Formal derivative of a constant (or 1)

Question

I'm just trying to get a feeling for the formal derivative defined as it is here. What would the derivative of a constant, say $a_0 \in K$ be formaly? I mean should one see a $X^0$ besides it? But how would that make sense?

Answer 1

As defined on the referenced site under "Definition", the (formal) derivative of $$f(x)=a_nx^n+\cdots +a_1x+a_0 $$ is $$f'(x)=na_nx^{n-1}+\cdots +a_1, $$ so in particular the formal derivative of $a_0$ is $0$.

Likewise, according to the section "Axiomatic definition well suited for noncommutative rings", property 1), we have $r'=0$ for all $r\in R$, so the same result.

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If we wanted to be very formal, we'd have to be clearer about what an element of $K[X]$ is formally instead of just a something that looks like $a_nX^n+\cdot +a_1X+a_0$. Using the universal property definitoin, $K[X]$ is just any ring together with a ring homomorphism $\iota_X\colon K\to K[X]$ and a map $\iota_X\colon \{X\}\to K[X]$ such that for every ring homomoprhism $f\colon K\to A$ and map $g\colon \{X\}\to A$, there exists a unique ring homomorphism $h\colon K[X]\to A$ with $h\circ \iota_K=f$ and $h\circ\iota_X=g$. We can model $K[X]$ (and $\iota_K$ and $\iota_X$) for example as follows: Let $K[X]$ be the set of maps $p\colon \Bbb N_0\to K$ such tat $p(n)=0$ for almost all $n$; turn this into a ring by pointwise addition $$(p+q)(n):=p(n)+q(n) $$ and the Cauchy product $$(p\cdot q)(n)=\sum_{i=0}^np(i)q(n-i). $$ Define $$\iota_K(a)(n)=\begin{cases}a&n=0\\0&n>0\end{cases}$$ and $$\iota_X(n)=\begin{cases}1&n=1\\0&n\ne 1\end{cases}.$$ With this, simply $ p'(n)=(n+1)p(n+1)$ and so $(\iota_K(a))'$ is the zero-map $\iota_K(0)$.