I am reading on p.128 of Fulton and Harris's Representation Theory the proof of the following fact about lie algebras
Let $\mathfrak{g}$ be a complex Lie algebra, and set $\mathfrak{g}_{ss} := \mathfrak{g}/$Rad$(\mathfrak{g})$. Then every irreducible representation of $\mathfrak{g}$ is of the form $V = V_0 \otimes L$ where $V_0$ is an irreducible representation of $\mathfrak{g}_{ss}$ (i.e., a representation of $\mathfrak{g}$ which is trivial on Rad($\mathfrak{g}$)) and $L$ is a one-dimensional representation.
The part I do not understand is this. We had found using previous results a linear functional $\lambda$: Rad$(\mathfrak{g})\rightarrow \mathbb{C}$ such that Rad($\mathfrak{g}$) acts on $V$ by the eigenvalues $\lambda$, and had extended it to a linear functional $\tilde{\lambda}:\mathfrak{g}\rightarrow \mathbb{C}$ which vanished on $[\mathfrak{g},\mathfrak{g}]$. This $\tilde{\lambda}$ gives a one-dimensional representation $L$ (and by duality $L^*$)
It is then said that $V \otimes L^*$ is trivial for Rad($\mathfrak{g}$). I think this would mean that for any $X \in$ Rad($\mathfrak{g}$), we have $$ X(v\otimes \mu) = Xv \otimes \lambda(X)\mu = 0 $$
Why should this last quantity be 0?
Here you first need to understand how $X\in\mathrm{Rad}(\mathfrak{g})$ act on $L^*=\mathrm{Hom}(L,\mathbb{C})$: $$(Xf)(u)=-f(Xu)=-\lambda(X)f(u),$$ so $X$ acts on $L^*$ by $-\lambda$.
Next, the action of $\mathfrak{g}$ on $V\otimes L^*$ is not the diagonal action. It is the action induced by the coproduct on $\mathfrak{g}$: $\Delta(X)=X\otimes 1+1\otimes X$. Therefore, on $V\otimes L^*$ and $X\in\mathrm{Rad}(\mathfrak{g})$, we have $$X(v\otimes f)=(Xv)\otimes f + v\otimes(Xf)=\lambda(X)(v\otimes f)-\lambda(X)(v\otimes f)=0.$$