An augmented coideal ideal of an enveloping algebra is generated by its primitive part

By [A1] and [L1], choose a basis of $\mathfrak{l}$; by [L2], give it a total well-order. Apply [L3]. In particular $\mathfrak{l}$ embeds in $U(\mathfrak{l})$, and the associated graded algebra of the filtration [L5] is $S(\mathfrak{l})$.

By [L4], $\mathrm{\Delta }$ is filtered for the total-degree filtration on $U(\mathfrak{l})\otimes U(\mathfrak{l})$: this holds on each degree-one generator because $\mathrm{\Delta }(x)=x\otimes 1+1\otimes x$, and hence on products because $\mathrm{\Delta }$ is an algebra map. Its associated graded coproduct on $S(\mathfrak{l})$ is the algebra map making every $x\in \mathfrak{l}$ primitive, since the two maps agree on the algebra generators.

Put $J_n:=J\cap F_{n}U(\mathfrak{l})$ and $K:=\mathrm{gr}J$. The ideal property makes $K$ a graded ideal of $S(\mathfrak{l})$. To see its coideal property, use AC and [L1] to choose a complement $W_n$ of $J_{n-1}$ in $J_n$; its image in $F_n\mathrm{/}{F}_{n-1}$ is a subspace, so choose a complement there and lift it to $C_n\subset F_n$. Thus $F_n=F_{n-1}\oplus W_n\oplus C_n$ and $J_n=J_{n-1}\oplus W_n$. The total-degree filtration then has associated graded pieces $W_p\otimes (W_q\oplus C_q)$ and $(W_p\oplus C_p)\otimes W_q$ for $J\otimes U+U\otimes J$. Taking highest-degree symbols in the assumed containment gives ${\mathrm{\Delta }}_{S(\mathfrak{l})}(K)\subseteq K\otimes S(\mathfrak{l})+S(\mathfrak{l})\otimes K$.

Since $\epsilon (J)=0$ and $F_{0}U(\mathfrak{l})=k1$, we have $J_0=0$. PBW gives $F_{1}U(\mathfrak{l})=k1\oplus \mathfrak{l}$, so $K_0=0$ and $K_1=\mathfrak{j}$. Also $\mathfrak{j}$ is a Lie ideal: for $x\in \mathfrak{l}$ and $a\in \mathfrak{j}$, the commutator $xa-ax=[x,a]$ lies in $J$ because $J$ is two-sided, and lies in $\mathfrak{l}$ by the enveloping relation.

Let $I_{\mathfrak{j}}$ be the ideal of $S(\mathfrak{l})$ generated by $\mathfrak{j}$, and let $\pi :S(\mathfrak{l})\to S(\mathfrak{l}\mathrm{/}\mathfrak{j})$ be the map induced by the vector-space quotient. The universal properties in [L6] construct inverse algebra maps between $S(\mathfrak{l})\mathrm{/}{I}_{\mathfrak{j}}$ and $S(\mathfrak{l}\mathrm{/}\mathfrak{j})$: maps from either algebra to a commutative unital $k$-algebra $A$ correspond exactly to linear maps $\mathfrak{l}\to A$ vanishing on $\mathfrak{j}$. The maps are inverse because they agree with the identity on the algebra generators. Hence $\ker \pi =I_{\mathfrak{j}}$.

We prove by induction on $n$ that $K_n\subseteq I_{\mathfrak{j}}\cap S^n(\mathfrak{l})$. The claim holds in degrees $0$ and $1$ by step 1.4. For $n\ge 2$, take $z\in K_n$. Its reduced coproduct has, in bidegree $(p,q)$ with $p,q>0$ and $p+q=n$, a component in $K_p\otimes S^q(\mathfrak{l})+S^p(\mathfrak{l})\otimes K_q$ by step 1.3; both $p,q<n$, so the induction hypothesis makes its image under $\pi \otimes \pi$ zero. Thus $\pi (z)$ has zero reduced coproduct and is primitive in $S(\mathfrak{l}\mathrm{/}\mathfrak{j})$. The $(n-1,1)$ component of the coproduct of a homogeneous degree-$n$ element $f$ is the sum obtained by placing each of its $n$ factors in the second tensor slot; multiplying the two slots gives $nf$. If $f$ is primitive this component is zero, so [L7] forces $f=0$. Consequently $\pi (z)=0$ and $z\in I_{\mathfrak{j}}$.

The reverse inclusion $I_{\mathfrak{j}}\subseteq K$ follows because $K$ is an ideal and contains $K_1=\mathfrak{j}$. Hence $\mathrm{gr}J=K=I_{\mathfrak{j}}$.

Since $J$ is two-sided, the left and right ideals $U(\mathfrak{l})\mathfrak{j}$ and $\mathfrak{j}U(\mathfrak{l})$ lie in $J$, so their associated graded spaces lie in $\mathrm{gr}J=I_{\mathfrak{j}}$ by step 3.1. Conversely, each degree-$n$ element of $I_{\mathfrak{j}}$ is a finite sum of products in $S^{n-1}(\mathfrak{l})\mathfrak{j}$, and PBW lifts that sum to an element of $U(\mathfrak{l})\mathfrak{j}\cap F_n$ with the same symbol. The right-handed products lift identically. Hence both associated graded spaces equal $I_{\mathfrak{j}}$.

For $x\in J\cap F_n$, step 4.1 supplies $y\in U(\mathfrak{l})\mathfrak{j}\cap F_n$ with the same degree-$n$ symbol as $x$. Then $x-y\in J\cap F_{n-1}$; descending induction, starting with $J\cap F_0=0$ from step 1.4, gives $x\in U(\mathfrak{l})\mathfrak{j}$. The identical argument with the right-generated ideal gives $x\in \mathfrak{j}U(\mathfrak{l})$. Thus $J=U(\mathfrak{l})\mathfrak{j}=\mathfrak{j}U(\mathfrak{l})$, and applying this equality to any other ideal satisfying the same two conditions and the same intersection proves the stated uniqueness.