# A proof that shows surjective homomorphic image of prime ideal is prime

Let $A, B$ be commutative rings with $1_{A}, 1_{B}$. Suppose that $\mathfrak{p} \neq (1)$ is a prime ideal in $A$ with $\mathfrak{p} \supseteq \ker{\varphi}$ where $\varphi: A \rightarrow B$ is a surjective homomorphism. I want to show $\varphi(\mathfrak{p})$ is a prime ideal.

I tried to solve it directly, but I got stuck, and I realized that it is easy to see from non-unital ring isomorphism $\mathfrak{p}/\ker{\varphi} \simeq \varphi(\mathfrak{p})$, but it is strange that I could not write more direct basic set-theoretic proof.

1. I would like to see if someone can write it down (though the word "direct" may be ambiguous).
2. If prime ideal does not contain kernel of surjective homomorphism, it seems like primeness is not preserved, but there is no reason for me to believe this yet, so I want to know if this is true and why / why not.

Add: For #2 above, now I think $\varphi(\mathfrak{p})$ is always prime as long as $\varphi$ is surjective since $\mathfrak{p}/(\ker{\varphi} \cap \mathfrak{p}) \simeq \varphi(\mathfrak{p})$. I know this may be a technical question, but it is important for me to make this accurate since many arguments that I come up with depend on this.

Add 3: What I considered was $\varphi|_{\mathfrak{p}}:\mathfrak{p} \rightarrow \varphi(\mathfrak{p})$. We can think of any ideal as ring without $1$, so this gives $\mathfrak{p}/(\ker\varphi \cap \mathfrak{p}) = \mathfrak{p}/\ker\varphi|_{\mathfrak{p}} \simeq \varphi(\mathfrak{p})$. The problem here is that $\mathfrak{p}/(\ker\varphi \cap \mathfrak{p})$ is not an ideal of $A/\ker\varphi$ unless $\ker\varphi \subseteq \mathfrak{p}$.

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Suppose $x,y \in B$ and $xy \in \phi(\mathfrak{p})$. Choose $a, b \in A$ such that $a \mapsto x$, $b \mapsto y$, and choose $c \in \mathfrak{p}$ such that $c \mapsto xy$. Then $ab -c \in ker(\phi)$, so $ab \in \mathfrak{p}$. Thus, either $a$ or $b$ is in $\mathfrak{p}$, which means either $x$ or $y$ is in $\phi(\mathfrak{p})$.

Why is surjectivity necessary? Take $\mathbb{Z} \rightarrow \mathbb{Q}$, where the map is inclusion. Why is $\mathfrak{p}$ containing the kernel necessary? Take $\mathbb{Z} \rightarrow \mathbb{Z}/(2)$; $(3)$ does not map to a prime ideal.

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Yet another, more abstract proof.

If $\phi : A \to B$ is a surjective homomorphism, then for every ideal $\mathfrak{b} \subseteq B$ we get an isomorphism $\overline{\phi} : A/\phi^{-1}(\mathfrak{b}) \to B/\mathfrak{b}$. In particular, if $\phi^{-1}(\mathfrak{b}) \subseteq A$ is a prime (primary, maximal, reduced, etc.) ideal, then the same holds for $\mathfrak{b}$. Now, if $\mathfrak{a} \subseteq A$ is an ideal, then $\phi^{-1}(\phi(\mathfrak{a}))=\mathfrak{a} + \ker(\phi)$. Hence, if $\mathfrak{a}+\ker(\phi)$ is prime, then also $\phi(\mathfrak{a})$ is prime. The claim is for the special case $\ker(\phi)\subseteq \mathfrak{a}$.

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What is $\mathfrak{q}$? Did you mean to say <<if $\phi^{-1}(\mathfrak{q})\subseteq A$ is a prime...>> ? –  user38268 Feb 2 '13 at 15:57
@MartinBrandenburg: Nice argument! –  Manos Feb 2 '13 at 15:59

Firstly it is clear that $\varphi(\mathfrak{p})$ is an ideal. Now suppose $xy \in \varphi(\mathfrak{p})$. By surjectivity of $\varphi$ we can write $x = \varphi(x')$ and $y = \varphi(y')$ for some $x',y' \in A$. Now $xy$ can be written as $\varphi(z)$ for some $z \in \mathfrak{p}$. Thus we can write

$$\varphi(z) = \varphi(x')\varphi(y') = \varphi(x'y')$$

and hence $z - x'y' \in \ker \varphi$. Since $\mathfrak{p}$ contains $\ker\varphi$ this implies that $x'y' \in \mathfrak{p}$ since $z \in\mathfrak{p}$. By primality of $\mathfrak{p}$ this implies that $x'$ or $y'$ is in $\mathfrak{p}$ and so $x$ or $y$ is in $\varphi(\mathfrak{p})$. Thus $\varphi(\mathfrak{p})$ is a prime ideal.

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@GilYoungCheong: No, it will not work. –  Manos Feb 2 '13 at 15:47
@BenjaLim I see. But what goes wrong when we think about ideals as rings without $1$ and apply isomorphism theorem (see Add 3)? P.S. I erased my comments here, since it was getting long. –  GYC Feb 2 '13 at 16:13
@GilYoungCheong Right the application of that isomorphism is valid. However how does that say anything about $\mathfrak{p}/(\ker \varphi \cap \mathfrak{p})$ being an ideal of $B$? –  user38268 Feb 2 '13 at 16:42
@BenjaLim I totally agree. We can't unless $\ker\varphi \subseteq \mathfrak{p}$ (in that case the larger ring is $A/\ker\varphi$). What you point out is essential reason that my argument does not work when it comes to $\mathfrak{p}$ that does not contain the kernel. –  GYC Feb 2 '13 at 21:21
Adding to BenjaLim's answer, if $\mathfrak{p}$ does not contain $\operatorname{ker} \phi$, then $\phi(P)$ is not in general a prime ideal because $\phi^{-1}(\phi(P)) = \operatorname{ker} \phi + P$ and the latter need not be a prime ideal, since $\operatorname{ker} \phi + P \neq P$ and the inverse image of a prime ideal is always a prime ideal. (I am assuming that $\phi$ is still surjective)
Let $\phi:A \rightarrow B$ be surjective homomorphism of rings. Then $B \cong A/ \operatorname{ker} \phi$. We know that the prime ideals of $A / \operatorname{ker} \phi$ are in one to one correspondence with the prime ideals of $A$ that contain $\operatorname{ker} \phi$. In view of $B \cong A/\operatorname{ker} \phi$ you can think of the prime ideals of $B$ as the prime ideals of $A/\operatorname{ker} \phi$.
I am aware of that, i am still assuming that $\phi$ is surjective. –  Manos Feb 2 '13 at 15:53