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\begin{document}

\section*{Assignment 4: Due April 2, 2019}
Let's do some basic things with division rings.  Recall that a division ring is an associative ring in which every nonzero element has an inverse.  Some theorems that you might not know, but that are useful are the Artin-Wedderburn theorem.  We'll use a very special case that you might have seen if you took representation theory: if $F$ is an algebraically closed field and $A$ is a finite-dimensional simple $F$-algebra then $A\cong M_n(F)$ for some $n$. 
\begin{enumerate}
\item[1] Let $D$ be a division ring. Show that the centre $Z=\{x\in D\colon zx=xz\}$ is a field. 
\item[2] Let $D$ be a division ring and let $Z$ be its centre. We say that $D$ is finite-dimensional if $[D:Z]<\infty$; i.e., $D$ is finite-dimensional as a $Z$-vector space. Show that if $D$ is a finite-dimensional division ring and $F$ is a field extension of $Z$ then $B:=D\otimes_Z F$ is a simple $F$-algebra whose centre is $F\cong Z\otimes_Z F$ and ${\rm dim}_F(B)={\rm dim}_Z(D)$.  (Hint: to show that it is simple, first do the part about the centre.  Now Let $I$ be a nonzero ideal and pick nonzero $x=\sum_{i=1}^s a_i\otimes \lambda_i\in I$ with $s$ minimal, where the $a_i\in D$ and $\lambda_i\in F$.  First show by minimality of $s$ that you may assume that the $a_i$ are linearly independent over $Z$ and the $\lambda_i$ are linearly independent over $Z$ and that $a_1=1$ and $\lambda_1=1$.  Next show that if $s=1$ then $x=1$ under these assumptions, so if you can show that $s=1$, you're done.  So now assume $s>1$.  Show that if $x (b\otimes \gamma) - (b\otimes\gamma)x$ has shorter length, so show that $x$ is central and use your description of the centre to finish things off.)
\item[3] Use the Artin-Wedderburn theorem to show that if $D$ is a finite-dimensional division ring then $[D:Z]$ is a perfect square.  (Hint: Look at $D\otimes_Z \bar{Z}$, where $\bar{Z}$ is the algebraic closure of $Z$.)
\item[4] Let $K$ be a field and assume that there is $\omega\in K$ with $\omega$ an $n$-th root of unity.  Let $a,b\in K\setminus \{0\}$. Define a ring $R=K\{x,y\}/(x^n-a, y^n-b, xy-\omega yx)$.  Show that $R$ is an $n^2$-dimensional $K$-algebra.  (Hint: put a dlex order on monomials by declaring that $x>y$ and show that the relations given yield a Gr\"obner-Shirshov basis and that the monomials that do not have initial terms as subwords are those of the form $y^i x^j$ with $i,j<n$.  Show also that $R$ is simple.
\item[5] Let $K$ be a field, let $n=2$, $\omega=-1$, $a=b=-1$.  Show that the $R$ from the preceding example is a division ring if zero cannot be written as a non-trivial sum of at most four squares in $K$ (e.g., fields such at $\mathbb{Q}$ and $\mathbb{R}$) and show that it is isomorphic to $M_2(K)$ if $0$ can be written as a sum of at most four nonzero squares (e.g., fields like $\mathbb{Q}(i)$ and $\mathbb{C}$).  In the former case, we call $R$ the division ring of quaternions over $K$.  (Hint: Show that $1, i:=x$, $j:=y$, and $k:=xy$ is a $K$-basis for $R$ now show that if $u:=a+bi+cj+dk$ is a zero divisor then so is $uu^*$, where $u^*=a-bi-cj-dk$ and that $uu^*=u^*u=a^2+b^2+c^2+d^2$.)  
\item[6] Let $H=\{a+bi+cj+dk \colon 2a,2b,2c,2d\in \mathbb{Z}, 2a\equiv 2b\equiv 2c\equiv 2d~(\bmod~2)\}$.  Show that $H$ is a subring of the division ring of quaternions over $\mathbb{Q}$ and that if $u\in H$ then $N(u):=uu^*=u^*u$ is a positive integer and is nonzero whenever $u$ is nonzero, and that $N(uv)=N(u)N(v)$.  
\item[7] Show that $H$ has a left-division algorithm given as follows: If $a,b\in H$ with $b\neq 0$ then there exist $q,r\in H$ such that $a=qb+r$ with $N(r)<N(b)$.  (Hint: this is a bit tricky, but it is much easier when $b$ is a positive integer, so do this case first.  Now let $n=bb^*>0$ and do the case you've just done to get $ab^* = qn +r$ with $N(r)<N(n)=n^2$.  Now here's the fun part: $r=ab^*-qn = ab^*-qbb^* = (a-qb)b^*$.  Let $r'=a-qb$.  Show that $N(r')<N(b)$!)
\item[8] Use the preceding result to show that every left ideal of $H$ can be generated by a single element as a left ideal.
\item[9] Show that if $a,b\in H$ are nonzero and are such that $N(a)=N(b)$ and $a=ub$ with $u\in H$, then $u$ is a unit of $H$.  
\item[10] Show that every positive integer can be written as a sum of $4$ squares (including $0$ as a square).  (Hint: using questions $5$ and $6$ that it is enough to prove that every prime number is a sum of four squares.  Let $p\ge 3$ be prime (I assume you can write $p=2$ as a sum of four squares) and let $R_p$ be the algebra produced in question 4 with $K=\mathbb{F}_p$, $n=2$, $\omega=-1$, $a=b=-1$.  Show $R_p$ is not commutative if $p>2$ and use Wedderburn's theorem to show that $R_p$ cannot be a division ring.  OK, so now let $I$ denote the two-sided ideal of $H$ given by $ \{a+bi+cj+dk \colon 2a,2b,2c,2d\in \mathbb{Z}, 2a\equiv 2b\equiv 2c\equiv 2d~(\bmod~2), p|2a, p|2b, p|2c, p|2d\}$.  Show that $I$ is indeed an ideal and that $H/I$ is isomorphic to $R_p$.  What next?  Since $R_p$ is not a division ring it has a nonzero proper left ideal $J$.  Then by correspondence, there is a proper left ideal $J'$ of $H$ that properly contains $I$.  Show that every element $u\in J'$ must have the property that $N(u)$ is a multiple of $p$ since otherwise we could generate the unit ideal.  By the preceding question, $J'$ can be generated by a single element $a:=A+Bi+Cj+Dk\in J'\setminus I$ with $N(a)$ a multiple of $p$.  Since $p\in J'$ we have $p=ba$ for some $b\in H$.  Then $N(p)=N(b)N(a)$. 
Now $N(p)=p^2$ and $N(b)$ and $N(a)$ are integers and $N(a)$ is a multiple of $p$, so $N(a)\in \{p,p^2\}$.  Show using question 9 that if $N(a)=p^2$ then $p=ba$ with $b$ a unit of $H$ and this cannot occur since $J'$ properly contains $I$.  Conclude that $N(a)=p$.  What does this mean? It means $p=A^2+B^2+C^2+D^2$.  There's just one problem: $A,B,C,D$ are not necessarily integers---they are only half integers and $2A,2B,2C,2D$ have the same parity.  But we're close.  But let $A'=2A$, $B'=2B$, $C'=2C$, $D'=2D$, so now we have integers and we have $4p=(A')^2+(B')^2+(C')^2+(D')^2$ and $A', B', C', D'$ have the same parity.  OK, now let $X=(A'-B')/2$, $Y=(A'+B')/2$, $Z=(C'-D')/2$, $W=(C'+D')/2$.  What happens?)
\end{enumerate}
\end{document}
