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This question regards the proof of proposition 8 in chapter 8.1 of Bosch's 'Algebraic Geometry and Commutative Algebra'.

Let $A, B$ be $R$-algebras, $f_0: A \to B$ a morphism of $R$-algebras (that is, an $R$-linear homomorphism of rings), and $\mathfrak{I} \subseteq B$ and ideal satisfying $\mathfrak{I}^2=0$.

Let $f_1: A \to B$ be an $R$-linear map satisfying $f_1(xy) = f_1(x)f_1(y)$, for $x, y \in A$, and $f_0 \equiv f_1 \mbox{ mod } \mathfrak{I}$. I want to understand how this implies that $f_1(1) = 1$. The author argues as follows:

"Using the geometric series in conjunction with $\mathfrak{I}^2=0$, the congruence $f_1(1) \equiv f_0(1) = 1 \mbox{ mod } \mathfrak{I}$ shows that $f_1(1)$ is a unit in $B$. The latter must be idempotent and, thus, corresponds with $1$ if $f_1$ is multiplicative."

What does he mean by 'geometric series'? Why must $f_1(1)$ be idempotent? Can someone help me?

user26857
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user480840
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2 Answers2

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I think an easier proof is to note that $f_1 - f_0 \in \mathfrak{I}$. Then $0 = (f_1(1) - f_0(1))^2 = (f_1(1) - 1)^2 = f_1(1)^2 - 2 f_1(1) + 1$, so $1 = 2 f_1(1) - f_1(1)^2$. Applying the multiplicative property gives us $f_1(1)^2 = f_1(1) f_1(1) = f_1(1 \cdot 1) = f_1(1)$, so we have $1 = 2 f_1(1) - f_1(1)^2 = 2 f_1(1) - f_1(1) = f_1(1)$.

Note that in general, if $a$ is nilpotent then $1 - a$ is a unit. This is because if $a^n = 0$, we have $1 = 1^n - a^n = (1 - a) (1 + ... + a^{n - 1})$ (this is the geometric series argument that the author was probably alluding to). So since $f_1(x) = f_0(x) - (f_0(x) - f_1(x))$, and $f_0(x) - f_1(x)$ is nilpotent, it follows that $f_1(x)$ is a unit whenever $f_0(x) = 1$.

In particular, $f_1(1)$ must be a unit since $f_0(1) = 1$.

And we have shown that $f_1(1)^2 = f_1(1)$. Since $f_1(1)$ is a unit, we can divide both sides by $f_1(1)$ to get that $f_1(1) = 1$.

Mark Saving
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You have $f_1(1) = f_0(1)+x = 1+x$ for some $x\in\mathfrak I$. Then $$ f_1(1)(1-x) = 1-x^2 = 1, $$ since $\mathfrak I^2=0$, so $f_1(1)$ is a unit. This can be seen as an application of the geometric series, since in general $$ (1+x)^{-1} = \sum_{n=0}^\infty (-x)^n $$ if $x^n=0$ for large enough $n$.

By multiplicativity, $$ f_1(1)=f_1(1\cdot1) = f_1(1)^2, $$ showing idempotence. And the only idempotent unit is $1$.

Milten
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