Let \[ M &= \{(w, x, y, z) \in \ZZ^4 \suchthat w + x + y + z \in 2\ZZ\} \\ N &= \ts{ (w, x, y, z) \in \ZZ^4 \suchthat 4\divides (w - x),~ 4\divides (x - y),~ 4\divides ( y - z) } .\]
Show that \(N\) is a \(\ZZ\dash\)submodule of \(M\) .
Find vectors \(u_1 , u_2 , u_3 , u_4 \in \ZZ^4\) and integers \(d_1 , d_2 , d_3 , d_4\) such that \[ \{ u_1 , u_2 , u_3 , u_4 \} && \text{is a free basis for }M \\ \{ d_1 u_1,~ d_2 u_2,~ d_3 u_3,~ d_4 u_4 \} && \text{is a free basis for }N \]
Use the previous part to describe \(M/N\) as a direct sum of cyclic \(\ZZ\dash\)modules.
Solution
(a) \(N\) is a \(\ZZ\)-submodule of \(M\): - Both \(M\) and \(N\) are subgroups of the free abelian group \(\ZZ^4\). - \(M\) is the kernel of the group homomorphism \(\phi: \ZZ^4 \to \ZZ_2\) given by \((w,x,y,z) \mapsto (w+x+y+z) \bmod 2\). - \(N\) consists of vectors where \(w \equiv x \equiv y \equiv z \pmod 4\). If \((w,x,y,z) \in N\), then \(w = x + 4a, y = x + 4b, z = x + 4c\) for some \(a, b, c \in \ZZ\). Then \(w + x + y + z = 4x + 4(a+b+c) = 4(x+a+b+c) \in 4\ZZ \subseteq 2\ZZ\). Thus \((w, x, y, z) \in M\), which shows \(N \subseteq M\). - Since \(N\) is the kernel of the homomorphism \(\psi: \ZZ^4 \to (\ZZ_4)^3\) defined by \((w-x, x-y, y-z) \bmod 4\), \(N\) is a subgroup (i.e. \(\ZZ\)-submodule) of \(\ZZ^4\), and hence of \(M\).
(b) Smith Normal Form basis:
Basis for \(M\): \(M\) is generated by vectors with even coordinate sum. A convenient basis for \(M\) is:
- \(u_1 = (1, 1, 1, 1)\) (sum = 4)
- \(u_2 = (1, -1, 0, 0)\) (sum = 0)
- \(u_3 = (0, 1, -1, 0)\) (sum = 0)
- \(u_4 = (0, 0, 1, -1)\) (sum = 0)
To verify \(\{u_1, u_2, u_3, u_4\}\) is a basis of \(M\): Any \((w,x,y,z) \in M\) has \(w+x+y+z = 2k\). Then \((w,x,y,z) = w(1,0,0,-1) + x(0,1,0,-1) + y(0,0,1,-1) + (w+x+y+z)(0,0,0,1)\), and since \(w+x+y+z=2k\), this is spanned over \(\ZZ\) by differences of standard basis elements together with \((1,1,1,1)\) and \((2,0,0,0)\). The determinant of the matrix with columns \(u_1, u_2, u_3, u_4\) is \(1 \cdot 1 - (-1) \cdot (-1) \cdot (-1) = 4\), which equals the index \([\ZZ^4 : M] = 2\) times 2.
Expressing \(N\) in this basis: Any \(v \in N\) has \(w \equiv x \equiv y \equiv z \pmod 4\). Let \(w = c + 4a_1, x = c + 4a_2, y = c + 4a_3, z = c + 4a_4\). Then: \[ v = c(1,1,1,1) + 4a_1 e_1 + 4a_2 e_2 + 4a_3 e_3 + 4a_4 e_4. \] Since \(4e_1 = (1,1,1,1) + 3u_2 + 2u_3 + u_4 - (0,0,0,0)\), in terms of \(u_1, u_2, u_3, u_4\):
- \(N\) contains \(u_1 = (1,1,1,1)\) (since all coordinates are equal, differences are \(0 \equiv 0 \pmod 4\)).
- For differences, \(v \in N\) has \(w-x \in 4\ZZ, x-y \in 4\ZZ, y-z \in 4\ZZ\).
- Thus the generators along \(u_2, u_3, u_4\) are \(4u_2, 4u_3, 4u_4\).
Therefore, a free basis for \(N\) aligned with \(\{u_1, u_2, u_3, u_4\}\) is: \[ \{1 \cdot u_1, \ 4 \cdot u_2, \ 4 \cdot u_3, \ 4 \cdot u_4\}. \] Thus \(d_1 = 1, \ d_2 = 4, \ d_3 = 4, \ d_4 = 4\).
(c) Quotient module \(M/N\): From the aligned bases: \[ M / N \cong \frac{\ZZ u_1 \oplus \ZZ u_2 \oplus \ZZ u_3 \oplus \ZZ u_4}{\ZZ (1 u_1) \oplus \ZZ (4 u_2) \oplus \ZZ (4 u_3) \oplus \ZZ (4 u_4)} \] \[ \cong (\ZZ / 1\ZZ) \oplus (\ZZ / 4\ZZ) \oplus (\ZZ / 4\ZZ) \oplus (\ZZ / 4\ZZ) \cong \ZZ_4 \oplus \ZZ_4 \oplus \ZZ_4. \]