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I am trying a question identical to the one asked here;

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

Restating;

''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 have a proof which is quite elementary as follows;

Suppose $st \in \phi(\mathfrak{p})$ for $s,t \in B$. Then $\phi$ is onto so we have $x,y \in A$ s.t. $\phi(x)=s, \phi(y)=t$. So we have $\phi(xy)=\phi(x)\phi(y)=st$, which was assumed to be in $\phi(\mathfrak{p})$. So $xy \in \mathfrak{p}$ and $\mathfrak{p}$ is prime so either $x$ or $y$ is in $\mathfrak{p}$. Then one of $\phi(x)=t$ or $\phi(y)=s$ is in $\phi(\mathfrak{p})$.

A couple of things are bothering me;

-I never use unitality - EDIT; I do indeed.

-I never use the condition that the prime ideal contains ker($\phi$).

I've read the linked post several times and it's still unclear to me what's going on here.

EDIT: So I guess I never checked that $\phi(\mathfrak{p})$ is not all of $B$ but this is clear since that implies that $1_A$ in $\mathfrak{p}$?

Thanks in advance.

Muselive
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    To answer your last question, yes: if $1_B\in \phi(\mathfrak{p})$, then there exists $x\in\mathfrak{p}$ such that $\phi(x)=1_B$. Then $x-1_A\in\ker(\phi)$, hence $1_A\in\mathfrak{p}$, which is impossible. – Arturo Magidin Sep 16 '20 at 00:06

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From $\phi(xy) \in \phi(\mathfrak{p})$ you cannot immediately deduce $xy \in \mathfrak{p}$; it only means that there is some $z \in \mathfrak{p}$ with $\phi(z) = \phi(xy)$. Now you should use the assumption that $\ker(\phi) \subseteq \mathfrak{p}$ to show that this implies $xy \in \mathfrak{p}$.

Dániel G.
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  • Ok so I got stuck trying to make some slick argument using P+ (I'm curious if this works), its actually a lot simpler. To confirm we can say that $\phi(xy-z) \in ker(\phi) \subset \mathfrak{p}$. $\mathfrak{p}$ is an ideal so $xy+z-z=xy \in \mathfrak{p}$ – Muselive Sep 15 '20 at 16:53
  • A typo, it should be just $xy-z \in kerf$ – Muselive Sep 15 '20 at 19:39
  • That's right. This is the same argument as in the accepted answer in the linked question. – Dániel G. Sep 15 '20 at 19:53