00:02
Hello, let's have a look at the question.
00:04
So now here, let us consider phi for zmn, which belongs to zm plus zn.
00:17
Now we can take that phi for 1 is equals to 1 .1.
00:26
Clearly we can say that phi for x1 is equal to x1.
00:35
And we can write this as x into 1 .1 or this could also be written as x comma x.
00:48
Now here this is for all x belongs to z.
00:54
Now we can take for x and y which belongs to z mn the value for fy x minus y is equals to f f f f f f f is equals to f f x minus y or we can write this as phi x minus y and into one also this can be written as phi 1 1 into x minus y or we can write this as x into 5 1 minus y into 5 1 now this could be written as 5 x minus 5 y into 5 y into 5 1.
01:33
Now this could be written as 5 x minus 5 y or lambda belongs to zmn, we can write that phi into lambda into x is equals to phi lambda x.
02:01
And this can be written as lambda x into phi 1 and this is equals to lambda phi x.
02:09
Therefore we can say that phi is a zmn module homomorphism.
02:23
Now the kernel of phi is equals to x which belongs to zmn and phi x is equals to 0 .0...