Lemma 10.131.6. In diagram (10.131.4.1), suppose that $S \to S'$ is surjective with kernel $I \subset S$. Then $\Omega _{S/R} \to \Omega _{S'/R'}$ is surjective with kernel generated as an $S$-module by the elements $\text{d}a$, where $a \in S$ is such that $\varphi (a) \in \beta (R')$. (This includes in particular the elements $\text{d}(i)$, $i \in I$.)

First proof. Consider the map of presentations (10.131.4.2). Clearly the right vertical map of free modules is surjective. Thus the map is surjective. Suppose that some element $\eta$ of $\Omega _{S/R}$ maps to zero in $\Omega _{S'/R'}$. Write $\eta$ as the image of $\sum s_ i[a_ i]$ for some $s_ i, a_ i \in S$. Then we see that $\sum \varphi (s_ i)[\varphi (a_ i)]$ is the image of an element

$\theta = \sum s_ j'[a_ j', b_ j'] + \sum s_ k'[f_ k', g_ k'] + \sum s_ l'[r_ l']$

in the upper left corner of the diagram. Since $\varphi$ is surjective, the terms $s_ j'[a_ j', b_ j']$ and $s_ k'[f_ k', g_ k']$ are in the image of elements in the lower right corner. Thus, modifying $\eta$ and $\theta$ by subtracting the images of these elements, we may assume $\theta = \sum s_ l'[r_ l']$. In other words, we see $\sum \varphi (s_ i)[\varphi (a_ i)]$ is of the form $\sum s'_ l [\beta (r'_ l)]$. Next, we may assume that we have some $a' \in S'$ such that $a' = \varphi (a_ i)$ for all $i$ and $a' = \beta (r_ l')$ for all $l$. This is clear from the direct sum decomposition of the upper right corner of the diagram. Choose $a \in S$ with $\varphi (a) = a'$. Then we can write $a_ i = a + x_ i$ for some $x_ i \in I$. Thus we may assume that all $a_ i$ are equal to $a$ by using the relations that are allowed. But then we may assume our element is of the form $s[a]$. We still know that $\varphi (s)[a'] = \sum \varphi (s_ l')[\beta (r_ l')]$. Hence either $\varphi (s) = 0$ and we're done, or $a' = \varphi (a)$ is in the image of $\beta$ and we're done as well. $\square$

Second proof. We will use the universal property of modules of differentials given in Lemma 10.131.3 without further mention.

In (10.131.4.1) let $R'' = S \times _{S'} R'$. Then we have following diagram:

$\xymatrix{ S \ar[r] & S \ar[r] & S' \\ R \ar[r] \ar[u] & R'' \ar[r] \ar[u] & R' \ar[u] }$

Let $M$ be an $S$-module. It follows immediately from the definitions that an $R$-derivation $D : S \to M$ is an $R''$-derivation if and only if it annihilates the elements in the image of $R'' \to S$. The universal property translates this into the statement that the natural map $\Omega _{S/R} \to \Omega _{S/R''}$ is surjective with kernel generated as an $S$-module by the image of $R''$.

From the previous paragraph we see that it suffices to show that $\Omega _{S/R} \to \Omega _{S'/R'}$ is an isomorphism when $S \to S'$ is surjective and $R = S \times _{S'} R'$. Let $M'$ be an $S'$-module. Observe that any $R'$-derivation $D' : S' \to M'$ gives an $R$-derivation by precomposing with $S \to S'$. Conversely, suppose $M$ is an $S$-module and $D : S \to M$ is an $R$-derivation. If $i \in I$, then there exist an $a \in R$ with $\alpha (a) = i$ (as $R = S \times _{S'} R'$). It follows that $D(i) = 0$ and hence $0 = D(is) = iD(s)$ for all $s \in S$. Thus the image of $D$ is contained in the submodule $M' \subset M$ of elements annihilated by $I$ and moreover the induced map $S \to M'$ factors through an $R'$-derivation $S' \to M'$. It is an exercise to use the universal property to see that this means $\Omega _{S/R} \to \Omega _{S'/R'}$ is an isomorphism; details omitted. $\square$

Comment #6681 by Frank on

Any hint for the diagram chasing in the first step here? I tried for several times but I cannot get the result because the presentation is not exact on the left.

