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Jacobson's book on "Lie algebras" has the following definition of enveloping algebra generated by a subset (Definition 2, Chap II) : Start with an unital associative algebra $A$ (over a field $F$) and $S \subset A$. The enveloping algebra $S^{\ast}$ is simply the associative subalgebra of $A$ containing $1_A$ generated by $S$. The way I see it as $S^{\ast} = \sum_{n \geq 0} S^n$ where $S^n$ is the $F$-submodule generated by the set $\{ s_1 \dotsc s_n ~:~ s_1, \dotsc, s_n \in S \}$ of $n$-fold monomials (of course $S^{0} := F$). Since we require $1_A \in S^{\ast}$, we need $n=0$ in the sum.

Now while describing the subsequent properties, it says for $w \in A$ we have $\{ w \}^{\ast}$ is the algebra of polynomials in $w$ with constant term $0$. How can we have the constant terms $0$ if I need $1_A \in \{ w \}^{\ast}$?

Siddhartha
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  • I think(?) I found the problem : while discussing the enveloping sub-algebra (containing 1) generated by $S$ the book uses the notation $S^{\ast}(S^{\dagger})$, elsewhere it simply uses $S^{\ast}$. Maybe this is the difference between whether to take $n=0$ or not in the sum of the definition. There is no a priori clarification of these, and so I am not sure. – Siddhartha Jan 21 '19 at 14:12
  • This has nothing to do with Lie algebras. It's only about associative algebras. You have added "universal" in the title but it's not called "universal" in Jacobson. Anyway, what Jacobson calls "enveloping algebra" should have been called "enveloping subalgebra", and is now rather referred to as "unital subalgebra generated by $S$". – YCor Jan 27 '19 at 04:13

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You've misread Jacobson (p32 in his Dover book).

In a unital associative algebra $A$, he defines $S^*$ as the non-unital subalgebra generated by $S$ and calls it "enveloping associative algebra of $S$ in $A$", and $S^\dagger$ as the unital subalgebra generated by $S$ and calls it "enveloping algebra of $S$ in $A$". This is consistent with his description of $S^*$ as polynomials with 0 constant term.

(This is terrible terminology! But mathematicians from the XXth century built an irreversibly messy terminology with rings and algebras...)

YCor
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  • I see it now : the notation $S^{\ast}(S^{\dagger})$ simply means these are two separate cases. It should be that way as I should have guessed, otherwise the celebrated nilpotency can never be achieved. – Siddhartha Jan 27 '19 at 12:36