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Could you please help me to understand the following "example" from my lecture script in the abstract algebra?

Example 12.34. Let $R = K[[x]]$ be a formal power series ring over a field $K$. For $0 \ne f = \sum_{i=0}^\infty a_ix^i \in R$ we can define $$subdeg(f) = \min\{i \in \mathbb{N}_0 \mid a_i \ne 0\}.$$ As $f$ is a unit iff $a_0 \ne 0$, we conclude $f \sim x^{subdeg(f)}$. With $subdeg(0) := \infty$ we have $$f \mid g \Leftrightarrow subdeg(f) \le subdeg(g)$$ and $$f \sim g \Leftrightarrow subdeg(f) = subdeg(g)$$ for all $f, g \in R$. As $subdeg(fg) = subdeg(f) + subdeg(g)$, we have $$f \text{ is irreducible } \Leftrightarrow subdeg(f) = 1 \Leftrightarrow f \text{ is prime element}.$$

The background: According to the script,

  1. We call $a$ and $b$ associated (notated $a \sim b$), if $a \mid b$ and $b \mid a$. Alternatively, we say that there is a $c \in R^\times$ with $b = ca$ and the ideals $(a)$ and $(b)$ are equal.

  2. For a commutative ring $R$ we define $(a) := Ra = \{ra \mid r \in R\}$.

  3. $R^\times := \{a \in R \mid a \text{ is a unit.}\}$.

  4. We call $a$ irreducible, if $a \ne 0, a \notin R^\times$ and for all $b,c \in R$ with $a=bc$ it is $b \sim a$ or $c \sim a$.

  5. We call $a$ a prime element, if $a \ne 0, a \notin R^\times$ and for all $b,c \in R$ with $a \mid bc$ it is $a \mid b$ or $a \mid c$. Alternatively we demand that the principal ideal $(a)$ is a prime ideal.

  6. We say that $(a)$ is a prime ideal in $R$, if $R/(a)$ is an integrity domain.

The first thing I do not understand is the statement marked bold: $$f \in (K[[x]])^\times \Leftrightarrow a_0 \ne 0.$$ Why does it hold? I tried to figure out the inverse for $\sum_{i=0}^\infty a_ix^i$ with $a_0 \ne 0$, but did not succeed.

The second thing is the conclusion that follows from this statement. So, let $K = \mathbb{R}$ and $R=\mathbb{R}[[x]]$ with $f = 4x^3 + x \in \mathbb{R}[[x]]$. Then the $subdeg(f)$ is 1, and the statement says $4x^3 + x \sim x$. However, I cannot understand how this must be true. Although $x \mid 4x^3 + x$, but $4x^3 + x \nmid x$, so they must be not associated. Now I can try to modify $f$ and get $f'=4x^3 + x + 2$. Then $subdeg(f') = 0$ and $4x^3 + x + 2 \sim 1$. But then I have the same problem. Even though $1 \mid 4x^3 + x + 2$, how can $4x^3 + x + 2 \mid 1$ hold?

This script often uses examples as a place for new definitions, which serve for confusions instead of insights. This example was intended to be an example for irreducible and prime elements, but it serves for more confusion introducing the new unproven truth in the form of the bold statement. However, I do not understand the intention of the example, the purpose of $subdeg$ definition, and cannot follow after the part marked bold. Could you please explain me, what the professor tried to say, why the statement $f \in (K[[x]])^\times \Leftrightarrow a_0 \ne 0$ holds, and how can either $4x^3 + x \mid x$ or $4x^3 + x + 2 \mid 1$ be true?

jupiter_jazz
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  • "The first thing I do not understand is ..." - well, for an answer see this duplicate. Because $a_0\in K$ is a unit iff $a_0\neq 0$, and $(K[[x||)^{\times}=K^{\times}$. – Dietrich Burde Sep 16 '22 at 16:40
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    See the linked dupe for characterization of power series units. $\ x+4x^3 = x(1+4x^2)\mid x,$ because $u = 1+4x^2$ is a unit (invertible), i.e. $u^{-1}\in K[[x]],,$ so $,uf\mid f, $ by $,u^{-1}(uf) = f.\ \ $ – Bill Dubuque Sep 16 '22 at 17:27
  • Thank you for the answers, they helped me a lot! Could you please point me out where is my error in the inverse calculation?: The inverse of $1 + 4x^2$ should satisfy $a_0b_0 + (a_1b_0 + a_0b_1)x + (a_2b_0 + a_1b_1 + a_0b_2)x^2 = 1$, hence $1 + 0x + 0x^2 = 1$. So, 1) $b_0$ must be $1$. 2) $a_1 = 0$, hence $a_1b_0 + a_0b_1 = 0 \Leftrightarrow b_1 = 0$. 3) $a_2b_0 + a_1b_1 + a_0b_2 = 0 \Leftrightarrow b_2 = -4$. So $b_0 = 1, b_1 = 0, b_2 = -4$, but $-4x^2 + 1$ is not the inverse of $4x^2+1$ in $(\mathbb{R}[[x]])^\times$. – jupiter_jazz Sep 17 '22 at 11:47
  • ok, I got it. It was not an error, but an incomplete calculation. The inverse of $4x^2 + 1$ in $(\mathbb{R}[[x]])^\times$ is $\sum_{i=0}^\infty(-4)^ix^{2i}$. – jupiter_jazz Sep 17 '22 at 12:27
  • I do not understand the last line from the script, as well. So for the case you could comment on that, that might help a lot. What I get is the following: 1) $f$ is irreducible $\Rightarrow$ for all $g,h$ with $f = gh$ holds $g \sim f$ or $h \sim f$. 2) E.g. $f \sim h$. 3) Then $subdeg(f) = subdeg(gh) = subdeg(g) + subdeg(h) = subdeg(h)$, so $subdeg(g) = 0$. But why $subdeg(f) = 1$? – jupiter_jazz Sep 17 '22 at 12:43

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