^2-2 X-8$.
    1. Show that $f$ is irreducible over 𝐐.

      Suppose $f$ is reducible in 𝐐, by Gauss’s lemma, it is product of two monic polynomials in 𝐙. As $\deg f=3$, it has a root $pβˆˆπ™$. $$p^3-p^2-2p-8=0β‡’p|8$$We verify none of divisors of 8 is a root of $f$, contradiction.

    2. Let $Ξ±$ be a root of $f$ and let $K=𝐐(Ξ±)$. Show that $\frac12Ξ±(Ξ±+1) ∈ π’ͺ_K$.

      First show that $4/Ξ± ∈ π’ͺ_K$. \[0=\frac{-8}{Ξ±^3}(Ξ±^3-Ξ±^2-2Ξ±-8)=-8+2(\frac4Ξ±)+(\frac4Ξ±)^2+(\frac4Ξ±)^3\] so $4/Ξ±$ is a root of the monic polynomial $-8+2x+x^2+x^3$, so$$\frac12Ξ±(Ξ±+1)=\frac1{2Ξ±}(Ξ±^3+Ξ±^2)=\frac1{2Ξ±}((Ξ±^2+2Ξ±+8)+Ξ±^2)=Ξ±+1+\frac4α∈π’ͺ_K$$

    3. Calculate $\disc_{K/𝐐}\left(1, Ξ±, \frac12Ξ±(Ξ±+1)\right)$, and hence conclude that $e_1, e_2, e_3$ is an integral basis for $π’ͺ_K$, where $e_1=1, e_2=Ξ±$ and $e_3=\frac12Ξ±(Ξ±+1)$.
      From previous question $\frac4Ξ±=e_3-e_2-e_1$, by 1.23 $$\disc_{K/𝐐}(e_1,e_2,e_3)=\disc_{K/𝐐}\left(e_1,e_2,\frac4Ξ±\right)$$ since the change of basis matrix has determinant 1. By a computation similar to Q5, \[\disc_{K/𝐐}\left(1, Ξ±, \frac4Ξ±\right)=\begin{vmatrix}\tr_{K/𝐐}(1)&\tr_{K/𝐐}(Ξ±)&\tr_{K/𝐐}(\frac4Ξ±)\\\tr_{K/𝐐}(Ξ±)&\tr_{K/𝐐}(Ξ±^2)&\tr_{K/𝐐}(4)\\\tr_{K/𝐐}(\frac4Ξ±)&\tr_{K/𝐐}(4)&\tr_{K/𝐐}(\frac{16}{Ξ±^2})\end{vmatrix}=-503\] is squarefree, by 2.18 $e_1, e_2, e_3$ is an integral basis for $π’ͺ_K$.
  5. Suppose that $[K: 𝐐]=n$ and that, of the $n$ embeddings $Οƒ_i: K β†’ 𝐂$, $r_1$ of them are real and there are $r_2$ complex conjugate pairs, where $r_1+2 r_2=n$. Show $\operatorname{sgn}Ξ”_K=(-1)^{r_2}$.
    By 1.22 $$Ξ”_K∈𝐐$$ By 1.19 $$\sqrt{Ξ”_K}=\det[Οƒ_i(Ξ±^j)]$$ If we call complex conjugation $Ο„$, we know that it commutes with this determinant because the determinant is a polynomial in the entries of the matrix, so $$Ο„(\det[Οƒ_i(Ξ±^j)])=\det[τσ_i(Ξ±^j)]$$ This permutes the $Οƒ$'s, i.e. the rows of the matrix. So whether or not the sign of the determinant is changed depends whether the permutation is odd or even. By definition, $Ο„$ exchanges each pair of complex-conjugate embeddings, and so $Ο„$ acts by swapping $r_2$ rows, which from linear algebra corresponds to changing the sign by $(-1)^{r_2}$.
    When $r_2$ is even, $Ο„(\sqrt{Ξ”_K})=\sqrt{Ξ”_K}β‡’Ξ”_K>0$; when $r_2$ is odd, $Ο„(\sqrt{Ξ”_K})=-\sqrt{Ξ”_K}β‡’Ξ”_K<0$.
  6. Let $K=𝐐(Ξ±)$, where $Ξ±=2^{1/3}$. Show that $1, Ξ±, Ξ±^2$ is an integral basis for $π’ͺ_K$, evaluate $Ξ”_K$.

