# Find all of homomorphisms $Φ$ from $\mathbb C$[$x$]/$I$ to $\mathbb C$ that satisfies specific condition

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.

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

## 2 Answers

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?