Ring homomorphisms from the real polynomial ring to the real field Let $\mathbb{R}[x]$ be the ring of polynomials with real coefficients in the determinate $x$. Each $\mathbb{R}$-algebra homomorphism from $\mathbb{R}[x]$ to $\mathbb{R}$ has the form 
$$
f(x) \mapsto f(\lambda)
$$ 
for some $\lambda \in \mathbb{R}$.
Is there a (commutative unital) ring homomorphism $\varphi: \mathbb{R}[x]\to \mathbb{R}$ not of the above form?  
 A: Let $\eta : \mathbb{R}[x] \to \mathbb{R}$ be a ring homomorphism. Since $\eta (1) =1$, so $\eta (n) = n, \forall n\in \mathbb{Z}.$  Thus $$\forall \ \frac{q}{p} \in \mathbb{Q}, q= \eta (q) = \eta (p\cdot \frac{q}{p} ) =\eta(p)\cdot \eta (\frac{q}{p} ) = p\cdot \eta (\frac{q}{p}) \Rightarrow \eta (\frac{q}{p} )= \frac{q}{p}.$$
Since $\eta (r) \neq 0$ for all $0\neq r\in \mathbb{R}$, so $\eta (r)= {(\eta (\sqrt{r}))}^2 > 0$.
Now we claim that $\eta(r)=r, \forall r\in \mathbb{R}.$
If $\eta(r) > r$, then we can find $x\in \mathbb{Q}$ such that $\eta(r) > x > r$, thus $\eta(x-r)= \eta(x)-\eta(r)=x-\eta(r) >0$, contradiction to $\eta(r)-x >0 $.
Similarly, if $\eta(r) < r$, then we can find $y\in \mathbb{Q}$ such that $\eta(r) < y < r$, thus $\eta(r-y)= \eta(r)-\eta(y)=\eta(r) -y >0$, contradiction to $\eta(r)-y < 0$.
Hence $\eta(r)=r, \forall r\in \mathbb{R}.$
Let $\lambda=\eta (x)$, we have  $$\eta (f(x))= \sum_{i=1}^{n} \eta(a_i) \cdot \eta (x^i) =\sum_{i=1}^{n} a_i t^i = f(\lambda), \forall f(x)=\sum_{i=1}^{n} a_i x^i \in \mathbb{R}[x].$$
