Proof of $A\left [ \left [ x \right ] \right ]$ PID implies $A$ is a field

abstract-algebraformal-power-seriesprincipal-ideal-domains

It is a well known result that $A$ (commutative ring with 1) is a field if and only if $A\left [ x \right ]$ is PID. Now, while reading about the formal power series ring $A\left [ \left [ x \right ] \right ]$, I found this result: Let $A$ be a commutative ring with 1. if $A$ is a field, then $A\left [ \left [ x \right ] \right ]$ is a PID and a local ring. Now, $A\left [ x \right ]$ PID implies that A is a field, which made me wonder if the same is true for $A\left [ \left [ x \right ] \right ]$. In other words:

  • If $A\left [ \left [ x \right ] \right ]$ is a PID and a local ring, then $A$ is a field. proof:
    Let $m$ be the maximal ideal of $A\left [ \left [ x \right ] \right ]$. Then, it is true that $m=A\left [ \left [ x \right ] \right ]-units\left ( A\left [ \left [ x \right ] \right ] \right )$. Now, an element $\sum_{0}^{\infty }a_nx^n $ in $A\left [ \left [ x \right ] \right ]$ is a unit if and only if $a_0$ is a unit in $A$. Then, in order to check that $A$ is a field, i would need to check that $m=\left ( x \right )$. Now, because $A\left [ \left [ x \right ] \right ]$ is local, i should only check that $\left ( x \right )$ is maximal. Now, because we are in a PID, this ideal is maximal if and only if $x$ is irreducible in $A\left [ \left [ x \right ] \right ]$. But i think i could show that, because: if it were $x=f.g$, with $f =\sum_{0}^{\infty }a_nx^n $, $g = \sum_{0}^{\infty }b_m x^m $, then $a_0=b_0=0$ is impossible, because then it would be $f.g=x^2.h$, with h in $A\left [ \left [ x \right ] \right ]$. It is also impossible that $a_0 \neq 0 \neq b_0$, because then the independent term of $f.g$ would be $a_0.b_0 \neq 0$ (because $A\left [ \left [ x \right ] \right ]$ is a PID, so it is an integral domain). Then, only one of the two elements $f,g$ can have a non zero independent term. WLOG, let it be $a_0$. then, the first term of $f.g$ should be $a_0.b_1x$, that should be equal to $x$. then, $a_0.b_1 = 1$, and $a_0$ is a unit, so $f$ is a unit, and $x$ is irreducible in $A\left [ \left [ x \right ] \right ]$.

    Then, the non units of $A\left [ \left [ x \right ] \right ]$ would be the ones that have $a_0 = 0$ (independent term equal to zero), so every $a\in A$ not equal to zero should be a unit, so A would be a field.

So then:

  1. Is my reasoning ok?

  2. Is it redundant to ask $A\left [ \left [ x \right ] \right ]$ to be local? Does being $A\left [ \left [ x \right ] \right ]$ a PID imply that $A\left [ \left [ x \right ] \right ]$ is local? (This question arises mainly because i initially found that $A$ is a field $\rightarrow$ $A\left [ \left [ x \right ] \right ]$ is a PID)

(Note: I also tried a proof without asking that the power series ring is local, by adapting the reasoning made for the polynomial ring A[x]. However, in certain part the author uses the fact that $f.g = a$ implies that the degree of $f$ is zero. However, i think this is false for the power series ring, because $\left ( 1-x \right ).\left ( 1+x+x^2 +… \right ) = 1$.)

Best Answer

You are overcomplicating things.

If $A[[X]]$ is a PID, then it is a domain, and thus so is $A$ (as a subring of a domain). In this case $(X)$ is a non zero prime ideal: indeed, taking constant terms induces a ring isomorphism $A[[X]]/(X)\simeq A,$ which is a domain.

But in a PID, nonzero prime ideals are maximal. Hence $(X)$ is maximal and the corresponding quotient, namely $A$, is a field. Note that this proof works if we replace $A[[X]]$ by $A[X]$.

Now it is known that $K[[X]]$ is local with unique maximal ideal $(X)$ for any field $K$ (it follows directly from the fact that any power series with nonzero constant term is invertible in this case)

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