# Existence of homomorphism $\phi:\mathbb{Z}[[X]]\to\mathbb{Q}$ such that $\phi(X)=3/4$?

I'm curious as to whether there is a ring homomorphism from $$\mathbb{Z} [[X]]$$,the ring of formal power series in indeterminate $$X$$, to the rationals such that for every polynomial $$p(X)$$ in the ring of power series, $$\phi(p(X))=p(3/4)$$.

• Yes I mean that p is a formal polynomial Feb 21, 2020 at 2:13
• Right. Sorry I persisted in being ambiguous hahah Feb 21, 2020 at 2:39

The central binomial coefficients have generating function $$p(X)=\sum_{n=0}^{\infty} \binom{2n}n X^n$$ which satisfies $$p(X)^2(1-4X)=1$$ in $$\mathbb Z[[X]].$$
So a $$\phi$$ is a homomorphism sending $$X\mapsto \frac{3}{4}$$ would have to have $$\alpha=\phi(p(X))$$ satisfying $$\alpha^2 \left(1-4\frac{3}{4}\right)=1$$ Or $$\alpha^2=\frac{-1}{2}.$$ There is no such $$\alpha,$$ rational or real.
The generating function $$F \in \mathbb{Z}[[X]]$$ of Catalan numbers $$(C_n)_{n=0}^\infty$$ is given by $$F(X) = \sum_{n=0}^\infty C_nX^n = \frac{1-\sqrt{1-4X}}{2X}$$ and hence it satisfies $$(1-2XF(X))^2 = 1-4X.$$ If such a homomorphism $$\phi$$ would exist, we would have $$(1-2\phi(X)\phi(F(X)))^2 = 1-4\phi(X) \implies \left(\underbrace{1-\frac32\phi(F(X))}_{\in\mathbb{Q}}\right)^2 = 1-4\cdot \frac34 = -2$$ which is a contradiction.