Exercise 3.1.8 (Substitution of polynomials)

By the universal property, there is a unique homomorphism $𝜃:R[t]\to R[u]$ such that $𝜃(a)=a$ for all $a\in R$ and $𝜃(t)=u^2+c$, so that

If $p={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=0}^d{a}_i{t}^i$, then the degree of every monomial in $𝜃(p)={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=0}^d{a}_i{(u^2+c)}^i$ is even, so that for instance $u$ is not in the image of $𝜃$. This shows that $𝜃$ is not surjective, thus $𝜃$ is not an isomorphism.

Note: if some polynomial $p={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=0}^d{a}_i{t}^i\ne 0$, with $d=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(p)\ge 0,a_d\ne 0$, is in the kernel of $𝜃$, then

Therefore $a_d=0$ : this is a contradiction, thus $\ker <!--\; FUNCTION\; APPLICATION\; -->(𝜃)=\{0\}$. This proves that $𝜃$ is injective.