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Find $Ker(\phi)$ and $\phi(-3,2)$ for the given homomorphism $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ where $\phi(1,0) = 3$ and $\phi(0,-1) = -5$

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I have no idea why how they came up with the mapping $\phi(m,n)$ themselves nor do I fully understand why the kernel is correct. My thinking for this problem was

$Ker(\phi) = \{(m,n) \in \mathbb{Z} \times \mathbb{Z}: \phi(m,n)=0\}$ so the problem now lies in finding $m$ and $n$.

What is the thinking behind the solution? I really don't understand how they come up with the function themselves

EDIT: okay I understand they are basically setting the function to 0 and that's how they came up with that kernel, but still no explanation on how they came up with the function

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    I'm curious as to where you came across this question. I used part of it on an exam I gave earlier this year.2012-11-19
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    I found it in my textbook by Fraleigh2012-11-19
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    Thanks. That's probably where I found it, too.2012-11-19

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A typical element of ${\bf Z}\times{\bf Z}$ is $(a,b)$, which can be written as $$(a,b)=a(1,0)+b(0,1)$$ Since you are told what $\phi(1,0)$ is and you are told what $\phi(0,1)$ is, and you are told $\phi$ is a homomorphism, you can figure out what $\phi(a,b)$ is, and then you can figure out what $a,b$ have to do for $(a,b)$ to be in the kernel.

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    What is binary operation on the sets? How do we know it is addition here?2012-11-19
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    Also how does homomorphism imply that $(a,b) = a(1,0) + b(0,1) = a\phi(1,0) + b\phi(0,1)$? Is the way you have written it often obvious and thus omitted?2012-11-19
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    The integers form a group under addition. In the absence of any other information on the group operation, that makes addition the default operation. You have written $a(1,0)+b(0,1)=a\phi(1,0)+b\phi(0,1)$ which is certainly false as the left side is in ${\bf Z}\times{\bf Z}$ whereas the right side is in $\bf Z$. What did you mean to ask?2012-11-19
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    How does $(a,b)=a(1,0)+b(0,1) \implies \phi(a,b) = 3a - 5b$? Like they have in the solutions2012-11-19
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    $\phi$ is a homomorphism. So, $\phi(x+y)=\phi(x)+\phi(y)$ for all $x$ and $y$ in the domain. If $n$ is a positive integer, then $\phi(nx)=\phi(x+x+\cdots+x)=\phi(x)+\phi(x)+\cdots+\phi(x)=n\phi(x)$. If $n$ is a negative integer, then $\phi(nx)=-\phi(-nx)=-(-n)\phi(x)=n\phi(x)$. These are basic properties of homomorphisms.2012-11-19
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    I have a poor grasp of the basics unfortunately. Why can't I let $x = (1,0)$ here? So $\phi((1,0) - (0,1)) = \phi(1,0) - \phi(0,1)$2012-11-19
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    You can let $x=(1,0)$, and it is true that $\phi((1,0)-(0,1))=\phi(1,0)-\phi(0,1)$. But I'm not sure what you bring up this particular calculation.2012-11-19