This is excerpt from W. Scharlau: Quadratic and Hermitian Forms, page 36.
A second invariant is the determinant. Since $\det(\phi \perp \psi) = \det(\phi) \det(\psi)$ we get by $1.1$ a group homomorphism $$ \det \colon \hat{W}(K) \to K^\bullet/K^{\bullet 2}. $$ We will see in the next section that over finite fields, for example, dimension and determinant are sufficient to classify quadratic spaces. This can be expressed as follows: there is a group isomorphism $$ (\dim, \det) \colon \hat{W}(K) \to \mathbb{Z} \times K^\bullet / K^{\bullet 2}. $$
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I know that there is a homomorphic map $\dim \colon \widehat W(K) \to \mathbb{Z}$. It is the same as the construction of integers from the natural number and $0$. Also there is homomorphic map from $\det \colon \widehat W(K) \to K^\bullet/K^{\bullet 2}$. Also how one can show homomorphism by using $(\dim,\det)$ map. I think homomorphism is true since the direct product of homomorphism is a homomorphism. But how to write it explicitly? If we know it is homomorphism then we have to show for $(0,0)$ element dimension and $\det$ is $0$ to show kernel is trivial. How to show that?
Given any quadratic form, there exists a diagonal representation, let's call these representations $D=\langle d_1,\ldots,d_n\rangle$ where $\dim(D)=n.$ Crucially, we may always choose $d_j\in K^\ast/{K^\ast}^2.$ This defines a homomorphism,
$$\dim : \widehat{W}(K)\longrightarrow \Bbb{Z}$$ by $$\dim(D)\mapsto \text{length}(D).$$ This is certainly a homomorphism, since orthogonal sums of quadratic forms obey the following additive law $$\underbrace{\langle a_1,\ldots,a_k\rangle}_{\dim = k}\perp\underbrace{ \langle b_1,\ldots,b_\ell\rangle}_{\dim =\ell}\cong\underbrace{ \langle a_1,\ldots,a_k,b_1,\ldots,b_\ell\rangle}_{\dim = k +\ell}.$$
Similarly $$\det(\langle d_1,\ldots,d_n\rangle)=\prod_{j=1}^n d_j\in K^\ast/{K^\ast}^2$$
So the map $$(\dim,\det)(D)=(\dim(D),\det(D))=\left(\text{length}(D),\prod_{j=1}^n d_j\right)$$ is the explicit homomorphism.