Comment #6683 by on

Yes, this is horrible (see below) and we should do this another way! I will rewrite the proof the next time I go through all the comments.

Diagram chase (maybe not literally a diagram chase). Suppose that some element $\eta$ of $\Omega_{S/R}$ maps to zero in $\Omega_{S'/R'}$. Write $\eta$ as the image of $\sum s_i[a_i]$ for some $s_i, a_i \in S$. Then we see that $\sum \varphi(s_i)[\varphi(a_i)]$ is the image of some huge sum of terms $s'[a', b']$ and $s'[f', g']$ and $s'[r']$. Since $\varphi$ is surjective, we may assume the terms $s'[a', b']$ and $s'[f', g']$ are not there by modifying our choices. Then we see that $\sum \var(s_i)[\varphi(a_i)]$ is of the form $\sum s'_j [r'_j]$. I guess now you still have to do a little bit here. First, you can assume that $s' = \varphi(a_i)$ is a fixed element of $S'$ by grouping the sum into a sum of sums. Then you can write $a_i = a + x_i$ for some $x_i \in I$. Thus we may assume that all $a_i$ are equal to $a \in S$ by using the relations that are allowed. But then we may assume our element is of the form $s[a]$ and the result is clear in that case.

Comment #6684 by on

Undefined control sequence should read $\sum \varphi(s_i)[\varphi(a_i)]$.

Comment #7704 by Ryo Suzuki on

This lemma can be proved by universal property.

It suffices to show that $\Omega_{S/R} \cong \Omega_{S'/R'}$ when $S\to S'$ is surjective and diagram (10.131.4.1) is pullback.

To show the surjectivity of $\Omega_{S/R} \to \Omega_{S'/R'}$, suppose $d\colon S'\to M$ is a $R'$-derivation and assume $d\circ \phi = 0$. In this case, $d = 0$ because $\phi$ is surjective.

Next, let $d\colon S\to M$ be a $R$-derivation. If $a\in I$, there exists $s\in R$ such that $\alpha(s) = a$. It is because $R = R'\otimes_{S'} S$. Hence $d$ factors through $S'$. By taking $M=\Omega_{S/R}$, we see that $S\to \Omega_{S/R}$ factors through $\Omega_{S'/R'}$. This means that $\Omega_{S/R} \to \Omega_{S'/R'}$ has a retraction. In particular $\Omega_{S/R} \to \Omega_{S'/R'}$ is injective.

Comment #7710 by on

See Lemma 10.131.12 for the statement of the tensor product algebra whose proof also uses a universal property. Are you suggesting to deduce this lemma from that one? What is your justification for the sentence: "It suffices to show that ... when ... is surjective and diagram ... is pullback."?

Comment #7713 by Ryo Suzuki on

In diagram 10.129.1, let $R''$ be a pullback $S\times_{S'} R'$. (not $\otimes$) Then we have following diagram:

For any $R''$-derivation $d\colon S\to M$ which is $0$ as $R$-derivation is $0$ also as $R'$-derivation. Hence $\Omega_{S/R}\to \Omega_{S/R''}$ is surjective. Moreover, the kernel of this map is generated as $S$-module by the image of $R''$, by definition of derivation. So, it suffice to show that $\Omega_{S/R''} = \Omega_{S'/R'}$.

Comment #7714 by Ryo Suzuki on

Hmm... The diagram is not shown propery. I try again.

Comment #7715 by Ryo Suzuki on

I guess preview is not working propery. At first I wrote as this:

\xymatrix{ S \ar[r] & S\ar[r] & S' \\\\ R \ar[r] \ar[u] & R'' \ar[r] \ar[u] & R' }

This works propery in preview, but doesn't work in comment. Then I wrote as this:

\xymatrix{ S \ar[r] & S\ar[r] & S' \\ R \ar[r] \ar[u] & R'' \ar[r] \ar[u] & R' }

This doesn't work propery in preview, but does work in comment.

Comment #8855 by Et on

At the end of the first proof, it would be helpful to say that we are done in the case $\varphi(s)=0$ thanks to the Leibnitz rule.

Comment #9234 by on

Dear Et, I think that if $\varphi(s) = 0$ in the last sentence of the first proof then $\varphi(s)[a'] = 0[a'] = 0$ in the direct sum so we're done by fiat. I did remove a superfluous sentence here.

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