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I want to demonstrate that the ring $\mathbb{K}[x]/\mathbb{K}\cong \mathbb{K}[x]$, in the sense of additive group, where $\mathbb{K}$ is a field.

I try to proof it with the homomorphism $\alpha:\mathbb{K}[x]\rightarrow \mathbb{K}[x]$ defined by $\alpha(f(x))=f'(x)$, and then conclude with the first theorem of homomorphism.

It is right?

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    $\begingroup$ As abelian groups, $\mathbb{K}[x] \cong \mathbb{K}^{(\Bbb N)}$, and $\mathbb{K}^{(\Bbb N)} / \Bbb K \cong \mathbb{K}^{(\Bbb N)}$. $\endgroup$ – Watson Feb 10 '17 at 12:38
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    $\begingroup$ $f'$ seems overly complicated. Plus some weird things happen on fields on non-$0$ characteristic (for example, the derivative of $X^p$ is $0$ in $\mathbb F_p$). Just shifting should be enough since you don't use the ring structure. $\endgroup$ – xavierm02 Feb 10 '17 at 12:44
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Yes, your approach is correct if $\mathbb{K}$ is a field of characteristic zero. The kernel of $\alpha$, the map defined by differentiation, is the subgroup of (the abelian group) $\mathbb{K}[x]$ consisting of the constant polynomials. You might improve your proof by explicitly mentioning this.

However if $\mathbb{K}$ were a field with prime $p$ characteristic, there would be additional elements of the kernel of $\alpha$. The statement would be true, but your proof would not work. (Think about the infinite dimensional vector space, as mentioned in the Comments.)

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  • $\begingroup$ If $K=\Bbb F_p$, then $X^p$ is sent to $0$. $\endgroup$ – Watson Feb 10 '17 at 12:44
  • $\begingroup$ @Watson: Yes, just added that... thanks. $\endgroup$ – hardmath Feb 10 '17 at 12:47

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