$\mathbb{H}$ is a division ring that is not commutative, hence not a field: $q^{-1}=\bar{q}\mathrm{/}N(q)$ for $q\mathrm{\ne }0$, while $ij=k$ and $ji=-k$

The additive group. Addition on $\mathbb{H}$ is defined coordinatewise from addition on $\mathbb{R}$, so it is associative and commutative, $0_{\mathbb{H}}$ is a two-sided identity, and $(-x_0,-x_1,-x_2,-x_3)$ is a two-sided additive inverse of $x$. Hence $(\mathbb{H},+,0_{\mathbb{H}})$ is an abelian group and $-x=(-x_0,-x_1,-x_2,-x_3)$.

Coefficient form of the product. For $m,p,q\in 4$ let ${\epsilon }_m(p,q)$ be the real coefficient of the monomial $x_p{y}_q$ in the $m$-th coordinate of the defining product formula, and ${\epsilon }_m(p,q)=0$ when that monomial does not occur there; every such coefficient is $1$, $-1$ or $0$. Reading the four coordinates of the formula off one at a time, $[xy{]}_m={\sum }_{p<4}{\sum }_{q<4}{\epsilon }_m(p,q)x_p{y}_q$ for all $x,y\in \mathbb{H}$ and all $m\in 4$, the right-hand side being a sum of sixteen real numbers.

The cyclic symmetry of the table. Let $\gamma :4\to 4$ fix $0$ and send $1\mapsto 2\mapsto 3\mapsto 1$, and let $\mathrm{\Gamma }:\mathbb{H}\to \mathbb{H}$ be the coordinate permutation determined by $[\mathrm{\Gamma }x{]}_{\gamma (p)}=x_p$, that is $\mathrm{\Gamma }(x_0,x_1,x_2,x_3)=(x_0,x_3,x_1,x_2)$. Then $\mathrm{\Gamma }$ is a bijection, $\mathrm{\Gamma }(e_p)=e_{\gamma (p)}$, $\mathrm{\Gamma }$ fixes each $\widehat{\lambda }$, and $\mathrm{\Gamma }(\lambda x)=\lambda \mathrm{\Gamma }(x)$, all immediately from the definition of $\mathrm{\Gamma }$.

Claim 2: from the table, $ij=k=(0,0,0,1)$ and $ji=-k=(0,0,0,-1)$. By [L1], $1\mathrm{\ne }0$, so [L3] gives $1=1^2>0$; the ordered-field definition in [L2] then gives $-1<0$, hence $1\mathrm{\ne }-1$. Thus the two products differ, and multiplication on $\mathbb{H}$ is not commutative.

Claim 3, the norm identity. Evaluating the product formula at $y=\bar{x}$ gives coordinates $x_0{x}_0-x_1(-x_1)-x_2(-x_2)-x_3(-x_3)=N(x)$, then $x_0(-x_1)+x_1{x}_0+x_2(-x_3)-x_3(-x_2)=0$, then $x_0(-x_2)+x_2{x}_0+x_3(-x_1)-x_1(-x_3)=0$, then $x_0(-x_3)+x_3{x}_0+x_1(-x_2)-x_2(-x_1)=0$; so $x\bar{x}=\widehat{N(x)}$. Evaluating it at $x:=\bar{x}$, $y:=x$ gives $x_0{x}_0-(-x_1)x_1-(-x_2)x_2-(-x_3)x_3=N(x)$ and, in the same way, $0$ in each of the other three coordinates; so $\bar{x}x=\widehat{N(x)}$ as well.

Claim 3, positivity. Let $x\mathrm{\ne }{0}_{\mathbb{H}}$; then $x_p\mathrm{\ne }0$ for at least one $p\in 4$. Each $x_{p^{\mathrm{\prime }}}^2$ with $x_{p^{\mathrm{\prime }}}\mathrm{\ne }0$ is positive, and each $x_{p^{\mathrm{\prime }}}^2$ with $x_{p^{\mathrm{\prime }}}=0$ equals $0$; a sum in which at least one summand is positive and the rest are positive or $0$ is positive, since $P$ is closed under addition and $u+0=u$. Hence $N(x)>0$, and in particular $N(x)\mathrm{\ne }0$.

Real scalars pass through the product. For $\lambda \in \mathbb{R}$ the formula gives $\widehat{\lambda }x=x\widehat{\lambda }=(\lambda x_0,\lambda x_1,\lambda x_2,\lambda x_3)$, an element we abbreviate $\lambda x$; in particular $\widehat{\lambda }\widehat{\mu }=\widehat{\lambda \mu }$, $1_{\mathbb{H}}x=x{1}_{\mathbb{H}}=x$, and $(-1)x=-x$ by step 1.1.

