Isomorphism of $\Bbb C[x]$ and $x\cdot\Bbb C[x]$ as modules over $\Bbb C[x]$

Question

Let $R = \mathbb{C}[x]$. Are $\mathbb{C}[x]$ and $x \cdot \mathbb{C}[x]$ isomorphic as $R$ modules? I was thinking of the map $f(x) \mapsto x f(x)$ as a potential isomorphism. It is certainly a bijection, and it seems to satisfy the axioms for a module homomorphism.

Answer 1

Your suggested map is, indeed, an isomorphism of $R$-modules. To be precise, define $$\varphi\colon\Bbb C[x]\to x\cdot\Bbb C[x], f\mapsto x\cdot f$$ First off, check that it is a bijection

• Injectivity: Suppose $\varphi(f)=\varphi(g)$, then $$\varphi(f)=\varphi(g)\iff x\cdot f=x\cdot g\iff x\cdot(f-g)=0\iff f=g$$
• Surjectivity: Let $\tilde f\in x\cdot\Bbb C[x]$. Then $\tilde f=\sum_{k=1}^n a_kx^k$, where $a_k\in\Bbb C$. Let $f:=\sum_{k=0}^{n-1}a_{k+1}x^k$. Certainly $$\varphi(f)=x\cdot\sum_{k=0}^{n-1}a_{k+1}x^k=\sum_{k=0}^{n-1}a_{k+1}x^{k+1}=\sum_{k=1}^na_kx^k=\tilde f$$

So $\varphi$ defines a bijection between $\Bbb C[x]$ and $x\cdot\Bbb C[x]$. It remains to show that $\varphi$ satisfies the axioms of a $R$-module homomorphism. As a straightforward computation shows we have for $a,b\in R$ and $f,g\in\Bbb C[x]$ \begin{align*} \varphi(a\cdot f+b\cdot g)&=x\cdot(a\cdot f+b\cdot g)\\ &=x\cdot(a\cdot f)+x\cdot(b\cdot g)\\ &=(x\cdot a)\cdot f+(x\cdot b)\cdot g\\ &=(a\cdot x)\cdot f+(b\cdot x)\cdot g\\ &=a\cdot(x\cdot f)+b\cdot(x\cdot g)=a\cdot\varphi(f)+b\cdot\varphi(g) \end{align*} Therefore $\Bbb C[x]$ and $x\cdot\Bbb C[x]$ are isomorphic as $R$-modules and an isomorphisms given by $\varphi$.