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I have an algebraic question that I cannot solve. It is extracted from Adams and Margolis' paper on modules over the Steenrod Algebra. Here is the problem :

Let $K$ be a commutative ring with unit, $R$ a connected $K$-algebra (not commutative in my case), i.e., graded as $R \cong K \oplus R_1 \oplus R_2 \oplus \cdots$, and $M$ be a connected $R$-module, i.e., only positively graded (no condition on $M_0$, but I am okay to assume $M_0 \cong K$, only as a $K$-module).

Consider elements $\{m_i\} \in M$ such that $\{ 1 \otimes m_i \} \in K \otimes_R M$ form a $K$-basis.

(Note that $K \otimes_R M \cong M / I(R)M$ where $I(R)$ is the augmentation ideal for the canonical augmentation here, so another way to see it is that $\{ \overline{m_i} \}$ form a $K$-basis)

Then they claim that "the following facts are well known and easily proved:"

  1. The $\{ m_i \}$ $R$-generates $M$.
  2. If $M$ is $R$-free, then the $\{ m_i \}$ are an $R$-basis of $M$.

Thank you for your help.

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Bogdan
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    Regarding the comment above, you may wish to see: http://en.wikipedia.org/wiki/Nakayama_lemma#Graded_version – Manny Reyes Nov 08 '13 at 18:38
  • Well, if $m_i$ are not homogeneous I don't think your claims hold. –  Nov 08 '13 at 18:56
  • @MannyReyes Thank you, I was not aware of a non commutative Nakayama's Lemma. I'm looking into it. – Bogdan Nov 08 '13 at 18:59
  • @user Thanks, I believe you are right. However, I only consider homogeneous elements, ideals, etc in the graded setting, so this is not a problem for my application :). – Bogdan Nov 08 '13 at 19:01
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    Actually the comment of Manny Reyes was a complement of my first comment here where I suggested you for 1. to use graded Nakayama, but I've deleted that comment since I've realized that you didn't assume $m_i$ homogeneous and then can't apply graded Nakayama. –  Nov 08 '13 at 19:19
  • The quoted paper is cryptical about the conditions when the two claims hold and as they are stated is hard to believe that really hold. Only the OP (who I assumed that read the paper) can say exactly what the paper needs. –  Nov 09 '13 at 08:46

1 Answers1

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  1. If $(m_i)$ are homogeneous, then use Nakayama Lemma, graded version.

  2. Since $M$ is free it has a basis $(x_j)_{j\in J}$ and this remains so in $K\otimes_RM$, therefore $|J|=|I|$. (For the sake of simplicity let's assume $J=I$.) Now let $f:M\to M$ given by $f(x_i)=m_i$. Then $f$ is surjective, and using this result $f$ is bijective, so $(m_i)$ is a basis. (Here I need $M$ finitely generated and I don't have a proof if remove this condition.)

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  • This looks good. I believe it is possible to allow $M$ to be of finite type in the case I am interested (when the ground algebra is finitely generated). Thanks. – Bogdan Nov 09 '13 at 23:20