00:01
Hi, today we are solving the question in which let us see the proof.
00:05
So, we know that n is a finitely generated free a module.
00:23
So, there exists a basis n1, n2 till nk.
00:34
For since g is surjective we can choose elements m1, m2 till mk in m such that g of mi is equals to ni for all i.
01:22
Now, see this let m be n any element of m.
01:48
Since the kernel of g is finitely generated and free, we can choose a basis k1, k2 till ki for kernel of g and elements a1, a2 till ai in a such that m is equals to a1 into m1 plus a2 into m2 plus akk.
02:47
So, ak into mk.
02:51
So, from here we have g of m is equals to g of a1 m1 plus a2 m2 plus till ak into mk and it is equals to a1 g of m1 plus a2 of m2 till ak into g of mk.
03:22
So, it is equals to a1 into n1 plus a2 into n2 plus till ak into nk which shows that m is a linear combination of m1 till mk with coefficients in a.
03:52
So, therefore m1 till mk generates m as an a module...