I know I'm supposed to be proving that $(\mathbb Z× \mathbb Z)/〈(0, 1)〉 \cong \mathbb Z$ by using the First Isomorphism Theorem. I'm having trouble defining a homomorphism $φ:\mathbb Z× \mathbb Z → \mathbb Z$ whose kernel is $〈(0, 1)〉$. We did a similar example in class where we proved $(\mathbb Z× \mathbb Z)/〈(1, 1)〉 \cong \mathbb Z$ and $φ:\mathbb Z× \mathbb Z → \mathbb Z$ defined by $φ((a,b))= a-b$, but I cant determine the homomorphism for this one.
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$\begingroup$ $\varphi(a,b) \mapsto a$. $\endgroup$ – Morgan Rodgers Jan 31 '17 at 21:10
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$\begingroup$ I actually had a similar suggestion, but then when I have to prove that it's a homomorphism, I get stuck. We need to prove φ((a,b)+(c,d)) = φ(a,b) + φ(c,d). So, φ((a,b)+(c,d)) = φ((a+c), (b+d)) = a+c. Then I get stuck. $\endgroup$ – user21 Jan 31 '17 at 21:14
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2$\begingroup$ You also have that $\varphi((a,b))+\varphi((c,d)) = a+c$. $\endgroup$ – Morgan Rodgers Jan 31 '17 at 21:22
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$\begingroup$ @johnie4usc Maybe what you're missing is that $(a,b) + (c,d) = (a+c,b+d)$. $\endgroup$ – Viktor Vaughn Feb 1 '17 at 0:16
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$\begingroup$ Yeah, you all are both right. I was missing it and it was right in front of me. $\endgroup$ – user21 Feb 1 '17 at 15:26
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Define $f:\mathbb{Z}\times \mathbb{Z}\rightarrow \mathbb{Z}:(x,y)\mapsto x$, the kernel is exactly generated by $(0,1)$ and $f$ is surjective. Thus the result indeed follows from the first isomorphism theorem.
answered Jan 31 '17 at 21:10
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$\begingroup$ I understand. I just get confused when proving its a homomorphism. I'm copy/pasting from my above comment: I actually had a similar suggestion, but then when I have to prove that it's a homomorphism, I get stuck. We need to prove φ((a,b)+(c,d)) = φ(a,b) + φ(c,d). So, φ((a,b)+(c,d)) = φ((a+c), (b+d)) = a+c. Then I get stuck. $\endgroup$ – user21 Jan 31 '17 at 21:15
