Let $\mathfrak{g}$ be a finite dimensional simple Lie algebra over $\mathbb{C}$.  

The polynomial current Lie algebra $\mathfrak{g}[t] = \mathfrak{g} \otimes \mathbb{C} [t]$
has the bracket 
$$[xt^r, yt^s] = [x,y] t^{r+s}$$
 for $x,y \in \mathfrak{g}$. It is graded with deg$(t) = 1$.


If we set $h=0$ in Drinfeld's first presentation of the Yangian (given in Theorem 12.1.1 of Chari and Pressley's [Guide to Quantum Groups](http://books.google.com/books?id=bn5GNkLfnsAC&lpg=PP1&dq=chari%2520pressley%2520guide%2520to%2520quantum%2520groups&pg=PP1#v=onepage&q=&f=false)) then we get a presentation of $U(\mathfrak{g}[t])$ 
where the generators are the elements $x \in \mathfrak{g}$ and $J(x) = xt$ of $\mathfrak{g}[t]$ with degree $=0,1$, and the relations all have degree of both sides less than $3$.




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Specifically we require that all the relation in $\mathfrak{g}$ are satisfied for the elements with degree 0, and 
(for all $x,y, x\_i, y\_i, z\_i \in \mathfrak{g}$ and complex numbers $\lambda, \mu$):


$$\lambda xt + \mu yt = (\lambda x + \mu y)t$$
$$[x, yt] = [x,y]t,$$
$$\sum\_i [x\_i, y\_i] = 0 \implies \sum\_i [x\_i t, y\_i t ] = 0$$
$$ \sum\_i [[x\_i, y\_i], z\_i] = 0 \implies \sum\_i [[x\_i t, y\_i t], z\_i t]=0$$
Then assuming that all the relations of degree less than or equal to $3$ hold is enough to get the remaining ones.
The elements $xt^2, xt^3, \ldots$ are defined inductively.
This can be proved by induction, using the Serre presentation of the finite-dimensional Lie algebra and then checking all the required relations in several cases.
 But even in the $\mathfrak{sl}\_2$ case the argument is laborious. 



> 
> Is there a better way of seeing that one needs only relations of degree less than three in order to get the rest?
> 
> 
> 



Why are relations of degree 3 or less enough in a presentation of the polynomial current Lie algebra g[t]?
For a nilpotent Lie algebra L, generators could be described in terms of $H\_1(L) = L/[L,L]$, and relations in terms of $H\_2(L)$. While this is not applicable directly to $\mathfrak g \otimes \mathbb C[t]$, it is close enough: it could be decomposed, for example, as the semidirect sum $\mathfrak g \oplus (\mathfrak g \otimes t\mathbb C[t])$, or, better yet, as $(\mathfrak g\_- \oplus \mathfrak h) \oplus (\mathfrak g\_+ \oplus (\mathfrak g\otimes t\mathbb C[t]))$, where $\mathfrak g = \mathfrak g\_- \oplus \mathfrak h \oplus \mathfrak g+$ is the triangular decomposition. From here, I presume, one may glue the defining relations of the whole algebra from the defining relations of the second summand, and Serre defining relations for $\mathfrak g$. The second summand, $\mathfrak g\otimes t\mathbb C[t]$ or close to it, is, formally, still not nilpotent, but it is an $\mathbb N$-graded algebra with finite-dimensional components, and the isomorphism between the space of defining relations and $H\_2(L)$ still applies. So the whole thing boils down to computation of $H\_2(\mathfrak g\otimes t\mathbb C[t])$. The whole cohomology $H\_\*(\mathfrak g\otimes t\mathbb C[t])$ was computed in the celebrated 1976 paper by Garland and Lepowsky, or, one may use a more direct and pedestrian approach and derive it from the known formulae which describe the second (co)homology of the current Lie algebra $L\otimes A$ in terms of symmetric invariant bilinear forms of $L$, cyclic (co)homology of $A$, etc. I am not sure that, written accurately will all the details, this will provide a shorter way, but it is definitely a different one. The case of $sl(2)$ would be exceptional in a sense ($H\_2(sl(2)\otimes t\mathbb C[t])$ is bigger than in the case of generic $\mathfrak g$).


