# Quasi-local commutative ring with maximal ideal M

#### Peter

##### Well-known member
MHB Site Helper
I am reading R.Y. Sharp: Steps in Commutative Algebra.

In Chapter 3: Prime Ideals and Maximal Ideals, Exercise 3.19 reads as follows:

Let R be a quasi-local commutative ring with maximal ideal M.

Show that the ring $$\displaystyle R[[X_1, ... \ ... , X_n]]$$ of formal power series in indeterminates $$\displaystyle X_1, ... \ ... , X_n$$ with coefficients in R is again a quasi-local ring, and that its maximal ideal is generated by $$\displaystyle M \cup \{ X_1, ... \ ... , X_n \}$$

Can someone please help me get started on this problem?

Peter

Note: On page 41, Chapter 3, Sharp defines a quasi-local ring as follows:

3.12 Definition. A commutative ring R which has exactly one maximal ideal, M say, is said to be quasi-local.

#### Turgul

##### Member
It is worth noting that these days, what you are calling a quasi-local ring is often referred to as a simply a local ring. Classically one required the ring to also be noetherian to be called local (as opposed to quasi-local), but over the last 50 years, it has become standard to omit this demand.

There is one very important observation to be made about a (quasi-)local ring: any element not contained in the (unique) maximal ideal must be a unit (why?). Hence all things in the maximal ideal are the non-units of the ring and everything else is a unit. So it is good enough to understand when something is a unit if you want to understand the maximal ideal.

With that observation in mind, try to simplify the problem and consider the ring $R[[x_1]]$. What are the units of a power series ring in one variable?

#### Peter

##### Well-known member
MHB Site Helper
It is worth noting that these days, what you are calling a quasi-local ring is often referred to as a simply a local ring. Classically one required the ring to also be noetherian to be called local (as opposed to quasi-local), but over the last 50 years, it has become standard to omit this demand.

There is one very important observation to be made about a (quasi-)local ring: any element not contained in the (unique) maximal ideal must be a unit (why?). Hence all things in the maximal ideal are the non-units of the ring and everything else is a unit. So it is good enough to understand when something is a unit if you want to understand the maximal ideal.

With that observation in mind, try to simplify the problem and consider the ring $R[[x_1]]$. What are the units of a power series ring in one variable?
Thank you for the hint Turgul.

I suspect that the units of R[X] are simply the units of R, but I am not sure how to use this to show that R[X] is (quasi-)local.

Peter

#### Turgul

##### Member
The units of $R[x]$ are indeed the units of $R$, but we're not looking for that. We are looking for the units of $R[[x]]$. Note for example that $(1-x)(1+x+x^2+x^3+ \cdots) = (1+x+x^2+x^3+ \cdots) - x(1+x+x^2+x^3+ \cdots) = 1$.

In fact, as per my comment on the maximal ideal of local rings, if we are to believe the problem, the units of $R[[x]]$ are precisely those things not in the ideal generated by $M$ and $x$. What does that mean? It means that any power series which has a nonzero constant term which is also a unit in $R$ (ie the constant term is not in $M$) should be a unit.

Note that the observation about local rings goes the other way as well; if all elements of a ring outside of a given ideal happen to be units, then that ideal must be maximal and the ring local with that as the unique maximal ideal (why?).

But this means that if we were to show that any power series with constant term not in $M$ were in fact a unit of $R[[x]]$, then $R[[x]]$ would have to be a local ring with maximal ideal exactly those power series which happen to have constant term in $M$. But this is exactly the ideal generated by $M \cup \{x\}$.

#### Deveno

##### Well-known member
MHB Math Scholar
Thank you for the hint Turgul.

I suspect that the units of R[X] are simply the units of R, but I am not sure how to use this to show that R[X] is (quasi-)local.

Peter
Well, no, that's not right. We certainly can write any power series in $R[[x_1]]$ as:

$a - x_1f$, where $a \in R$, and $f$ is some other power series in $R[[x_1]]$.

