HW8

Let $\mathbb{Z}[\sqrt{2}]$ be the set of elements $a+b\sqrt{2}$ with $a,b\in \mathbb{Z}$. Show that it is a ring. Show that it is an Euclidean domain.

\begin{proof} i) Firstly, we show $\mathbb{Z}[\sqrt{2}]$ is a ring. For any $a_1+b_1\sqrt{2}$ and $a_2+b_2\sqrt{2}\in \mathbb{Z}[\sqrt{2}]$, we have that

$$\begin{array}{r}(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2})=(a_1{a}_2+2b1b_2)+(a_1{b}_2+a_2{b}_1)\sqrt{2}\in \mathbb{Z}[\sqrt{2}],\\ (a_1+b_1\sqrt{2})+(a_2+b_2\sqrt{2})=(a_2+b_2\sqrt{2})+(a_1+b_1\sqrt{2})\in \mathbb{Z}[\sqrt{2}],\end{array}$$

and $0\in \mathbb{Z}[\sqrt{2}]$ is the identity of $(\mathbb{Z}[\sqrt{2}],+)$. Therefore, $\mathbb{Z}[\sqrt{2}]$ is a ring.

ii) Define $r:\mathbb{Q}[\sqrt{2}]\to \mathbb{Q},a+b\sqrt{2}\mapsto |a^2-2b^2|$ and note that $r(\mathbb{Z}[\sqrt{2}])\subseteq \mathbb{Z}$. For any $a+b\sqrt{2}\in \mathbb{Z}[\sqrt{2}]$ and $c+d\sqrt{2}\in \mathbb{Z}[\sqrt{2}]$, we aim to find $q\in \mathbb{Z}[\sqrt{2}]$ such that

$$r(a+b\sqrt{2}-q(c+d\sqrt{2}))<r(a+b\sqrt{2}).\text{\hspace{1em}(*)}$$

Note that $r(a)r(b)=r(ab)$ for any $a,b\in \mathbb{Q}[\sqrt{2}]$, then $(\ast )$ is equivalent to $r(\frac{a+b\sqrt{2}}{c+d\sqrt{2}}-q)<1$.

Hence, to show $\mathbb{Z}[\sqrt{2}]$ is Euclidean, it suffices to show for any $u+v\sqrt{2}\in \mathbb{Q}\sqrt{2}$, there exists $m+n\sqrt{2}\in \mathbb{Z}\sqrt{2}$ satisfying $|(u-m{)}^2-2(v-n{)}^2|<1$. We only need to choose $m,n\in \mathbb{Z}$ such that $|u-m|<1/2$ and $|v-n|<1/2$. Now we finish the proof. \end{proof}

$\mathbb{Z}[\sqrt{-2}]$ is an Euclidean domain.

\begin{proof} Similarly, define $r:\mathbb{Q}[\sqrt{-2}]\to \mathbb{Q},a+b\sqrt{-2}\mapsto a^2+2b^2$ and note that $r(\mathbb{Z}[\sqrt{-2}])\subseteq \mathbb{Z}$, $r(a)r(b)=r(ab)$ for all $a,b\in \mathbb{Q}[\sqrt{2}]$. For any $a+b\sqrt{-2}\in \mathbb{Z}[\sqrt{-2}]$ and $c+d\sqrt{-2}\in \mathbb{Z}[\sqrt{-2}]$, we aim to find $q\in \mathbb{Z}\sqrt{-2}$ such that $r(a+b\sqrt{2}-q(c+d\sqrt{2}))<r(a+b\sqrt{2})$, that is, $r(\frac{a+b\sqrt{2}}{c+d\sqrt{2}}-q)<1$.

