These are some brief notes from Michael Penn's youtube video. Essentially we go through the proof that the only associative commutative finite dimensional algebras over \(\mathbb{R}\) with no ZeroDivisors are \(\mathbb{R}\), \(\mathbb{C}\) and \(\mathbb{H}\) (the Quaternions).
Desirable Properties
\(\mathbb{C} = \{ a+bi | a,b in \mathbb{R} \}\)
- it is a two dimensional extension of \(\mathbb{R}\)
- it is a field (every nonzero element has a multiplicative inverse)
- it has on zero divisors (i.e. if \(ab=0\) then \(a=0\) or \(b=0\))
- multiplication is commutative and associative
Claim: The only finite dimensional associative algebras over \(\mathbb{R}\) without zero-divisors are \(\mathbb{R}\), \(\mathbb{C}\) and \(\mathbb{H}\). If we require commutativity then only \(\mathbb{R}\) and \(\mathbb{C}\). Note that none of these are 3-dimensional.
Proof. Let \(A\) be an \(n\)-dimensional algebra over \(\mathbb{R}\) without zero divisors.
- \(A\) has a multiplicative identity.
- Take any \(0\not=a\in A\) and define \(l_a:A\rightarrow A\), \(l_a(x)=ax\). So \(l_a\) is a linear transformation from \(A\) to \(A\). Since there are no zero divisors, the null space is \(\{0\}\). (To see this, if \(b\in\mathrm{Nul}(l_a)\) then \(l_a(b)=0\) so \(ab=0\) and since \(a\not=0\) we get \(b=0\).)
- So there is \(e\in A\) with \(l_a(e)=a\). Then \(ae=a\). We want to show that \(e\) is a multiplicative. Take some other \(b\in A\). Then \(ab=aeb\) so \(a(b-eb)=0\) and since \(a=0\) and there are no zero divisors, then \(b-eb=0\) and so \(b=eb\), so \(e\) is a left-multiplicative inverse.
- A similar argument shows that there is an \(e'\) with \(b=be'\) for all \(b\in A\). But then \(e=ee'=e'\) so \(e\) is a two-sided identity.
- So we can define \(1=e\), and then \(\mathbb{R}=\mathrm{span}(\{1\})=\mathbb{R}\subseteq A\).
- Every nonzero element of \(A\) has a multiplicative inverse.
- Take \(l_a\) as before. Then since \(l_a\) is surjective, there is \(b\in A\) with \(l_a(b)=1\). Then \(ab=1\), so \(a\) has a multiplicative inverse. We can use a similar argument with \(r_a\) to see that \(a\) has a right-inverse. To see it is two-sided, let \(c\) be such that \(bc=1\). Then \(c=c(ab)=(ca)b=b\).
- If \(\mathrm{dim}(A)=1\) then \(A\simeq\mathbb{R}\)
and we are done. So assume that \(\mathrm{dim}(A)\geq 2\).
- Take \(x\in A\setminus\mathbb{R}\). Look at the subalgebra generated by \(x\). We have \(\{a_0+a_1x+\cdots+a_mx^m|a_k\in\mathbb{R}\). Now the maximum degree of an irreducible polynomial over \(\mathbb{R}\) is 2. So since \(x\) can't be the root of a linear polynomial (since \(x\not\in A\)), it is a root of an irreducible quadratic, we have \(x^2+2bx+c=0\) with \(b^2-c<0\) so that there are no real roots. Then \(\mathrm{dim}(\mathbb{R}(x))=2\). So \(\mathbb{R}(x)=\mathbb{R}\oplus\mathbb{R}x\). Set \(y=x+b\). Then \(x=y-b\). So then \(0=x^2+2bx+c=(y-b)^2+2b(y-b)+c=y^2-2by+b^2+2by-2b^2+c=y^2-b^2+c<0\). Let's set \(i=\frac{1}{\sqrt{c-b^2}}y\) (ok since \(c-b^2>0\)). Then \(\mathbb{R}(x)=\mathbb{R}(i)\simeq\\mathbb{C}\).
- So if \(\mathrm{A}=2\) then \(A\simeq\mathbb{C}\).
- So let's assume \(\mathrm{dim}A\geq 3\).
- Define \(\sigma:A\rightarrow A\), \(\sigma(x)=-ixi-i^{-1}xi\).
- Then \(\sigma(xy)=-ixyi=(-ixi)(-iyi)=\sigma(x)\sigma(y)\).
- Now take \(x\in A\setminus\mathbb{C}\).
- Then \(\mathbb{C}\subseteq\mathbb{C}(x)\subseteq A\).
- We claim that \(xi\not= ix\).
- Suppose \(xi=ix\). Then \((xi)^m=x^mi^m\) by commutativity.
- Then let \(p(z)\in\mathbb{C}[z]\) be satisfied by \(x\), i.e. \(p(x)=0\).
- Now over \(\mathbb{C}\) this factors into linear factors \((z-\alpha_1)...(z-\alpha_m)\) and so \(x=\alpha_j\) for some \(j\).
- So \(x\in\mathbb{C}\) contradicting \(x\not\in\mathbb{C}\).
- Note: \(\mathbb{R}(i)=\mathbb{C}=\{x\in A|\sigma(x)=x\}\). (If \(x\not\in\mathbb{C}\) then \(\sigma(x)=x\) implies \(xi=ix\)).
- So set \(B=\{x\in A|\sigma(x)=-x\}\).
- If \(y\in A\) then \(y=\frac{1}{2}(y-\sigma(y))+\frac{1}{2}(y+\sigma(y))\). But \(\sigma(y-\sigma(y))= \sigma(y)-\sigma(\sigma(y))=\sigma(y)-y\) so \(y-\sigma(y)\in B\). And \(\sigma(y+\sigma(y))=\sigma(y)+\sigma(\sigma(y))=\sigma(y)+y\) so \(y+\sigma(y)\in\mathbb{C}\).
- So \(A=B\oplus\mathbb{C}\).
- Now take \(b\in B\). Then \(b\in A\setminus\mathbb{R}\). So \(j\in B, j^2=-1\).
- Now consider \(T_j:A\rightarrow A\), \(T_j(x)=jx\). Then for \(x\in B\) we have \(\sigma(jx)=\sigma(j)\sigma(x)=(-j)(-x)=jx\). So \(T_j(x)\in\mathbb{C}\). And if \(x\in\mathbb{C}\), then \(\sigma(jx)=\sigma(j)\sigma(x)=-jx\) and so \(T_j(x)\in B\). Also \(T_j^4=\mathrm{id}\) and so \(T_j\) is invertible. Thus \(B\) and \(\mathbb{C}\) have the same dimension.
- So \(\mathrm{dim}(A)=\mathrm{dim}(B)+\mathrm{dim}(\mathbb{C})=2+2=4\). And so \(A\) cannot have dimension \(3\).
- Finally we want to show that \(A=\mathbb{H}\).
Now we have three independent vectors over \(\mathbb{R}\):
\(1\), \(i\) and \(j\).
- Now \(\sigma(j)=-j\), so \(-iji=-j\) so \(ji=-ij\). But then \((ij)(ij)=i(ji)j=-i(ij)j=-(ii)(jj)=-(-1)(-1)=-1\).
- So define \(k=ij\). This is independent of \(1,i,j\), so \(1,i,j,k\) span \(A\). Also \(ijk=kk=-1\).
- So \(A=\{a+bi+cj+dj|a,b,c,d\in\mathbb{R},i^2=j^2=k^2=ijk=-1\}\) which is precisely the definition of \(\mathbb{H}\).