If $\phi: R \rightarrow S$ is a ring homomorphism, then $\phi':R/ker(\phi) \rightarrow S$, defined as $\phi'(r+ker(\phi)) = \phi(r)$ $\forall r \in R$ is an injective homomorphism.

This is true because $r_1+ker(\phi) = r_2+ker(\phi) \iff r_1-r_2\in ker(\phi) \iff \phi(r_1-r_2)=0 \iff \phi(r_1)=\phi(r_2)\iff \phi'(r_1+ker(\phi))=\phi'(r_2+ker(\phi))$, and so $\phi'$ is well-defined and is injective. (It is also a homomorphism - not relevant here). Also, $Im(\phi) = Im(\phi')$ - I can't seem to prove this.

Now, being $\phi$ an homomorphism, $ker(\phi)={0} \iff \phi$ is injective.

The question is: if I have an homomorphism from the quotient ring $R[X]/⟨f(X)⟩$ to a certain $S$ ring, where $degree(f(X)) = n$ is an irreducible polynomial in $R[X]$, what can I conclude about those two being isomorphic or not?

P.S.: I had this question whilst reading Is this quotient Ring Isomorphic to the Complex Numbers and the answer by Bill Dubuque (how can we know that the $f$ mapping is onto?)

  • $\begingroup$ Note that the result you are referring to is not about $S$ but about $\mathbb{R}[X]/I$ where $I$ is generated by an irreducible polynomial $\endgroup$
    – JJR
    May 25, 2017 at 9:49

2 Answers 2


Intuitively, you should think that whenever you mod out by the kernel of a homomorphism, you are not changing the image.

To see that the image of $\phi$ is the same as the image of $\phi'$ more formally, pick an element $s\in S$ in the image of $\phi$. Then there is some $r\in R$ such that $\phi(r)=s$.

Okay great, so what. Now what do you think in $R/\text{ker}(\phi)$ will map to this $s$? The most obvious thing: just the coset of $r$ above. More concretely, we compute $$\phi'(r+\text{ker}(\phi))=\phi(r)=s $$ so that the $\text{im}(\phi')$ is atleast as big as that of $\phi$

And the image of $\phi'$ can't be bigger because any $s\in S$ hit by some $r+\text{ker}(\phi)$ under $\phi'$ is actually hit by a representing $r$ under $\phi$!


Intuitively, when you mod out by an ideal, what you're really saying is that you're setting everything in the "denominator" equal to $0$. So, in $R/\text{ker}(\phi)$, everything in the kernel of $\phi$ is grouped into one term, which has $0$ as a representative.

Next we ask: what is the kernel of $\phi'$? We ask this because if the kernel of $\phi'$ is $0$ (the $0$ in $R/\text{ker}(\phi)$, not the $0$ in $R$), then $\phi'$ is injective. But the kernel of $\phi'$ is the set of all $r+\text{ker}(\phi) \in R/\text{ker}(\phi)$ such that $\phi'(r+\text{ker}(\phi))=0$, which is only the case when $\phi(r)=0$; i.e. the kernel of $\phi'$ is the set $\{\text{ker}(\phi)\}$ (note that this set contains a single element!). Since $\text{ker}(\phi)$ is the zero of $R/\text{ker}(\phi)$, the kernel of $\phi'$ is just that, so $\phi'$ is injective.

You can say that $\text{Im}(\phi')=\text{Im}(\phi)$ because $\phi(r)=\phi'(r+\text{ker}(\phi))$, so everything in the image of $\phi$ is $\phi'$ of something, and vice versa.

As for your final question about $R[X]/\langle f(X)\rangle$, suppose $\varphi :R[X]/\langle f(X)\rangle \to S$ is a homomorphism. When is a homomorphism an isomorphism? Strictly when it is bijective (injective and surjective).

Trivially $\varphi$ is surjective onto its image, so $\varphi$ is surjective when $\text{Im}(\varphi)=S$.

Next, $\varphi$ is injective if and only if its kernel is $0$. What is equal to $0$ in $R[X]/\langle f(X)\rangle$? Exactly the ideal $\langle f(X)\rangle$, since this is what modding by an ideal does. Thus, $\varphi$ is injective when $\text{ker}(\varphi)=\langle f(X)\rangle$.

In general, for a ring homomorphism $\varphi:R\to S$, we have that $R/\text{ker}(\varphi)\cong \text{Im}(\varphi)$. This is called the First (Ring) Isomorphism Theorem, and I highly recommend familiarizing yourself with it!


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