It suffices to show for any $u+v\sqrt{-2}\in \mathbb{Q}\sqrt{-2}$ there exists $m+n\sqrt{-2}\in \mathbb{Z}\sqrt{-2}$ satisfying $|(u-m{)}^2-2(v-n{)}^2|<1$, and we only need to choose $m,n\in \mathbb{Z}$ such that $|u-m|<1/2$ and $|v-n|<1/2$ as $1\cdot 1/4+2\cdot 1/4<1$. \end{proof}

Let $A=\mathbb{Z}[i]$. Let $M=Ae\oplus Af$. Let $M^{\prime }$ be the submodule generated by $2e$ and $(3+i)f$. Find a basis of $M^{\prime }$ as in Part(b) of our main theorem. you need to show your calculations.

\begin{proof} Note that $(2,3+i)=1+i$, and we can check

$$⟨2e,(3+i)f⟩=⟨(1+i)((1-i)e+(2-i)f),(4-2i)(-e-f)⟩$$

because RHS equals $\left[\begin{array}{cc}2e& (3+i)f\end{array}\right]\left[\begin{array}{cc}1& i-2\\ 1& i-1\end{array}\right]$ and the matrix is invertible.Furthermore, $M=⟨(1-i)e+(2-i)f,-e-f⟩$ by $\left[\begin{array}{cc}1-i& -1\\ 2-i& -1\end{array}\right]$ is invertible. Therefore, $M^{\prime }=⟨(1+i)e^{\prime },(4-2i)f^{\prime }⟩$ where $e^{\prime }=(1-i)e+(2-i)f$ and $f^{\prime }=-(e+f)$. \end{proof}

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Note: Unless specified otherwise $F$ is always an extension field of the field $K$

If $v$ is algebraic over $K(u)$ for some $u\in F$ and $v$ is transcendental over $K$, then $u$ is algebraic over $K(v)$.

\begin{proof} Since $v$ is algebraic over $K(u)$, there exists $f(x)={\sum }_{k=0}^n{a}_n{x}^n$ such that $a_n\in K(u)$ and $f(v)=0$. Assume that $a_k=b_k(u)/c_k(u)$ where $b_k(x),c_k(x)\in K[x]$ and $c_k(u)\ne 0$, then

$$f(v)=\sum _{k=0}^n\frac{b_k(u)}{c_k(u)}{v}^k=0\text{mbox}andso(\prod _{k=0}^n{c}_k(u))f(v)=\sum _{k=0}^n(b_k(u)\prod _{m\ne k}{c}_m(u))v^k=0.$$

Therefore, there exists a $h(x)={\sum }_{k=0}^{\ell }{d}_k{x}^k$ where $d_k\in K(v)$ and $h(u)=0$. Since $v$ is transcendental over $K$, not all $b_k(u)/c_k(u)\in K$ and so $\mathrm{deg}h⩾1$. Hence, $u$ is algebraic over $K(v)$. \end{proof}

If $F$ is algebraic over $K$ and $D$ is an integral domain such that $K\subset D\subset F$, then $D$ is a field.

\begin{proof} For each non-zero element $d\in D$, since $D\subseteq F$ and $F$ is algebraic over $K$, there exists a non-zero polynomial $f(x)={\sum }_{k=0}^n{a}_k{x}^k$ such that $a_k\in K$ and $f(d)=0$. If $a_0=0$, then $f(d)=d({\sum }_{k=1}^n{a}_k{d}^{k-1})=0$ and take $f(x)={\sum }_{k=1}^n{a}_k{x}^{k-1}$. Repeat this procedure, then we can assume that $a_0\ne 0$.

Note that $(-a_0{)}^{-1}d(a_1+a_{2}d+\cdots +a_n{d}^{n-1})=1$, so $d^{-1}=(-a_0{)}^{-1}(a_1+a_{2}d+\cdots +a_n{d}^{n-1})$. Since $a_k$ and $d$ are contained in $D$, $d^{-1}\in D$. For each $0\ne d\in D$ there exists $d^{-1}\in D$ and $D$ is an integral domain. Therefore, $D$ is a field. \end{proof}