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Let $(\mathsf{CRing})^{\text{op}}$ be the category opposite to the category $\mathsf{CRing}$ of commutative rings with one, and $\mathsf{Set}$ the category of sets. Recall that $\text{Spec}(A)$ is the set of prime ideals of $A$, and that a morphism $f:A\to B$ in $\mathsf{CRing}$ induces, in a functorial way, a map $\text{Spec}(B)\to\text{Spec}(A)$ sending $\mathfrak q$ to $f^{-1}(\mathfrak q)$.

(1a) Is the functor $\text{Spec}:(\mathsf{CRing})^{\text{op}}\to\mathsf{Set}$ right exact?

Here are four other equivalent forms of Question (1a):

(1b) Does the functor $\text{Spec}:(\mathsf{CRing})^{\text{op}}\to\mathsf{Set}$ commute with finite colimits?

Let $(A_i)_{i\in I}$ be a projective system of commutative rings indexed by a finite category $I$.

(1c) Is the natural map $$\operatorname*{colim}_i\text{Spec}(A_i)\to\text{Spec}\left(\lim_i A_i\right)$$ bijective?

The fourth and fifth forms will use the fact that our functor commutes with finite coproducts. (Recall that coproducts in $(\mathsf{CRing})^{\text{op}}$ correspond to products in $\mathsf{CRing}$.)

(1d) Does the functor $\text{Spec}:(\mathsf{CRing})^{\text{op}}\to\mathsf{Set}$ commute with cokernels?

Let $A\rightrightarrows B$ be two parallel morphisms in $\mathsf{CRing}$.

(1e) Is the natural map $$\operatorname{Coker}\Big(\text{Spec}(B)\rightrightarrows\text{Spec}(A)\Big)\to\text{Spec}\Big(\operatorname{Ker}(A\rightrightarrows B)\Big)$$ bijective?

Here is a second question:

(2) In the above setting, is the natural map $$\text{Spec}(A)\to\text{Spec}\Big(\operatorname{Ker}(A\rightrightarrows B)\Big)$$ surjective?

Of course, if the answer to (2) is "not necessarily", then the answer to (1) will be "no".

Martin Brandenburg asked a somewhat related question.

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No. For instance, let $A=B=\mathbb{C}[x]$ and let $f:A\to B$ be the identity map and $g:A\to B$ send $x$ to $x+1$. Then the equalizer of $f$ and $g$ is the set of polynomials $p$ such that $p(x)=p(x+1)$, which is just the constant polynomials $\mathbb{C}$ (since if such a $p$ were nonconstant, it would need to have infinitely many roots). But the coequalizer of spectra has more than one point, since it is just the quotient $\mathbb{C}/\mathbb{Z}$ (together with a generic point).

The answer to the second question is also no. For instance, let $A=\mathbb{C}[x,y,r,s]/(rx+sy-1)$ and $B=\mathbb{C}[x,y,r,s,t,u]/(rx+sy-1,tx+uy-1)$. There is an obvious inclusion map $f:A\to B$ but there is also a map $g:A\to B$ which sends $r$ and $s$ to $t$ and $u$. I claim that the equalizer of $f$ and $g$ is just $C=\mathbb{C}[x,y]$. Given this claim, we see that the prime ideal $P=(x,y)\subset C$ does not extend to any prime ideal of $A$, so the map $\operatorname{Spec} A\to\operatorname{Spec} C$ is not surjective.

To prove the claim, let $a\in A$ be such that $f(a)=g(a)$. We can write $a$ as a sum of monomials which do not contain $rx$ (since we can replace $rx$ with $1-sy$). First suppose that no monomial of $a$ contains $r$. Then $a$ is a polynomial in just $s$, $x$, and $y$; suppose $s$ does appear in $a$. We can pick values $x_0,y_0\in\mathbb{C}$ with $x_0\neq 0$ such that when we substitute $x_0$ for $x$ and $y_0$ for $y$, $a$ becomes a nonconstant polynomial $p(s)$ in $s$. For any $\alpha,\beta\in\mathbb{C}$, we can now consider the homomorphism $e_{\alpha,\beta}:B\to\mathbb{C}$ which sends $y$ to $y_0$, $s$ to $\alpha$, $u$ to $\beta$, $x$ to $x_0$, $r$ to $\frac{1-\alpha y_0}{x_0}$, and $t$ to $\frac{1-\beta y_0}{x_0}$. We then have $e_{\alpha,\beta}(f(a))=p(\alpha)$ and $e_{\alpha,\beta}(g(a))=p(\beta)$. Since $p$ is nonconstant, we can choose $\alpha$ and $\beta$ such that $p(\alpha)\neq p(\beta)$ and reach a contradiction.

So, we may assume that $r$ does appear in $a$; write $a=\sum_{i=0}^n a_i x^i+\sum_{j=1}^m b_jr^j$ where the $a_i$ and $b_j$ are in $\mathbb{C}[y,s]$. Replacing every appearance of $r$ by $\frac{1-sy}{x}$, we may consider $a$ as a Laurent polynomial in $x$ with coefficients in $\mathbb{C}[s,y]$ (the division by $x$ is not an issue because we will eventually be substituting a nonzero complex number for $x$). Since $r$ appears in $a$, some $b_j$ is nonzero. Thus one of the coefficients of this Laurent polynomial involves $s$, since we replaced $r$ with $\frac{1-sy}{x}$. So, we can also consider $a$ as a nonconstant polynomial in $s$ with coefficients in $\mathbb{C}[x,1/x,y]$. We can now choose $x_0,y_0\in\mathbb{C}$ with $x_0\neq 0$ such that when we substitute them for $x$ and $y$, $a$ becomes a nonconstant polynomial in $s$, and reach a contradiction as in the previous case.

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  • $\begingroup$ Thanks a lot for your answer! I'm studying it. It seems to me you're also answering no to the second question question, since you write "The answer to the second question is also no". But then why do you write before "The answer to the second question is yes, though"? Is it a part of the answer you wanted to remove, and forgot to do so? $\endgroup$ – Pierre-Yves Gaillard Jan 24 '19 at 20:06
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    $\begingroup$ Oh dear, that's exactly what happened. I started proving the answer was yes and then realized it didn't work, and turned it into a counterexample, but then forgot to remove what I had written originally. :) $\endgroup$ – Eric Wofsey Jan 24 '19 at 20:11

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