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I am reading up the proof of Hensel's lemma here. On page 2, after equation 2, the author concludes that the degree of $\delta h_k$ is less than $n$ since the degree of $\Delta$ and $\epsilon g_k$ is less than $n$. I am not sure I understand this. We only know that $\Delta \cong \epsilon g_k +\delta h_k \mod{\mathfrak{m}^{k+1}[x]}$, so how do we get this equality of degrees?

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The examples given after the proof in that link are not terribly compelling. His version of Hensel's lemma involves the lifting of fairly general factors over A/m to a factorization over A but he only illustrates it with simple linear factors, for which the simpler version of Hensel's lemma about lifting simple roots would be adequate. –  KCd Sep 8 '11 at 0:36

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I'm not sure what he has in mind, but one way to proceed is to apply division algorithm to $\Delta - \epsilon g_k$ divided by $h_k$ in $(A/m^{k+1})[x]$ in equation 2. Equation 2 tells you it's divisible and uniqueness tells you $\delta$ has degree $< r$.

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Thanks for the prompt response. I understand your argument if $\delta$ were to be monic. I don't see this (if true) and otherwise, how are we sure that the degree is not reduced while reducing modulo $m^{k+1}$. –  BMI Sep 8 '11 at 1:10
    
1. I'm dividing by $h_k$ which is monic. –  Soarer Sep 8 '11 at 1:27
    
2. $\delta$ is not really fixed as in the notes. Rather, all you need is one $\delta$ that satisfies equation 2 and has degree less than $r$. –  Soarer Sep 8 '11 at 1:28
    
I understand you are dividing by $h_k$. I am just saying when considering this equation in $A/m^{k+1}$, we are reducing the coefficients of all polynomials in equation 2 modulo $m^{k+1}$. So in the reduced ring we conclude the image of $\delta$ has degree less than $r$. How do we conclude that $delta$ itself has degree less than $r$ in case it is not monic. –  BMI Sep 8 '11 at 1:31
    
You are thinking about $\delta$ being a chosen element in $A[x]$ that satisfies equation 2. I'm saying that do NOT do so, instead define $\delta$ as a lift of the unique element in $(A/m^{k+1})[x]$ that satisfies equation 2. Division algorithm says that the unique element in $(A/m^{k+1})[x]$ is of degree less than $r$, so obviously you can lift it to $A[x]$ with same degree. Then show that this lift lies in $m^k[x]$ which finishes the proof. –  Soarer Sep 8 '11 at 1:44

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