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I want to know how can I find ALL homomorphism that satisfies the condition mentioned in the question above. Please give me a method or answer in detail.

  • $\begingroup$ Can you show $\phi(f(x)) = f(1), f \in \mathbb{C}[x]$ isn't an homomorphism $\endgroup$ – reuns Feb 11 at 23:54

Because $\phi(a)=a$ for all $a\in\Bbb{C}$ such a homomorphism is surjective, hence its image is a field, hence its kernel is a maximal ideal. The maximal ideals of the quotient $\Bbb{C}[x]/I$ correspond bijectively to the maximal ideals of $\Bbb{C}[x]$ that contain $I$. The maximal ideals of $\Bbb{C}[x]$ are the principal ideals generated by the linear polynomials. An ideal $(x+\alpha)$ contains $I=(x^2(x+1))$ if and only if $x+\alpha$ divides $x^2(x+1)$, i.e. if and only if $\alpha\in\{0,1\}$. What are the corresponding ring homomorphisms?


Here is another approach:

  1. By the universal property of the quotient, giving a ring homomorphism $\phi:\Bbb{C}[x]/I\rightarrow \mathbb{C}$ is the same as giving a ring homomorphism $\psi:\Bbb{C}[x]\rightarrow \mathbb{C}$ with $\psi(I)=0$.

  2. By the universal property of the polynomial ring, giving a ring homomorphism $\psi:\Bbb{C}[x]\rightarrow \mathbb{C}$ is the same as giving a ring homomorphism $\rho : \mathbb{C} \rightarrow \mathbb{C}$ and specifying an element $\bar{x}\in\mathbb{C}$ (which will then be the image of $x$).

So combining these two and using that you want $\phi(a)=a$ for all $a\in \mathbb{C}$ such a ring homomorphism is uniquely determined by specifying an element $\bar{x}\in\mathbb{C}$ such that $\bar{x}^2(\bar{x}+1)=0$. What are the two corresponding ring homomorphisms?


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