# Prime ideals and maximal ideals of the Pullback of rings

Let $$A,B,C$$ be commutative Noetherian rings with given surjective ring homomorphisms $$f:A\twoheadrightarrow C$$ and $$g: B \twoheadrightarrow C$$. Let $$A\times_C B:=\{(a,b)\in A \times B : f(a)=g(b)\}$$ ( with the subring structure induced from $$A \times B$$ ) be the Pullback (https://en.wikipedia.org/wiki/Pullback_(category_theory) ).

Is there a good description of the prime ideals or maximal ideals of $$A\times_C B$$ in terms of ideals of $$A,B$$ and $$C$$ ?

• it is the pushout of the prime spectra of $A$ and $B$ along the prime spectrum of $C$. Commented Jun 10, 2019 at 21:37
• @quantum: push out of the corresponding spectrum in which category ? certainly it can't be in the category of sets ... Commented Jun 10, 2019 at 23:10
• Not category of sets but of topological spaces. Anyway the underlying set would be this (I believe). It shouldn't be too far from and the rough reason is because the images and preimages of a ring from a (commutative unitary) ring homomorphism are all rings. Commented Jun 12, 2019 at 6:32

## 1 Answer

$$\newcommand\spec{\mathrm{Spec}}$$Hint: You may be inspired by pure product, which is a special type of fiber product when $$C=0$$ (the zero ring). In this case the prime ideals are all the elements in the product of $$\spec A\cup \{A\}$$ and $$\spec B \cup \{B\}$$ without $$A\times B$$.

Since your homomorphism $$f:A\rightarrow C$$ and $$g:B\rightarrow C$$ are surjective it is easier to describe this (this is a general construction and we did not need $$A$$ and $$B$$ Noetherian), since the projections $$A\times_C B \rightarrow A$$ and $$A\times_C B \rightarrow B$$ of prime ideals in $$A\times_C B$$ are either prime ideals or the whole ring.

Things become more messier if $$f$$ or $$g$$ were not surjective.

• Maybe I misunderstood your hint, but it seems to me that prime ideals should be of the form $P_1 \times P_2$ with $P_1$ a prime ideal of $A$ and $P_2$ prime in $B$. Is my understanding correct? Commented Jul 13, 2021 at 12:50
• If my understanding is correct, than your hint is wrong: consider the ideal $(0)$ in $\mathbb Z/2\mathbb Z$, which is prime since the ring is a field. In the product $\mathbb Z/2\mathbb Z\times\mathbb Z/2\mathbb Z$ the product ideal $(0)\times (0)=((0,0))$ is not prime, since the ring is not an integral domain. Commented Jul 14, 2021 at 7:57
• I did not say that if $P_1$ is a prime ideal of $A$ and $P_2$ is a prime ideal of $B$ then $P_1\times P_2$ is the prime ideal of $A\times B$. What I said is: The canonical projections of a prime ideal $P$ in $A\times B$ is either a prime ideal or the whole ring it is projected to. So for $A\times B$ the prime ideals are of the form $P_1\times B$ or $A\times P_2$ and their canonical projections are $P_1$ or $A$ (projection on $A$) and $B$ or $P_2$ (projection on $B$). Commented Jul 21, 2021 at 15:51
• I can see that the prime ideals of $A\times B$ are of the form $\mathfrak p\times B,A\times\mathfrak q$ with $\mathfrak p,\mathfrak q$ prime; that argument uses that $(1,0),(0,1)\in A\times B$, which is not necessarily the case for $A\times_C B$. So how can we conclude from the information about the projections $A\times_C B\to A$ and $A\times_C B\to B$ how the primes of $A\times_C B$ look like? Commented Apr 28, 2023 at 20:30
• @Sha Can you provide an example of the fiber product for which $f$ and $g$ are surjective and the projection of a prime ideal is not a prime? The projections of a prime ideal of the fiber product is a prime because we assumed $f$ and $g$ are surjections. The proof is very similar to the projection of a prime ideal of $A\times B$. Commented Jun 5, 2023 at 12:18