So if:

$(a - x_1f)(b - x_1g) = ab - x_1(f+g) + x_1^2(fg) = ab - x_1(f+g - x_1(fg)) = 1$

We must have:

$ab = 1$ and $f+g - x_1(fg) = 0$.

The first equation tells us the constant term of a unit must be a unit of $R$.

On the other hand, if $a$ is a unit, so that there is $b \in R$ with $ab = 1$, then taking:

$g = -b(f + bx_1f^2 + bx_1^2f^3 + \cdots)$, we have:

$(a - x_1f)(b - x_1g) = (a - x_1f)(b(1 + bx_1f + (bx_1f)^2 + (bx_1f)^3 + \cdots))$

$= ab + (ab^2x_1f - bx_1f) + (ab^3x_1^2f^2 - b^2x_1^2f^2) + (ab^4x_1^3f^3 - b^3x_1^3f^3) + \cdots$

$= 1 + 0 + 0 + 0 + \cdots = 1$

So the units of $R[[x_1]]$ are precisely the power series with constant term in $U(R)$.

#### Peter

##### Well-known member
MHB Site Helper
Well, no, that's not right. We certainly can write any power series in $R[[x_1]]$ as:

$a - x_1f$, where $a \in R$, and $f$ is some other power series in $R[[x_1]]$.

So if:

$(a - x_1f)(b - x_1g) = ab - x_1(f+g) + x_1^2(fg) = ab - x_1(f+g - x_1(fg)) = 1$

We must have:

$ab = 1$ and $f+g - x_1(fg) = 0$.

The first equation tells us the constant term of a unit must be a unit of $R$.

On the other hand, if $a$ is a unit, so that there is $b \in R$ with $ab = 1$, then taking:

$g = -b(f + bx_1f^2 + bx_1^2f^3 + \cdots)$, we have:

$(a - x_1f)(b - x_1g) = (a - x_1f)(b(1 + bx_1f + (bx_1f)^2 + (bx_1f)^3 + \cdots))$

$= ab + (ab^2x_1f - bx_1f) + (ab^3x_1^2f^2 - b^2x_1^2f^2) + (ab^4x_1^3f^3 - b^3x_1^3f^3) + \cdots$

$= 1 + 0 + 0 + 0 + \cdots = 1$

So the units of $R[[x_1]]$ are precisely the power series with constant term in $U(R)$.
Thanks Deveno

Thanks to you and Turgul I am getting a much better idea of the ring of formal power series.

Just a couple of questions regarding your analysis above:

Question 1

You write:

"So if:

$(a - x_1f)(b - x_1g) = ab - x_1(f+g) + x_1^2(fg) = ab - x_1(f+g - x_1(fg)) = 1$

We must have:

$ab = 1$ and $f+g - x_1(fg) = 0$."

How can we be sure ab = 1 as a and b may be zero divisors since we are not given that R is an integral domain.

Question 2

In showing that

$$\displaystyle (a - x_1f)(b - x_1g) = 1$$ you assume $$\displaystyle g = -b(f + bx_1f^2 + bx_1^2f^3 + \cdots)$$

How on earth did you determine the expression for g? What were your considerations and figuring?

By the way, do you know of a text that treats the manipulation of power series in the way you and Turgul were doing in your posts? Most analysis books that I know just treat convergence issues and most algebra books assume that one is already familiar with such things and just treat abstract aspects of the ring R[[X]]. ... ...

... ... now must get back to thinking about a solution to the exercise ...

Peter

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#### Peter

##### Well-known member
MHB Site Helper
Thanks Deveno

Thanks to you and Turgul I am getting a much better idea of the ring of formal power series.

Just a couple of questions regarding your analysis above:

Question 1

You write:

"So if:

$(a - x_1f)(b - x_1g) = ab - x_1(f+g) + x_1^2(fg) = ab - x_1(f+g - x_1(fg)) = 1$

We must have:

$ab = 1$ and $f+g - x_1(fg) = 0$."

