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Is $\mathbb{Z}^2/\langle (1,1),(1,-1) \rangle$ free over $\mathbb{Z}$? No?

Suppose $G={m_{\alpha}}$ is a basis. But $m_{\alpha}=[z_{\alpha}]$ so $(1,1)[z_{\alpha}]=[(1,1)z_{\alpha}]=[0]$. Is this correct?

To clarify: Let us do another example. Let $R=k[x,y]$ and $M=R/\langle x^2+y^2 \rangle$. Write $[r]=r+I$ where $I \subset R$ is an ideal. Now suppose $G=\{m_{\alpha}\}$ is a basis. However $m_{\alpha}=[r_{\alpha}]$ so $(x^2+y^2).[r_{\alpha}]=[0]=0_M$. So $G$ is not a basis.

But then why is $\mathbb{Z}^2/\langle (1,1) \rangle$ free?

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    Can you clarify exactly what you mean by $\langle (1,1),(1,-1) \rangle$? In particular, is this just the subgroup generated by $(1,1)$ and $(1,-1)$ or the ideal in the ring $\mathbb{Z}^2$ generated by $(1,1)$ and $(1,-1)$? – 2017-02-06

2 Answers 2

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You seem to be assuming that $\langle (1,1),(-1,1)\rangle$ refers to the ideal in the ring $\mathbb{Z}^2$ generated by the elements $(1,1)$ and $(-1,1)$. I suspect this is not the intended interpretation: rather, it is intended to just mean the $\mathbb{Z}$-submodule of $\mathbb{Z}^2$ generated by the elements $(1,1)$ and $(-1,1)$ (so the quotient will just be a $\mathbb{Z}$-module and not a $\mathbb{Z}^2$-module). With this interpretation, your argument fails immediately, since you do not know that $(1,1)[x]=[(1,1)x]$ (the quotient map is only a map of $\mathbb{Z}$-modules, not a map of $\mathbb{Z}^2$-modules). To get a correct solution, I would suggest trying to compute exactly what elements there are in the set $\langle (1,1),(-1,1)\rangle$ to explicitly determine what the cosets are.

Moreover, even if you interpret $\langle (1,1),(-1,1)\rangle$ to refer to the ideal, your argument is not correct. It shows that if there is a basis for the quotient, each element of the basis is annihilated by $(1,1)$. But this is no contradiction--it just means that the basis cannot have any elements! It is still entirely possible that the quotient is a free $\mathbb{Z}$-module, as long as its basis is the empty set.

In fact, this is exactly what happens. Since $(1,1)$ is the unit of the ring $\mathbb{Z}$, $(1,1)[x]=[x]$ for any $x$ so your argument shows not just that a basis cannot have any elements, but that any element of the quotient at all is $0$. So the quotient is just $\{0\}$, which is the free $\mathbb{Z}$-module generated by the empty set. (To be clear, this is interpreting $\langle (1,1),(-1,1)\rangle$ as the ideal--if you interpret it as a $\mathbb{Z}$-submodule, you'll get a different answer.)

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    Also I'm assuming the ring is commutative with identity. And I know a basis cannot have 0 in it! – 2017-02-06
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    Yes, a basis cannot have $0$ in it. But you haven't proved your basis has $0$ in it. You've just proved that if $x$ is an element of your basis, then $x=0$. But maybe no such $x$ exists, because your basis is the empty set. – 2017-02-06
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    But why does this argument work for the other example I gave? That was given to me. – 2017-02-06
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    It doesn't work for the other example. To complete the proof in the other example, you need to prove that the empty set is not a basis in that case. – 2017-02-06
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    So my original question :it is a free module? – 2017-02-06
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    Yes (assuming $\langle (1,1),(-1,1)\rangle$ is the ideal). Since $(1,1)$ is the unit of the ring $\mathbb{Z}^2$, $[x]=(1,1)[x]=[(1,1)x]=[0]$ for all $x$ so the quotient module is trivial. – 2017-02-06
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    May I ask why the empty set is not a basis for the other example? – 2017-04-29
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    If the empty set is a basis that means the module consists only of the zero element. But that is not true in that case, since the ideal $\langle x^2+y^2\rangle$ is not all of $R$ (for instance, every nonzero element of this ideal has degree at least 2). – 2017-04-29
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    Also, would $\mathbb{Z}^2/\langle (1,1)\rangle$ be free by the same reason as posted in the comment – 2017-04-29
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Note that $(1,0)+\langle (1,1),(1,-1)\rangle$ is a non-zero element, but if you double it, you end up with $(2,0)=(1,1)+(1,-1)$ (plus $\langle (1,1),(1,-1)\rangle$).

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    $2[(1,0)]=[(2,0)]=[(1,1)+(1,-1)]=[(1,1)]+[(1,-1)]=0$? – 2017-02-06
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    Does this fully answer the question? Or would I need to do it for a general element/basis? – 2017-05-07