Verma module associated to a Lie algebra Let $L$ be a Lie algebra with Cartan decomposition $L=H \oplus(\sum_{\alpha \in \Phi}L_{\alpha})$. Let $B=\sum_{\alpha \in \Phi^+}L_{\alpha}$ and $N=\sum_{\alpha \in \Phi^-}L_{\alpha}$ so that $L=B \oplus N$. By definition Verma module is $\Delta(\lambda)=\mathcal{U}(L)\otimes_{\mathcal{U}(B)}\mathbb{C}_{\lambda}$ where $\mathcal{U}(L)$ is the universal enveloping algebra and $\mathbb{C}_{\lambda}$ is a one dimensional $\mathcal{U}(B)$ module. Now we can regard $\Delta(\lambda)$ as a $\mathcal{U}(N)$ module also and as $\mathcal{U}(N)$ modules I was able to prove that $\Delta(\lambda)$ is isomorphic to $\mathcal{U}(N) \otimes _{\mathbb{C}}\mathbb{C}_{\lambda}$.From this I want to conclude that $\Delta(\lambda)$ is a $\mathcal{U}(N)$  free module of rank 1. Also the map from $\mathcal{U}(N) \to \Delta(\lambda)$ mapping $n \to n.v_{\lambda }$ is an isomorphism where $v_{\lambda}$ is the canonical generator of $\Delta(\lambda)=\mathcal{U}(L)\otimes_{\mathcal{U}(B)}\mathbb{C}_{\lambda}$. 
 A: We need the following properties of tensor products:

*

*Let $R$ be a ring and $M$ a left $R$-module.
There exists a unique homomorpism of abelian groups
$$
  R ⊗_R M \to M \,, \quad r ⊗ m \mapsto rm \,,
$$
and this homomorphism is an isomorphism of left $R$-modules.


*Let $R$, $S$ and $T$ be three rings.
Let $M$ be an $R$-$S$-bimodule, $N$ an $S$-$T$-bimodule, and $P$ a left $T$-module.
There exists a unique homomorphism of abelian groups
$$
  (M ⊗_S N) ⊗_T P \to M ⊗_S (N ⊗_T P) \,,
  \quad
  (m ⊗ n) ⊗ p \mapsto m ⊗ (n ⊗ p) \,,
$$
and this homomorphism is an isomorphism of left $R$-modules.


*Let $S$ be a ring, $R$ a subring of $S$, and $F$ a free $R$-module.
The left $S$-module $S ⊗_R F$ is again free.
More precisely, if $(x_i)_{i ∈ I}$ is an $R$-basis of $F$, then $(1 ⊗ x_i)_{i ∈ I}$ is an $S$-basis of $S ⊗_R F$.
This entails that $F$ and $S ⊗_R F$ have the same rank.
It follows from the decomposition $L = N ⊕ B$ and the PBW-theorem that the map
$$
  \mathcal{U}(N) ⊗_ℂ \mathcal{U}(B) \to \mathcal{U}(L) \,,
  \quad
  x ⊗ y \mapsto xy
$$
is an isomorphism of vector spaces.
It is also a homomorphism of $\mathcal{U}(N)$-$\mathcal{U}(B)$-bimodules, and thus an isomorphism of $\mathcal{U}(N)$-$\mathcal{U}(B)$-bimodules.
We have therefore the isomorphism of left $\mathcal{U}(N)$-modules
\begin{align*}
  Δ(λ)
  &=
  \mathcal{U}(L) ⊗_{\mathcal{U}(B)} ℂ_λ
  \\
  &≅
  \Bigl( \mathcal{U}(N) ⊗_ℂ \mathcal{U}(B) \Bigr) ⊗_{\mathcal{U}(B)} ℂ_λ
  \\
  &≅
  \mathcal{U}(N) ⊗_ℂ \Bigl( \mathcal{U}(B) ⊗_{\mathcal{U}(B)} ℂ_λ \Bigr)
  \\
  &≅
  \mathcal{U}(N) ⊗_ℂ ℂ_λ
\end{align*}
by properties (1) and (2) of tensor products.
The vector space $ℂ_λ$ is one-dimensional, and has thus a basis consisting of a single vector $v_λ$.
It follows from property (3) of tensor products that $\mathcal{U}(N) ⊗_ℂ ℂ_λ$ is free as a left $\mathcal{U}(N)$-module, with a basis given by the single element $1 ⊗ v_λ$.
This means that $\mathcal{U}(N) ⊗_ℂ ℂ_λ$ is free of rank $1$ as a left $\mathcal{U}(N)$-module, and that the map
$$
  \mathcal{U}(N) \to \mathcal{U}(N) ⊗_ℂ ℂ_λ \,,
  \quad
  x \mapsto x ⋅ (1 ⊗ v_λ) = x ⊗ v_λ
$$
is an isomorphism of left $\mathcal{U}(N)$-modules.
The overall isomorphism of left $\mathcal{U}(N)$-modules
$$
  \mathcal{U}(N)
  ≅
  \mathcal{U}(N) ⊗_ℂ ℂ_λ
  ≅
  \dotsb
  ≅
  \mathcal{U}(L) ⊗_{\mathcal{U}(B)} ℂ_λ
$$
is given by
$$
  x
  \mapsto x ⊗ v_λ
  \mapsto x ⊗ (1 ⊗ v_λ)
  \mapsto (x ⊗ 1) ⊗ v_λ
  \mapsto (x ⋅ 1) ⊗ v_λ
  =       x ⊗ v_λ \,.
$$