The coefficients are the multiplication table: ${\epsilon }_m(p,q)=[e_p{e}_q{]}_m$ for all $m,p,q\in 4$. Fix $p$ and $q$ and substitute $x=e_p$, $y=e_q$ into step 1.2: then $x_{p^{\mathrm{\prime }}}=0$ for $p^{\mathrm{\prime }}\mathrm{\ne }p$ and $y_{q^{\mathrm{\prime }}}=0$ for $q^{\mathrm{\prime }}\mathrm{\ne }q$, and a real product with a factor $0$ is $0$, so every one of the sixteen summands vanishes except the one indexed by $(p,q)$, which equals ${\epsilon }_m(p,q)\cdot 1\cdot 1={\epsilon }_m(p,q)$.

Both distributive laws hold. By step 1.2 and distributivity in $\mathbb{R}$, $[(x+x^{\mathrm{\prime }})y{]}_m={\sum }_{p,q}{\epsilon }_m(p,q)(x_p+x_p^{\mathrm{\prime }})y_q={\sum }_{p,q}({\epsilon }_m(p,q)x_p{y}_q+{\epsilon }_m(p,q)x_p^{\mathrm{\prime }}{y}_q)=[xy{]}_m+[x^{\mathrm{\prime }}y{]}_m$, the last equality being a regrouping of a finite sum of thirty-two real terms; the same computation in the second argument gives $x(y+y^{\mathrm{\prime }})=xy+x{y}^{\mathrm{\prime }}$.

The nine table checks that establish $e_{\gamma (p)}{e}_{\gamma (q)}=\mathrm{\Gamma }(e_p{e}_q)$ for all $p,q\in \{1,2,3\}$, read off the table of The quaternions $\mathbb{H}$: real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1,i,j,k$: $jj=-1=\mathrm{\Gamma }(-1)=\mathrm{\Gamma }(ii)$; $jk=i=\mathrm{\Gamma }(k)=\mathrm{\Gamma }(ij)$; $ji=-k=\mathrm{\Gamma }(-j)=\mathrm{\Gamma }(ik)$; $kj=-i=\mathrm{\Gamma }(-k)=\mathrm{\Gamma }(ji)$; $kk=-1=\mathrm{\Gamma }(-1)=\mathrm{\Gamma }(jj)$; $ki=j=\mathrm{\Gamma }(i)=\mathrm{\Gamma }(jk)$; $ij=k=\mathrm{\Gamma }(j)=\mathrm{\Gamma }(ki)$; $ik=-j=\mathrm{\Gamma }(-i)=\mathrm{\Gamma }(kj)$; $ii=-1=\mathrm{\Gamma }(-1)=\mathrm{\Gamma }(kk)$. The cases with $p=0$ or $q=0$ are immediate, both sides being $e_{\gamma (q)}$ or $e_{\gamma (p)}$ respectively.

Claim 5: every field is a commutative ring by [L6], and the multiplication of $\mathbb{H}$ is not commutative by step 1.4; so $\mathbb{H}$ is not a field.

Real scalars pass through a triple product too: by step 1.2 and step 2.1, $[(\lambda x)y{]}_m={\sum }_{p,q}{\epsilon }_m(p,q)(\lambda x_p)y_q=\lambda [xy{]}_m$ and likewise $[x(\lambda y){]}_m=\lambda [xy{]}_m$, so $(\lambda x)y=\lambda (xy)=x(\lambda y)$. Taking $\lambda =-1$ gives $(-x)y=-(xy)=x(-y)$.

Reduction of associativity to the sixty-four basis triples. Applying step 1.2 twice and rearranging, $[(xy)z{]}_m={\sum }_{s,r}{\epsilon }_m(s,r)[xy{]}_s{z}_r={\sum }_{p,q,r}({\sum }_s{\epsilon }_m(s,r){\epsilon }_s(p,q))x_p{y}_q{z}_r$, and applying step 2.2 twice, $[(e_p{e}_q)e_r{]}_m={\sum }_s{\epsilon }_m(s,r)[e_p{e}_q{]}_s={\sum }_s{\epsilon }_m(s,r){\epsilon }_s(p,q)$; hence $[(xy)z{]}_m={\sum }_{p,q,r<4}{x}_p{y}_q{z}_r[(e_p{e}_q)e_r{]}_m$. The same computation with the other bracketing gives $[x(yz){]}_m={\sum }_{p,q,r<4}{x}_p{y}_q{z}_r[e_p(e_q{e}_r){]}_m$. Therefore, if $(e_p{e}_q)e_r=e_p(e_q{e}_r)$ holds for all $p,q,r\in 4$, then $(xy)z=x(yz)$ for all $x,y,z\in \mathbb{H}$.

Basis triples containing the index $0$. Since $e_0=1_{\mathbb{H}}$ is a two-sided identity by step 2.1, each of $(e_0{e}_q)e_r=e_q{e}_r=e_0(e_q{e}_r)$, $(e_p{e}_0)e_r=e_p{e}_r=e_p(e_0{e}_r)$ and $(e_p{e}_q)e_0=e_p{e}_q=e_p(e_q{e}_0)$ holds. So only the twenty-seven triples with $p,q,r\in \{1,2,3\}$ remain.

