$\det \varphi$ is not a zero divisor $\implies$ Endomorphism $\varphi$ of a free $R$-Module injective

Question

Let $R$ be a commutative Ring with $1$, $M$ a free $R$-Module of rank $2$ and $\varphi \in \operatorname{End}_R (M)$. Show that: If $\det \varphi$ is not a zero divisor in $R$, then $\varphi$ is injective.

Ok, so assume $\varphi$ is not injective. Then there exists a $x \in \ker \varphi$, $x\neq 0$ such that $\varphi(x) = 0$. Since $M$ is free, then there exist $r_1,r_2 \in R$ with $x = r_1m_1 + r_2m_2$ where not both $r_1, r_2$ are zero. Since $(m_1,m_2)$ is a basis of $M$, I know that $(m_1 \wedge m_2)$ is a basis of $\bigwedge^2 M$. From here, I don't really know how to proceed, advice is greatly appreciated.

score 1 · Accepted Answer · answered Jul 03 '20 at 12:47

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Hint : If $\varphi(x) = 0$ then $0=\varphi(x)\wedge \varphi(m_2) = r_1 \varphi(m_1)\wedge \varphi(m_2) = (\det(\varphi) r_1) m_1\wedge m_2$.

Similarly, $0= (\det(\varphi)r_2) m_1\wedge m_2$.

answered Jul 03 '20 at 12:47

Maxime Ramzi

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May be a dumb question but can you explain why $\varphi(x) \wedge \varphi(m_1) = 0$? – Shuster Jul 03 '20 at 12:50
Because $\varphi(x) = 0$ – Maxime Ramzi Jul 03 '20 at 12:52
Does it really hold that for all $m \in M$: $0 \wedge m = 0$? If so, how? – Shuster Jul 03 '20 at 12:56
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Yes of course it does, for the same reason that $0\otimes m = 0$ (by the way you can deduce it from that). For instance, $0\wedge m = (0+0)\wedge m = 0\wedge m + 0\wedge m$, and then if $x+x = x$, $x=0$ – Maxime Ramzi Jul 03 '20 at 12:57
Alright thank you! – Shuster Jul 03 '20 at 12:59

$\det \varphi$ is not a zero divisor $\implies$ Endomorphism $\varphi$ of a free $R$-Module injective

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