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I have a simple question, this is an isomorphism I don't understand What does the following line mean? This can be interpreted as taking the class of f(z) in the ring $\mathbb{C}[z]/(z-a) \cong \mathbb{C}[z] = \mathbb{C}.$ I don't understand the isomorphism, how and why does this happen?

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    I've removed [tag:algebra] tag, since we don't use algebra tag anymore, see [meta](http://meta.math.stackexchange.com/questions/473/the-use-of-the-algebra-tag/3081#3081) for details.2012-10-30

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You have some things wrong. I assume your question is give a $a \in \Bbb{C}$, why is $\Bbb{C}[z]/(z-a) \cong = \Bbb{C}$?

Well first of all we have the evaluation homomorphism that sends the indeterminate $z$ to $a \in \Bbb{C}$. Since $\Bbb{C}$ is a field, the division algorithm tells us that the kernel of this homomorphism is precisely $(z-a)$. Now the homomorphism is also surjective because for any $w \in \Bbb{C}$, I can write $w = w + a - a$ and so $w$ is the image of the polynomial $z+ (w - a)$.

By the First Isomorphism Theorem it follows that $\Bbb{C}[z]/(z-a) \cong \Bbb{C}$.