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 $\ell_1,\ldots,\ell_n\in L\subseteq V$ forming a basis for $V$ as a $K$-vector space (where $n=\dim_K V$).

Now suppose that we have $l_1,\ldots,l_r\in L$ 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 $c_i\in K$, we can multiply by a common denominator $d\ne 0\in A$, so that $c_id\in A$. 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 $d\ne 0$, $c_i=0$ for all $i$.

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

  • $\begingroup$ @mathsstudent I don't usually participate in chat, if you have another question, feel free to ask it, and drop me a comment linking to it and I'll check it out. $\endgroup$ – jgon Mar 27 at 16:26
  • $\begingroup$ @mathsstudent Yes $\endgroup$ – jgon Mar 27 at 16:29
  • $\begingroup$ I ask doubt below sir. $\endgroup$ – maths student Mar 27 at 16:38
  • $\begingroup$ math.stackexchange.com/questions/3164762/… I have ask it? $\endgroup$ – maths student Mar 27 at 16:49
  • $\begingroup$ I have got corollary 3.9 can you please explain to me how to prove note below corollary 3.9. $\endgroup$ – maths student Mar 27 at 17:06

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