# A question on Deligne-Serre lifting lemma

Good day, everyone. I have been studying the paper "Formes modulaires de poid 1" written by Deligne and Serre. I am confused by lemma 6.11, or the so-called Deligne-Serre lifting lemma in the paper.

The original proof is somewhat short and rough, so I have found another note which gives a more detailed statement at page 7 in the following website https://arxiv.org/pdf/0910.1408.pdf

However, there is still some problem I cannot figure out. The main question is as follows:

Let $\mathcal{O}$ be a discrete valuation ring, $\mathfrak{m}$ be its maximal ideal, and $k$, $K$ be residue field and field of fractions respectively. Let $M$ be a finitely generated free $\mathcal{O}$-module, and $\mathcal{T}$ a family of mutual commuting elements in End($M$), and we denote the $\mathcal{O}$-algebra generated by $\mathcal{T}$ by $\mathbb{T}$.

At the last part of this note, we have found a minimal prime $\mathfrak{p}$ in $\mathbb{T}$ which annihilates $f$ in the module $M / \mathfrak{m}M$, hence $\mathfrak{p} \in Ass_\mathbb{T}(M / \mathfrak{m}M) \subset Supp_\mathbb{T}(M / \mathfrak{m}M) \subset Supp_\mathbb{T}(M)$. Thus, $Ann_\mathbb{T}(M)\subset \mathfrak{p}$. Now we need to tensor all things with $K$, and let $\mathcal{P}$ be the ideal generated by $\mathfrak{p}$ in $K \otimes \mathbb{T}$. Then it claims that $Ann_{K \otimes \mathbb{T}}(K \otimes M) \subset \mathcal{P}$ and $\mathcal{P} \in Supp_{K\otimes\mathbb{T}}(K\otimes M)$. But how can I ensure that that $\mathcal{P}$ is still a prime ideal? Moreover, I need it to be a minimal prime so that I can deduce $\mathcal{P} \in Ass_{K\otimes M}(K\otimes M)$. Is such $\mathcal{P}$ always minimal?

• Sorry, thatis a mistype, it should be in $K \otimes \mathbb{T}$. Apr 1, 2018 at 3:16
• FYI Rodney Keaton's master's thesis also has an exposition of the Deligne-Serre lifting lemma. Apr 2, 2018 at 12:06

The quotient $\mathbb{T} / \mathfrak{p}$ being an integral extension of $R$, after base change, $\mathbb{T}\otimes_R K / \mathcal{P}$ is a field extension of $K$. Thus $\mathcal{P}$ is a prime ideal. If $\mathcal{P}$ were not minimal, $\mathfrak{p} = \mathcal{P}\cap \mathbb{T}$ wouldn't be minimal either.