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Show that $\mathbb{C}[x]/\langle x^2-x\rangle \cong \mathbb{C} \times \mathbb{C}$.

Since we're imposing the condition $x^2 \equiv x$ which is true when $x=0$ or $x=1$ on $\mathbb{C}[x]$ one candidate that came to my mind was to define $$\varphi: \Bbb C[x] \to \Bbb C \times \Bbb C, \ \varphi(p) = (p(1), p(0))$$ but I don't think that $\ker\varphi = \langle x^2-x\rangle$.

If $q \in \ker \varphi$, then $q(1) = 0$ and $q(0) = 0$. The latter implies that $q$ has a constant term $0$ and the former implies that the sum of the coefficients is $0$. I can't think of any reasonable way to show that $$q(x)=a(x)(x^2-x)$$ for some $a \in \Bbb C[x]$.

Is there another natural candidate for $\varphi$ here?

Kurosaki
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  • We have that $$\langle x^2-x\rangle = \langle x \rangle \cap \langle x-1\rangle.$$ Use that $x, x-1$ are coprime and the Chinese Remainder Theorem to conclude. – Severin Schraven Nov 09 '22 at 22:29
  • I haven't been introduced to the CRT when dealing with rings unfortunately. I know there is some theory on it with rings, but no idea about it yet. @SeverinSchraven – Kurosaki Nov 09 '22 at 22:30
  • It's a basic result about polynomials that if $q(a)=0$ then $q$ is divisible by $x-a$. So if $q(0)=q(1)=0$ then $q$ is divisible by both $x$ and $x-1$. These polynomials are coprime and so in this case $q$ is also divisible by their product, which is $x^2-x$. – Mark Nov 09 '22 at 22:31
  • That is nice! I forgot about this. Indeed the defined map does work in this case then? @Mark – Kurosaki Nov 09 '22 at 22:33
  • @Kurosaki Yes, it works. – Mark Nov 09 '22 at 22:34
  • At the point where you wrote "I can't think of any reasonable ..." I wanted to suggest that you take a random $q(x)$ (with constant term $0$ and coefficient sum $0$ as you said), say $3x^4-7x^3+6x^2-2x$, and try dividing it by $x^2-x$. Even though you now have an answer from other comments, this experiment, perhaps repeated for a few more choices of $q(x)$, is likely to be beneficial for your understanding of what is happening in your question. – Andreas Blass Nov 10 '22 at 00:59
  • The same methods in the linked dupe apply here (there are also many other dupes). – Bill Dubuque Nov 10 '22 at 13:10

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