Show that the constant polynomials define a subring of $R[X]$ that is isomorphic to $R$.

Let $\mathbb{R}$ be a commutative ring. Show that $f:\mathbb{R} \to \mathbb{R}[X]$, where $f(a)=(a,0,0,0 \ldots)$ is a injective ring homomoprhism. Furthermore show that the constant polynomials define a subring of $R[X]$ that is isomorphic to $\mathbb{R}$.

Here the notation sequence notation denotes a polynomial. For example $(0,0,2,0,-4,0,0, \dots)$ denotes the polynomial $-4X^4+2X^2$.

So let $a,b \in \mathbb{R}$, then $f(ab)=(ab,0,0,0 \ldots) = (a,0,0,0, \ldots)(b,0,0,0, \ldots) =f(a)f(b)$ and $f(a+b)=(a+b, 0,0,0, \ldots) = (a,0,0,0 \ldots) + (b,0,0,0, \ldots) = f(a)+f(b).$

Also suppose that $f(a)=f(b) \implies (a,0,0,0, \ldots)=(b,,0,0,0, \ldots) \implies a=b$. Which makes $f$ an injective ring homomoprhism.

Now if we restrict the codomain of $f$ to the image of $f$ we have that $f:\mathbb{R} \to f(\mathbb{R})$ is a bijective ring homomorphism in other words an isomorphism from $\mathbb{R}$ to $f(\mathbb{R}$?

Solution 1:

$\phi :R[X]\to R$ defined by

$\phi(f(x)) =c \space \text {[constant term of polynomial } f(x) ]$

Then, $\phi$ is a ring Homomorphism from $R[x]$ onto $R$.

$Ker(\phi) =\langle x\rangle $

Hence, $\frac{R[X]}{\langle x\rangle }\cong R$

Implies, $R$ is isomorphic to the set of all constant polynomials.