The signature is independent of the diagonalizing basis and unchanged by scalar extension from the rationals to the reals

Let $B$ be a symmetric form on a finite-dimensional space over an ordered field $F$, and let $K\supseteq F$ carry an extending order. By [F3] choose an orthogonal $F$-basis with diagonal entries $b_i$. Write $V=V_{+}\oplus V_{-}\oplus V_0$ for its positive, negative and zero coordinate subspaces, of dimensions $p,q,r$. The radical is $V_0$. A positive-definite subspace projects injectively to $V_{+}$: a vector with zero positive coordinates has $B(v,v)\le 0$, so cannot be a nonzero vector in such a subspace. Rank-nullity and the subspace dimension bound (Rank-nullity: ${\dim }_{F}V=\mathrm{nullity}T+\mathrm{rank}T$, If ${\dim }_{F}V=n$ and $U$ is a linear subspace of $V$, then $U$ is finite-dimensional, ${\dim }_{F}U\le n$, and ${\dim }_{F}U=n$ if and only if $U=V$) give dimension at most $p$, attained by $V_{+}$. Similarly the maximal negative dimension is $q$. Thus $(p,q,r)$ is intrinsic over any ordered field. By [F4], the extended basis has the same diagonal entries and signs over $K$, so these intrinsic dimensions, and hence the signature, are preserved.

Basis independence and well-posedness: by [F2] the diagonal entries' sign counts $(p,q,r)$ of any diagonalizing basis of $Q_M$ are intrinsic to $Q_M$, and by [F3] congruent forms have equal inertia, so $p$, $q$ are the same for every diagonalizing basis; since $Q_M$ is symmetric and nondegenerate by The middle-dimensional intersection form is symmetric and nondegenerate, $r=0$ and $\sigma (M)=p-q$ is the intrinsic inertia difference of $Q_M$ as claimed in The signature of a closed oriented manifold of dimension divisible by four.

Coefficient bridge: by [F5] applied with the divisible groups $F=\mathbb{Q}$ and $F=\mathbb{R}$, evaluation gives natural isomorphisms $H^{2k}(M;\mathbb{Q})\cong {\mathrm{Hom}}_{\mathbb{Z}}(H_{2k}(M;\mathbb{Z}),\mathbb{Q})$ and $H^{2k}(M;\mathbb{R})\cong {\mathrm{Hom}}_{\mathbb{Z}}(H_{2k}(M;\mathbb{Z}),\mathbb{R})$. Writing $H_{2k}(M;\mathbb{Z})={\mathbb{Z}}^b\oplus T$ with $T$ finite by [F6], both groups are ${\mathbb{Q}}^b$ and ${\mathbb{R}}^b$, and the map induced by the coefficient inclusion $\mathbb{Q}\hookrightarrow \mathbb{R}$ is the natural inclusion ${\mathbb{Q}}^b\hookrightarrow {\mathbb{R}}^b$. Hence the natural map $H^{2k}(M;\mathbb{Q}){\otimes }_{\mathbb{Q}}\mathbb{R}\to H^{2k}(M;\mathbb{R})$ is an isomorphism ${\mathbb{Q}}^b{\otimes }_{\mathbb{Q}}\mathbb{R}\cong {\mathbb{R}}^b$.

Form compatibility: for $x,y\in H^{2k}(M;\mathbb{Q})$, with images $x_{\mathbb{R}},y_{\mathbb{R}}$ under coefficient change, $Q_M(x_{\mathbb{R}},y_{\mathbb{R}})=Q_M^{\mathbb{Q}}(x,y)$. Indeed the cup product formula is coefficientwise by [F7], so $x_{\mathbb{R}}⌣y_{\mathbb{R}}$ is the coefficient-change image of $x⌣y$; the real fundamental class is the coefficient-change image of the integral one by [F8], and the Kronecker pairing is natural in coefficients by [F7]; evaluating the rational cup class on the integral fundamental class gives the rational number $Q_M^{\mathbb{Q}}(x,y)$, whose image in $\mathbb{R}$ is the real evaluation. Since the $x_{\mathbb{R}}$ span $H^{2k}(M;\mathbb{R})$ over $\mathbb{R}$ by step 1.3, the real form is the scalar extension $Q_M^{\mathbb{Q}}{\otimes }_{\mathbb{Q}}\mathbb{R}$ of the rational one.

Inertia agreement: by step 2.1 the real form $Q_M$ is the scalar extension of the rational form $Q_M^{\mathbb{Q}}$ along $\mathbb{Q}\subseteq \mathbb{R}$, so step 1.1 gives that their inertia triples agree; in particular the signatures over $\mathbb{Q}$ and over $\mathbb{R}$ coincide.

Steps 1.1 and 2.1 prove the basis-independence and well-posedness clause, and steps 1.1, 1.3 and 2.1 prove the scalar-extension clause for $M$; the general statement for arbitrary finite-dimensional symmetric forms over an ordered field is exactly step 1.1. If the vector space of the form is zero, its diagonal data are empty and its signature is $0$. For a zero-manifold the middle group need not vanish; both coefficient fields give its signed point count.