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I am having trouble coming up with an example to an old exam I am using to help study for some linear algebra qualifiers.

Let $R$ be commutative ring and set $M = R \times R$. Considering $M^2$ as a $R$ module is there an example of a surjective $R$-module homomorphism from $f: M^2 \rightarrow M$ so that the matrix corresponding to the linear transformation $f$ contains only entries which are divisors of zero.

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In your title you include the condition "surjective" but in your body you don't. Which do you mean? – Chris Eagle Nov 6 '11 at 10:25
Surely the OP means to keep surjectivity. Otherwise the question is kinda pointless. But there are several other typos, too. $M^2$ is an $R$-module. And I'm not sure whether the question is about a mapping from $R^4\to R^2$ or $R^2\to R$. Luckily my hint works all the same. BTW, I would use the term linear transformation only between vector spaces, and use $R$-module homomorphism here. – Jyrki Lahtonen Nov 6 '11 at 11:28
up vote 4 down vote accepted

Hint: Let $R=\mathbf{Z}/6\mathbf{Z}$.

Edit: Hint2: Consider mapping $R^2\to R$ that sends $(m,n)\mapsto(2m+3n)$.

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Thanks for y the our hint but I am still stuck on constructing the linear transformation. I am starting by considering $(\mathbb{Z}_6 \times \mathbb{Z}_6)^2$ and I want a map that sends all elements from the previous set onto $\mathbb{Z}_6 \times \mathbb{Z}_6$ – user7980 Nov 6 '11 at 23:23
So if $f :(\mathbb{Z}_6 \times \mathbb{Z}_6)^2 \rightarrow \mathbb{Z}_6 \times \mathbb{Z}_6)$ is surjective and we determine the matrix representation for $f$ by considering its action on $(0,1),(1,0)$. I don't see why the entries in the matrix representation corresponding to this maps action on the basis must all be zero divisors. – user7980 Nov 6 '11 at 23:26
They don't have to be zero divisors, but you're just looking for an example, so you are allowed to make them all zero divisors. If you do this in a sufficiently clever way, the determinant will not be a zero-divisor, so.... – Gerry Myerson Nov 7 '11 at 1:19
@user7980: What Gerry says. Now spelled out... – Jyrki Lahtonen Nov 7 '11 at 5:17

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