The multiplication of $\mathbb{H}$ commutes with $\mathrm{\Gamma }$: $\mathrm{\Gamma }(x)\mathrm{\Gamma }(y)=\mathrm{\Gamma }(xy)$ for all $x,y$. By step 1.2 and $[\mathrm{\Gamma }x{]}_p=x_{{\gamma }^{-1}(p)}$, $[\mathrm{\Gamma }(x)\mathrm{\Gamma }(y){]}_{\gamma (m)}={\sum }_{p,q}{\epsilon }_{\gamma (m)}(\gamma (p),\gamma (q))x_p{y}_q$, while $[\mathrm{\Gamma }(xy){]}_{\gamma (m)}=[xy{]}_m={\sum }_{p,q}{\epsilon }_m(p,q)x_p{y}_q$; by step 2.2 the two families of coefficients are $[e_{\gamma (p)}{e}_{\gamma (q)}{]}_{\gamma (m)}$ and $[e_p{e}_q{]}_m=[\mathrm{\Gamma }(e_p{e}_q){]}_{\gamma (m)}$, which agree by step 2.4.

Reduction of the twenty-seven triples to nine. Suppose $(e_p{e}_q)e_r=e_p(e_q{e}_r)$ for a triple $(p,q,r)$. Applying step 3.4, $(e_{\gamma (p)}{e}_{\gamma (q)})e_{\gamma (r)}=\mathrm{\Gamma }(e_p{e}_q)\mathrm{\Gamma }(e_r)=\mathrm{\Gamma }((e_p{e}_q)e_r)$ and $e_{\gamma (p)}(e_{\gamma (q)}{e}_{\gamma (r)})=\mathrm{\Gamma }(e_p)\mathrm{\Gamma }(e_q{e}_r)=\mathrm{\Gamma }(e_p(e_q{e}_r))$, so the identity holds for $(\gamma (p),\gamma (q),\gamma (r))$ as well. Since $\gamma$ restricted to $\{1,2,3\}$ is a cycle of length three, for each $p\in \{1,2,3\}$ there is exactly one $t\in \{0,1,2\}$ with ${\gamma }^t(p)=1$; hence every triple in $\{1,2,3{\}}^3$ is obtained by iterating $\gamma$ from a triple whose first entry is $1$, and it suffices to check the nine triples $(1,q,r)$ with $q,r\in \{1,2,3\}$.

The nine remaining checks, using the table of The quaternions $\mathbb{H}$: real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1,i,j,k$ and the sign rule of step 3.1: $(ii)i=(-1)i=-i$ and $i(ii)=i(-1)=-i$; $(ii)j=-j$ and $i(ij)=ik=-j$; $(ii)k=-k$ and $i(ik)=i(-j)=-(ij)=-k$; $(ij)i=ki=j$ and $i(ji)=i(-k)=-(ik)=j$; $(ij)j=kj=-i$ and $i(jj)=i(-1)=-i$; $(ij)k=kk=-1$ and $i(jk)=ii=-1$; $(ik)i=(-j)i=-(ji)=k$ and $i(ki)=ij=k$; $(ik)j=(-j)j=-(jj)=1$ and $i(kj)=i(-i)=-(ii)=1$; $(ik)k=(-j)k=-(jk)=-i$ and $i(kk)=i(-1)=-i$. All nine agree.

Multiplication on $\mathbb{H}$ is associative: by steps 3.3, 4.1 and 4.2 the identity $(e_p{e}_q)e_r=e_p(e_q{e}_r)$ holds for all sixty-four basis triples, and step 3.2 transfers it to all of $\mathbb{H}$.

Claim 1: by step 1.1 the additive structure is an abelian group; by step 5.1 and step 2.1 multiplication is associative with two-sided identity $1_{\mathbb{H}}$, so $(\mathbb{H},\cdot ,1_{\mathbb{H}})$ is a monoid; and both distributive laws hold by step 2.3. So $\mathbb{H}$ is a ring.

Claim 4. First $1_{\mathbb{H}}\mathrm{\ne }{0}_{\mathbb{H}}$, because $1\mathrm{\ne }0$ in $\mathbb{R}$. Let $x\mathrm{\ne }{0}_{\mathbb{H}}$ and put $\lambda :=N(x{)}^{-1}$, which exists by step 1.6, and $y:=\lambda \bar{x}=\widehat{N(x{)}^{-1}}\bar{x}$. Then $xy=x(\lambda \bar{x})=\lambda (x\bar{x})=\lambda \widehat{N(x)}=\widehat{\lambda N(x)}=\widehat{1}=1_{\mathbb{H}}$ by step 3.1, step 1.5 and step 2.1, and $yx=(\lambda \bar{x})x=\lambda (\bar{x}x)=1_{\mathbb{H}}$ in the same way. So $x$ is a unit of the ring $\mathbb{H}$ with $x^{-1}=\widehat{N(x{)}^{-1}}\bar{x}$, and $\mathbb{H}$ is a division ring; by [L5] its units form a group, and by the description of a division ring that group is $\mathbb{H}\setminus \{0_{\mathbb{H}}\}$.

Claims 1 to 5 are established: claim 1 in step 6.1, claim 2 in step 1.4, claim 3 in step 1.5 together with step 1.6, claim 4 in step 7.1 and claim 5 in step 2.5.