Answer to Question #203324 in Abstract Algebra for Raghav

Question #203324

Define f:Z→Zm × Zn :f (x)=(xmodm,xmod n),m,n∈N.

i) If (m, n) = (3, 4), find Ker f.

ii) If (m, n) = (6, 4), find Ker f.

iii) What can you generalize about Ker f from (i) and (ii)?


1
Expert's answer
2021-06-15T12:07:56-0400

Let "f: \\Z\\to \\Z_m \\times \\Z_n,\\ f (x)=(x\\mod m,\\ x \\mod n),\\ \\ m,n\\in\\N."


i) If "(m, n) = (3, 4)", then "\\ker f=\\{z\\in\\Z\\ :\\ f(z)=(0,0)\\}=\\{z\\in\\Z\\ :\\ (z\\mod 3,\\ z \\mod 4)=(0,0)\\}=\\{z\\in\\Z\\ :\\ z\\mod 3=0,\\ z \\mod 4=0\\}\n=\\{z\\in\\Z\\ :\\ 3|z,\\ 4|z \\}=\\{z\\in\\Z\\ :\\ 12|z \\}=12\\Z."


ii) If "(m, n) = (6, 4)", then "\\ker f=\\{z\\in\\Z\\ :\\ f(z)=(0,0)\\}=\\{z\\in\\Z\\ :\\ (z\\mod 6,\\ z \\mod 4)=(0,0)\\}=\\{z\\in\\Z\\ :\\ z\\mod 6=0,\\ z \\mod 4=0\\}\n=\\{z\\in\\Z\\ :\\ 6|z,\\ 4|z \\}=\\{z\\in\\Z\\ :\\ lcm(6,4)|z \\}=\\{z\\in\\Z\\ :\\ 12|z \\}=12\\Z."


iii) We see that "\\ker f" from (i) and (ii) coincide. We can generalize that for two pairs "(m_1,n_1)" and "(m_2,n_2)" the kernels of corresponding maps coincide if and only if "lcm(m_1,n_1)=lcm(m_2,n_2)."


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