Suppose that $k$ is a field with $char(k) \neq 2$. Let's agree that an "octonion algebra" over $k$ is an 8-dimensional unital $k$-algebra $A$, endowed with a quadratic form $N: A \rightarrow k$, whose associated bilinear form $T(x,y) = N(x+y) - N(x) - N(y)$ is nondegenerate, and which satisfies $N(xy) = N(x) N(y)$ for all $x,y \in A$.

When $x \in A - k$, the trace $Tr(x) = T(x,1)$ and norm $N(x)$ are determined by the algebra structure -- $x$ is the root of a quadratic polynomial with coefficients $-Tr(x)$ and $N(x)$. Hence the algebra structure determines the quadratic form $N$ in a convenient way. One can talk about the "isomorphism class of an octonion algebra" while carrying around the quadratic form or not -- it doesn't really matter.

It is an old theorem (Jacobson? Albert? I can't recall... check the "Book of Involutions" too) that the isomorphism class of the octonion algebra $k$ is determined by the isomorphism class of the quadratic form $N$. Now, essentially by the Cayley-Dickson doubling process, the norm form $N$ is a Pfister form, i.e. $N$ is isomorphic to $<1,-a> \otimes <1,-b> \otimes <1,-c>$ for some $a,b,c \in k$.

It is a fact about Pfister forms that when they are isotropic, they are split. So if the norm form represents zero, then $N$ is isomorphic to $<1,-1> \otimes <1,-1> \otimes <1,-1>$ and the octonion algebra is isomorphic to the split octonion algebra over $k$.

In this level of generality, the isomorphism classes of octonion algebras over $k$ are classified up to isomorphism by the Galois cohomology $H^3(k, \mu_2)$; you can see such a cohomology class too from the Pfister form perspective. The Pfister form $<1,-a> \otimes <1,-b> \otimes <1,-c>$ depends on the square-classes of $a, b, c$, giving three classes in $H^1(k, \mu_2)$, whose cup product is the element of $H^3(k, \mu_2)$ classifying the octonion algebra.

$\DeclareMathOperator\Tr{Tr} $Suppose that $k$ is a field with $\operatorname{char}(k) \neq 2$. Let's agree that an "octonion algebra" over $k$ is an 8-dimensional unital $k$-algebra $A$, endowed with a quadratic form $N: A \rightarrow k$, whose associated bilinear form $T(x,y) = N(x+y) - N(x) - N(y)$ is nondegenerate, and which satisfies $N(xy) = N(x) N(y)$ for all $x,y \in A$.

When $x \in A \setminus k$, the trace $\Tr(x) = T(x,1)$ and norm $N(x)$ are determined by the algebra structure: $x$ is the root of a quadratic polynomial with coefficients $-\Tr(x)$ and $N(x)$. Hence the algebra structure determines the quadratic form $N$ in a convenient way. One can talk about the "isomorphism class of an octonion algebra" while carrying around the quadratic form or not; it doesn't really matter.

It is an old theorem (Jacobson? Albert? I can't recall … check "The Book of Involutions" too) that the isomorphism class of the octonion algebra $k$ is determined by the isomorphism class of the quadratic form $N$. Now, essentially by the Cayley–Dickson doubling process, the norm form $N$ is a Pfister form, i.e., $N$ is isomorphic to $\langle1,-a\rangle \otimes \langle1,-b\rangle \otimes \langle1,-c\rangle$ for some $a,b,c \in k$.

It is a fact about Pfister forms that when they are isotropic, they are split. So if the norm form represents zero, then $N$ is isomorphic to $\langle1,-1\rangle \otimes \langle1,-1\rangle \otimes \langle1,-1\rangle$ and the octonion algebra is isomorphic to the split octonion algebra over $k$.

In this level of generality, the isomorphism classes of octonion algebras over $k$ are classified up to isomorphism by the Galois cohomology $H^3(k, \mu_2)$; you can see such a cohomology class too from the Pfister form perspective. The Pfister form $\langle1,-a\rangle \otimes \langle1,-b\rangle \otimes \langle1,-c\rangle$ depends only on the square-classes of $a, b, c$, giving three classes in $H^1(k, \mu_2)$ whose cup product is the element of $H^3(k, \mu_2)$ classifying the octonion algebra.

