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Find $\ker(\phi )$ given that $\phi$ is a homomorphism and

$\phi :\mathbb{Z\times Z \rightarrow Z\times Z}$

where $\phi(1, 0) \rightarrow(-2,3)$ and $\phi(0, 1) \rightarrow(-1,5)$.

One way that I think to do this is to use the fact that $\phi$ is a hom so we know that inverses must be mapped to inverses. Since the inverse in $\mathbb{Z\times Z}$ of $(1,0)$ is $(-1, 0)$ we then know that $\phi(-1, 0)$ must be the inverse of $(2, -3)$ which is $(-2, 3)$. But things break down from here.

So if someone could help me see where I went wrong and the correct way to figure out this problem I would really appreciate it.

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We want to know what elements are in the kernel of $\phi$. That means we are interested in when $\phi(a,b)=(0,0)$. However, since $(a,b)=a(1,0)+b(0,1)$, we get $$ (0,0)=\phi(a,b)\\ =a\phi(1,0)+b\phi(0,1)\\ =a(-2,3)+b(-1, 5) $$ Since the first components must be equal, and the second components must be equal, we get $$ \cases{0=-2a-3b\\0=3a+5b} $$ Solving this set of equations gives you exactly which $(a,b)\in\Bbb Z\times\Bbb Z$ are in the kernel.

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  • $\begingroup$ It is not obvious to me why $\phi(a(1,0))=a\phi(1,0)$ $\endgroup$ – Jac Frall Apr 9 at 14:36
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    $\begingroup$ @JacFrall Then remember what the $a$ means when we write it like that. If it's positive, it just means repeated addition. If it's negative, it means invert, then repeated addition. If you then use the homomorphism property of $\phi$ on this inversion and the repeated addition, you see how the equality follows. $\endgroup$ – Arthur Apr 9 at 14:45
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Hint: We have $$ \phi(a,b) =\phi[a(1,0)] + \phi[b(0,1)] = a \,\phi(1,0) + b\, \phi(0,1) = (-2a - b, 3a + 5b) $$

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Hint:

$\phi(a,b) = \phi(a\cdot (1,0) + b\cdot(0,1)) = \phi(a\cdot(1,0))+\phi(b\cdot(0,1))= a\phi(1,0) +b \phi(0,1)$

Of course the third equality needs a proof. You already showed that it holds for $a=-1$, now show for any positive integer and for any negative integer using a proof by induction.

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