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Let $A$ be euclidian ring and $K$ be its field of a fraction.Let $(V,B)$ be a nonzero IPS (inner product space) over $K$. A finitely generated A-submodule
$L ⊆ V $is said to be an $A$ -lattice in $V$ if $L$ contains a $K$-basis of $V$ . As we have
already observed, $L$ must be $A$-free since it is torsion-free, and it is easy to verify that
an $A$-basis for $L$ will automatically be a $K$-basis for $V$.

Attempt: I know that $V$ is finitely generated projective module since $(V, B)$ is inner product space and $B$ is regular bilinear form. Now question is what is the meaning of $L$ contains a $K$-basis of $V$ and how to verify $A$-basis for L will automatically be a $K$-basis for $V$ ; it certainly means we can extend from basis of $L$ to basis of $V$?

The meaning of $L$ containing a $K$-basis for $V$ is that there are $ell_1,ldots,ell_nin Lsubseteq V$ forming a basis for $V$ as a $K$-vector space (where $n=dim_K V$).

Now suppose that we have $l_1,ldots,l_rin L$ are a basis for $L$ as a free $A$-module. (Note the different style of $l$). We want to prove that $r=n$ and that the elements $l_i$ are linearly independent over $K$, so that they form a $K$-basis.

If $$sum_i=1^r c_il_i=0,$$
with $c_iin K$, we can multiply by a common denominator $dne 0in A$, so that $c_idin A$. Then we get
$$sum_i=1^r (c_id)l_i=0,$$
but the $l_i$ were an $A$-basis for $L$, so $c_id=0$ for all $i$, and since $dne 0$, $c_i=0$ for all $i$.

As for why $n=r$, observe that $V=KL=Kotimes_A L=Kotimes_A A^r = K^r$, so $r=dim K^r=dim V =n$.