Suppose that $k$ is a field with $char(k) \neq 2$. Let's agree that an "octonion algebra" over $k$ is an 8-dimensional unital $k$-algebra $A$, endowed with a quadratic form $N: A \rightarrow k$, whose associated bilinear form $T(x,y) = N(x+y) - N(x) - N(y)$ is nondegenerate, and which satisfies $N(xy) = N(x) N(y)$ for all $x,y \in A$.

When $x \in A - k$, the trace $Tr(x) = T(x,1)$ and norm $N(x)$ are determined by the algebra structure -- $x$ is the root of a quadratic polynomial with coefficients $-Tr(x)$ and $N(x)$. Hence the algebra structure determines the quadratic form $N$ in a convient way. One can talk about the "isomorphism class of an octonion algebra" while carrying around the quadratic form or not -- it doesn't really matter.

It is an old theorem (Jacobson? Albert? I can't recall... check the "Book of Involutions" too) that the isomorphism class of the octonion algebra $k$ is determined by the isomorphism class of the quadratic form $N$. Now, essentially by the Cayley-Dickson doubling process, the norm form $N$ is a Pfister form, i.e. $N$ is isomorphic to $<1,-a> \otimes <1,-b> \otimes <1,-c>$ for some $a,b,c \in k$.

It is a fact about Pfister forms that when they are isotropic, they are split. So if the norm form represents zero, then $N$ is isomorphic to $<1,-1> \otimes <1,-1> \otimes <1,-1>$ and the octonion algebra is isomorphic to the split octonion algebra over $k$.

In this level of generality, the isomorphism classes of octonion algebras over $k$ are classified up to isomorphism by the Galois cohomology $H^3(k, \mu_2)$; you can see such a cohomology class too from the Pfister form perspective. The Pfister form $<1,-a> \otimes <1,-b> \otimes <1,-c>$ depends on the square-classes of $a, b, c$, giving three classes in $H^1(k, \mu_2)$, whose cup product is the element of $H^3(k, \mu_2)$ classifying the octonion algebra.

Suppose that $k$ is a field with $char(k) \neq 2$. Let's agree that an "octonion algebra" over $k$ is an 8-dimensional unital $k$-algebra $A$, endowed with a quadratic form $N: A \rightarrow k$, whose associated bilinear form $T(x,y) = N(x+y) - N(x) - N(y)$ is nondegenerate, and which satisfies $N(xy) = N(x) N(y)$ for all $x,y \in A$.

When $x \in A - k$, the trace $Tr(x) = T(x,1)$ and norm $N(x)$ are determined by the algebra structure -- $x$ is the root of a quadratic polynomial with coefficients $-Tr(x)$ and $N(x)$. Hence the algebra structure determines the quadratic form $N$ in a convenient way. One can talk about the "isomorphism class of an octonion algebra" while carrying around the quadratic form or not -- it doesn't really matter.

It is an old theorem (Jacobson? Albert? I can't recall... check the "Book of Involutions" too) that the isomorphism class of the octonion algebra $k$ is determined by the isomorphism class of the quadratic form $N$. Now, essentially by the Cayley-Dickson doubling process, the norm form $N$ is a Pfister form, i.e. $N$ is isomorphic to $<1,-a> \otimes <1,-b> \otimes <1,-c>$ for some $a,b,c \in k$.

It is a fact about Pfister forms that when they are isotropic, they are split. So if the norm form represents zero, then $N$ is isomorphic to $<1,-1> \otimes <1,-1> \otimes <1,-1>$ and the octonion algebra is isomorphic to the split octonion algebra over $k$.

In this level of generality, the isomorphism classes of octonion algebras over $k$ are classified up to isomorphism by the Galois cohomology $H^3(k, \mu_2)$; you can see such a cohomology class too from the Pfister form perspective. The Pfister form $<1,-a> \otimes <1,-b> \otimes <1,-c>$ depends on the square-classes of $a, b, c$, giving three classes in $H^1(k, \mu_2)$, whose cup product is the element of $H^3(k, \mu_2)$ classifying the octonion algebra.