I computed homomorphism from formal additive group to formal multiplicative group, realizing that it's represented by $R=\Bbb Z[x_1,x_2,\Bbb…]/(x_ix_j- {i+j \choose i}x_{i+j}) $. Here $x_i\ (i \in \Bbb Z_{> 0})$ means free variables.
Intuitively, it's not noetherian. But how to prove it ?
I'll abuse notation and also write $x_i$ for the elements of $R$.
If $p$ is prime, $x_p$ is not in the ideal $\langle x_1,\dots,x_{p-1}\rangle$ of $R$ since the binomial coefficient $\binom{p}{k}$ is divisible by $p$ for all $1\leq k\leq p-1$. Consequently the increasing chain of ideals of $R$ $$ 0\subseteq\langle x_1\rangle\subseteq\langle x_1,x_2\rangle\subseteq\dots $$ has an infinite strictly increasing subchain $\langle x_1,\dots,x_{p_n-1}\rangle$.