Find all ring homomorphisms $\psi:\mathbb{C}[x]/\langle x^3 + x^2\rangle\to\mathbb{C}$ such that $\psi(a)=a$ for all $a\in\mathbb C$

Question

Here is the problem:

Let $\mathbb{C}[x]$ be the polynomial ring in one variable over $\mathbb{C}$, and $I = \langle x^2(x + 1)\rangle$ be the ideal generated by $x^2(x + 1)$. Determine all ring homomorphisms $\psi:\mathbb{C}[x]/I \longrightarrow \mathbb{C}$ such that $\psi (a) = a$ for all $a \in \mathbb{C}$

I considered

$$\mathbb{C}[x]/I \cong \mathbb{C}/\langle x^2 \rangle \times\mathbb{C}[x]/\langle x+1\rangle$$

I see that $\mathbb{C}[x]/\langle x+1 \rangle\cong \mathbb {C}$ as rings, if we consider $f( a+ \langle x+1\rangle) = a$. But I would like some hint on how to proceed from here?

In the end, I suppose I would want something like the following? $$\mathbb{C}[x]/I \cong^{iso} \mathbb{C}/\langle x^2 \rangle \times \mathbb{C}[x]/\langle x+1\rangle\cong^{iso} \text{something familiar}\times \mathbb{C}\cong^{hom}\mathbb{C}.$$

But what is $\mathbb{C}[x]/\langle x^2\rangle$? It's definitely not isomorphic to $\mathbb{C}^2$ as rings.

Also since the question asks me to determin all such homomorphisms, I suspect there should be something "cyclic", but I don't see it here.

update: So I considered $\alpha = x+I \in \mathbb{C}[x]/I$, and I see that $\psi^2(\alpha)(\psi(\alpha)+1)=0$. This tells me that we either have $\psi(x+I)=0$, or $\psi(x+I) = -1$. The second case is impossible if we want $\psi(a+I) = a$ for $a\in\mathbb{C}$.

In the first case, since $\psi(x+I)=0$, we can find that

$$\psi(ax^2 + bx + c + I) = \psi(c + I)$$

So we can define

$$\psi: \mathbb{C}[x]/I \longrightarrow \mathbb{C}$$

by $$\psi(ax^2 + bx + c + I ) = c$$

and we can verify that this is a homomorphism. That's it?

Answer 1

Let $\pi\colon\mathbb{C}[x]\to\mathbb{C}[x]/\langle x^3+x^2\rangle$ be the canonical projection.

Then a $\mathbb{C}$-homomorphism $\psi\colon \mathbb{C}[x]/I\to\mathbb{C}$ uniquely determines a $\mathbb{C}$-homomorphism $\alpha=\psi\circ\pi\colon \mathbb{C}[x]\to\mathbb{C}$ with $\ker\alpha\supseteq I$.

Conversely, any $\mathbb{C}$-homomorphism $\beta\colon\mathbb{C}[x]\to\mathbb{C}$ with $\ker\beta\supseteq I$ determines a $\mathbb{C}$-homomorphism $\mathbb{C}[x]/I\to\mathbb{C}$.

Since a $\mathbb{C}$-homomorphism $\beta\colon\mathbb{C}[x]\to\mathbb{C}$ is determined by $\beta(x)$, and $\ker\beta=\langle f(x)\rangle$ for a unique monic irreducible polynomial $f$ (since $\mathbb{C}$ is a field), we just need to see what monic irreducible polynomials $f(x)$ have the property that $$ \langle f(x)\rangle\supseteq\langle x^3+x^2\rangle $$ that is, $$ f(x)\mid x^3+x^2 $$ Can you tell what they are?

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We can see it in a different way. Elements of $\mathbb{C}[x]/I$ can be written in a unique way as $\alpha+\beta u+\gamma u^2$, where $u=x+I$ and $\alpha,\beta,\gamma\in\mathbb{C}$.

If $\psi\colon\mathbb{C}[x]/I\to\mathbb{C}$ is a $\mathbb{C}$-homomorphism, then we just need to determine $\psi(u)$.

Since $u^3+u^2=0$, we must have $\psi(u)^3+\psi(u)^2=0$, which implies $\psi(u)=0$ or $\psi(u)=-1$. Both choices actually correspond to a $\mathbb{C}$-homomorphism.