# Determinant Trick (Unfaithful module)

In the proof of one version of Nakayama's Lemma, one might use the Determinant Trick which is the following result.

Let $R$ be a commutative ring with $1$. Let $I$ be an ideal of $R$ and $M$ be a $R$-module. $\phi:M\to M$ be a $R$-module homomorphism and suppose that $\phi(M)\subset IM$. Then, there exists $a_i\in I^i$ such that $\phi^n+a_{1}\phi^{n-1}...+a_{n-1}\phi+a_n =0$

In the proof, of this one considers some generators $x_1,...x_n$ for $M$. So we can write $\phi(m_i)=\sum_{j=1}^{n} a_{ij} m_j$. Then, $(\phi_i\delta m_j-a_{ij})m_j=0$ Call $A_ij=\phi_i\delta_{ij}-a_{ij}$ Multiplying by the adjugate gives us $adj(A)Am_i=det(A) m_i$ for each $m_i$. Since these are the generators of $M$, we conclude that this is zero-endomorphism. My question is how doe that tell us that $det(A)=0$ Why can it not be a non-zero element that is a zero-endomorphism or in other words, $M$ is not a faithful module or that $det(A)\in Ann(M)$. Am I missing something?

My copy of "Undergraduate Commutative Algebra" by Miles Reid has no mention of the module being faithful.

The famous textbook "Introduction to Commutative Algebra" by Attiyah and MacDonald also has no mention of the the module being faithful.

And this is not a condition included in most texts I have seen.

However, in this set of notes, there is a mention that $R[\phi]$ has to be a faithful module. (Strictly stronger than just $R$ being faithful.)

https://faculty.math.illinois.edu/~r-ash/ComAlg/ComAlg2.pdf

• $\phi^k$makes sense only as elements of $\mathrm{End}(M)$, so the equation $\phi^n+a_1\phi^{n-1}+\cdots =0$ is an equation in this ring, meaningless in $R$. – Mohan Oct 11 '17 at 15:28
• We can append a new element $\phi$ to $R$ and make it into a commutative ring $R\subset R[\phi]\subset End(M)$. – daruma Oct 12 '17 at 13:02
• Why should the last be an inclusion? – Mohan Oct 12 '17 at 13:41
• Identify $R$ as subring of $R[\phi]$ by $r\mapsto r id_M$ and $\phi\in End(M)$ so $R[\phi]\subset End(M)$. These parts are mentioned in most literature. – daruma Oct 14 '17 at 11:18
• No, if an element in $R$ annihilates $M$, it will go to zero under this map, so in general, without assumptions, it is not an inclusion. – Mohan Oct 14 '17 at 13:37