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Let $R$ be a $k$-algebra and let $(R',\phi)$ be an extension of $R$ by a the $k$-algebra $R'$ by the ideal $I \subseteq R'$ such that $I^2=(0)$.

Let $B$ be another $k$-algebra and assume there exist two $k$-algebra morphisms $f_1, f_2: B \to R'$ such that $\phi f_1= \phi f_2$. Then clearly $f_1 - f_2: B \to I$. It is now claimed that $f_1 - f_2$ is a $k$-derivation of $B$ taking values in the $B$-module $I$.

I am really struggling to see this.

If we let $\psi=f_1 - f_2$, then naturally $I$ is a $B$-module (algebra) via the $k$-algebra morphism $\psi: B \to I$.

I feel like the trick to proving this statement is to somehow specify a particular $B$-module structure on $I$. I'm not sure though, I am at a loss because this seems like some sort of trick that I haven't seen.

$\textbf{My Attempt:}$ All, I know at this point is that if $b,b' \in B$, then $\psi(bb')=0$. Therefore, $\psi : B/B^2 \to I$. Now $B/B^2 = k + k \cdot \{ \text{generators of} B \}$. Since $\psi$ vanishes on $k$, $\psi: k \cdot \{ \text{generators of} \ B \} \to I$.

Then $\psi$ is sort of stupidly a derivation since $\psi (k g): k \psi(g)$ where $g$ is a generator of $B$. However, there are no products in this domain so there is nothing else to prove.

user7090
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  • In the commutative setting if $\varphi(a)=\frac{f_1(a)+f_2(a)}{2}$ is a morphism then $\psi(a)= f_1(a)-f_2(a)$ is a derivation since $\psi(ab)= \varphi(a) \psi(b)+\varphi(b) \psi(a)$ – reuns May 05 '19 at 19:26
  • But shouldn't a derivation be of the form $\psi(ab)=a \psi(b) + b \psi(a)$? – user7090 May 05 '19 at 19:50

1 Answers1

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You are right, there is another $B$-module structure on $I$!

In fact, a priori there are two different structures. Since $f_1$ and $f_2$ are $k$-algebra morphisms, there are induced actions on the ideal $I \subset R'$ given by

$b \cdot_1 m = f_1(b)m \qquad \mbox{and} \qquad b \cdot_2 m = f_2(b)m$

for $m \in I$ (the multiplication on the right-hand side of the equalities is of course in $R'$).

However, since $\phi f_1 = \phi f_2$, we have $(f_1-f_2)(B) \subset \ker \phi = I$, and as $I^2 = (0)$, it follows that $(f_1(b)-f_2(b))m = 0$, so that $f_1(b)m = f_2(b)m$.

Therefore, the two actions above coincide, and we get a single structure of $B$-module on $I$ via

$b \cdot m = f_1(b)m = f_2(b)m$.

Now verify that $\psi = f_1 - f_2 : B \rightarrow I$ is a $k$-derivation:

  • $\psi(b + b') = \psi(b) + \psi(b')$ is immediate.
  • For $x \in k$ (omitting the structure map $k \rightarrow B$ by abuse of notation), $\psi(x) = f_1(x) - f_2(x) = x (f_1(1) - f_2(1)) = 0$.
  • $\begin{align*} \psi(bb') &= f_1(bb') - f_2(bb')\\ &= f_1(b)f_1(b') - f_2(b)f_2(b') \\ &= f_1(b)\left( f_1(b)'-f_2(b') \right) + \left( f_1(b) - f_2(b) \right) f_2(b') \\ &= b \cdot \psi(b') + (b') \cdot \psi(b) \end{align*}$
Jane Doé
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    This is such a great answer. Thank you! I was missing exactly what you pointed about the two module structures on $B$. – user7090 May 05 '19 at 20:15