    Since $[K:𝐐]=3$, then $1, Ξ±, Ξ±^2$ are 𝐐-linearly independent, hence 𝐙-linearly independent, $$𝐙^3≅𝐙[Ξ±]=⟨1, Ξ±, Ξ±^2⟩_{𝐙}βŠ‚π’ͺ_K≅𝐙^3$$ we know the index ${|π’ͺ_K/𝐙[Ξ±]|}=r$ is finite. To show $π’ͺ_K=𝐙[Ξ±]$ we need to show $r=1$.

    By Q3 $Ξ”_K=-108$, so $r^2|108$, if $r>1$ then $2|r$ or $3|r$.

    If $2|r$ by Cauchy’s theorem $π’ͺ_K/𝐙[Ξ±]$ has an element of order 2, let $z=\frac a2+\frac b2Ξ±+\frac c2Ξ±^2∈π’ͺ_K$ for some $a,b,cβˆˆπ™$, by adding an integer we can assume $a,b,c∈\{0,1\}$. $$\tr_{K/𝐐}(Ξ±)=Ξ±+ωα+Ο‰^2Ξ±=0$$ $$\tr_{K/𝐐}(z)=\frac{3a}2βˆˆπ™β‡’a=0$$ Now $Ξ±Ξ²=\frac{b}2Ξ±^2+\frac{c}2Ξ±^3=\frac{b}2+c∈π’ͺ_K$ but $cβˆˆπ™$, so $\frac{b}2Ξ±^2∈π’ͺ_K$ $$N(Ξ±)=α⋅ωα⋅ω^2Ξ±=2$$ $$N\left(\frac{b}2Ξ±^2\right)=\left(\frac{b}2\right)^3N(Ξ±)^3=\frac{b^3}2βˆˆπ™β‡’b=0$$ Now $Ξ²=\frac{c}2Ξ±^2$ by the same argument $c=0$.

    If $3|r$ by Cauchy’s theorem $π’ͺ_K/𝐙[Ξ±]$ has an element of order 3, let $z=\frac a3+\frac b3Ξ±+\frac c3Ξ±^2∈π’ͺ_K$ for some $a,b,cβˆˆπ™$, modulo 3 we can assume $a,b,c∈F_3$.

    Let $S=\{(a,b,c)∈F_3^3:\frac a3+\frac b3Ξ±+\frac c3Ξ±^2∈π’ͺ_K\}$. Observe that $S⊊F_3^3$, as $(1,0,0)βˆ‰S$.

    As $z,zΞ±,zΞ±^2$ are three non-zero elements of $S$, they must be linearly dependent \begin{align*}0=\begin{vmatrix}a&b&c\\2c&a&b\\2b&2c&a\end{vmatrix}≑\begin{vmatrix}a&b&c\\-c&a&b\\-b&-c&a\end{vmatrix}&=a^3 - b^3 + c^3+ 3 a b c\\&≑a^3 - b^3 + c^3≑a-b+c\pmod3\end{align*} but $Ξ³=\frac{1-Ξ±+Ξ±^2}3=\frac1{1+Ξ±}βˆ‰π’ͺ_K$, as $$m_{𝐐,1+Ξ±}(x)=(x-1)^3-2=x^3-3x^2+3x-3$$ $$β‡’m_{𝐐,Ξ³}(x)=3 x^3 - 3 x^2 + 3 x - 1\text{ is not monic.}$$

    See MSE | kconrad