# Ring theory: Prove if $g \in \mathcal{B}$, then $g=f_1 h_1+...+f_nh_n$ for some elements $h_1,...,h_n$ in $A$.

This is problem about ring theory.

Problem: Let R be a commutative ring and $$A=R[[t]]$$, the power series over R. Let $$\mathcal{B}$$ be a prime ideal in A not containing t and $$\varphi:A \rightarrow R$$ the homomorphism induced by $$t \mapsto 0$$. Suppose that $$\varphi(\mathcal{B})=(a_1,...a_n).$$ Let $$f_i=a_i+\text{higher terms in t}$$ for $$i=1,...,n$$. Prove if $$g \in \mathcal{B}$$, then $$g=f_1 h_1+...+f_nh_n$$ for some elements $$h_1,...,h_n$$ in $$A$$.

Definition: Let R be a ring. Define the ring of power series over R to be the set $$R[[t]]=\{\sum_{i=0}^{\infty} a_it^i |a_i \in R \text{for all} i\}$$.

Lemma: Let R be a commutative ring and $$\mathcal{p} < R$$ an ideal. Then $$\mathcal{p}$$ is a prime ideal if and only if for all ideals $$\mathcal{A}$$ and $$\mathcal{B}$$ in $$R$$ satisfying $$\mathcal{A}\mathcal{B} \subset \mathcal{p}$$, either $$\mathcal{A} \subset \mathcal{p}$$ or $$\mathcal{B} \subset \mathcal{p}$$.

Discussion: Based on the lemma it is obvious that $$\{f_1...f_n\}$$ is a subset in prime ideal. So does it suffice to show that $$f_1,...,f_n$$ form a basis for $$g$$? I am stuck here. Appreciate the help!

• It is basically the proof of Hilbert's Basis Theorem (except it deals with the power series instead of polynomials). A proof of this can be found in Ch. VIII, Proposition 4.10 of the book "Algebra" by Hungerford (Springer edition). Mar 28 '21 at 18:35
• @MathIsNice1729 Thanks! Now I figured out. Mar 29 '21 at 0:24
• Hi Beacon, how do we answer this question if you figured it out? I'd say write an answer, and while you can't award it the bounty we can at least see if your attempt will need correction, or if there are extensions of the result, or something. (My need to answer this question is obvious : many people will have the same question as you but not all of them will get it from the hint. So you have to show them the way!) Mar 31 '21 at 10:40
• @TeresaLisbon I will do it soon. Thanks for telling. Mar 31 '21 at 16:48
• You are welcome, and I look forward to seeing your attempt. Mar 31 '21 at 16:49

Write $$S=Rf_1+...+Rf_n$$. Let $$\phi \in P$$, then $$\phi(0)=\sum_1^n \lambda_ia_i$$ where $$\lambda_i \in R$$ and $$\phi-\sum_1^n \lambda_if_i \in P \cap (t)$$. So $$\phi-\sum_1^n \lambda_if_i=t\psi$$ with $$\psi \in P$$ and every $$\phi \in P$$ is of the form $$\phi=\alpha+t\psi$$ where $$\alpha \in S$$ and $$\psi \in P$$.
Starting with $$\phi_0=g$$, we get $$\phi_0=\alpha_0+t\phi_1$$. Repeating the same with $$\phi_1$$ and so on we find $$g=\alpha_0+t\alpha_1+...+t^s \alpha_s+t^{s+1}\phi_{s+1}$$.
Since the t-valution of the terms tends to 0, we write $$g=\sum_0^{\infty}a_kt^k$$.
Collecting the terms belonging to the same $$f_i$$, we get $$g=\sum_{k=0}^{\infty}(\sum_{i=1}^n \lambda_{ki}f_i)t^k=\sum_{i=1}^n(\sum_{k=0}^{\infty} \lambda_{ki}t^k)f_i=\sum_{i=1}^n h_if_i$$.
So if $$t \notin P$$ and $$\phi(P) \in R$$ is finitely generated, then the same is true for P in A.