How can we be sure ab = 1 as a and b may be zero divisors since we are not given that R is an integral domain.

Question 2

In showing that

$$\displaystyle (a - x_1f)(b - x_1g) = 1$$ you assume $$\displaystyle g = -b(f + bx_1f^2 + bx_1^2f^3 + \cdots)$$

How on earth did you determine the expression for g? What were your considerations and figuring?

By the way, do you know of a text that treats the manipulation of power series in the way you and Turgul were doing in your posts? Most analysis books that I know just treat convergence issues and most algebra books assume that one is already familiar with such things and just treat abstract aspects of the ring R[[X]]. ... ...

... ... now must get back to thinking about a solution to the exercise ...

Peter
Just summarizing where I am with this interesting exercise ...

From Deveno's post plus some hints from Turgul, we have the following:

... the units of R[[X]] are precisely those power series with a with a constant term, say K, where k is a unit of R.

These elements of R[[X]] cannot possibly belong to the (only) maximal ideal, M' of R[[X]] because if a unit belonged to M' then M' = R[[X]].

Now working on Turgul's hints we have the following ideas ...

If we can show that all the elements outside M are units, then the maximal ideal M' would consist of precisely those power series with a constant term in M. (Any power series with a constant term outside of M would be a unit ...)

... ...

BUT

(1) how do we show that all elements of R outside M are units?

(2) how do we show that M' is the only maximal ideal in R[[X]]?

I would appreciate some further help.

Peter

#### Deveno

##### Well-known member
MHB Math Scholar
If it is *given* that $R$ is quasi-local, then if $d$ is any non-unit of $R$, we have that:

$(d)$ is a proper ideal of $R$ (why?), so $(d)$ is contained in a maximal ideal, so (because there is only one maximal ideal) $(d) \subseteq M$, so $d \in M$.

Thus $M$ contains every non-unit, and none of the units.

On the other hand, if an ideal contains every non-unit, and is not the entire ring, it must be maximal, and no other maximal ideal can possibly exist (since any proper ideal only has non-units in it, and is thus contained in our ideal of every non-unit).

The discussion up to now indicates we should look at the constant terms of elements of $R[[x_1]]$. So this is what you need to do:

A) Prove that the set of power series whose constant term is not a unit forms an ideal. By our discussion above, this will be a unique maximal ideal, showing that $R[[x_1]]$ is quasi-local. If you are perceptive enough, you will see that this boils down to a similar question about $R$, where the matter is already settled.

B) Next, consider the case $R[[x_1,x_2]]$. Once you have this settled, you can use an induction argument on the number of indeterminates.

#### Peter

##### Well-known member
MHB Site Helper
If it is *given* that $R$ is quasi-local, then if $d$ is any non-unit of $R$, we have that:

$(d)$ is a proper ideal of $R$ (why?), so $(d)$ is contained in a maximal ideal, so (because there is only one maximal ideal) $(d) \subseteq M$, so $d \in M$.

Thus $M$ contains every non-unit, and none of the units.

On the other hand, if an ideal contains every non-unit, and is not the entire ring, it must be maximal, and no other maximal ideal can possibly exist (since any proper ideal only has non-units in it, and is thus contained in our ideal of every non-unit).

The discussion up to now indicates we should look at the constant terms of elements of $R[[x_1]]$. So this is what you need to do:

A) Prove that the set of power series whose constant term is not a unit forms an ideal. By our discussion above, this will be a unique maximal ideal, showing that $R[[x_1]]$ is quasi-local. If you are perceptive enough, you will see that this boils down to a similar question about $R$, where the matter is already settled.

B) Next, consider the case $R[[x_1,x_2]]$. Once you have this settled, you can use an induction argument on the number of indeterminates.
Thanks for the help and guidance, Deveno

I wish to try to deal with your point "$(d)$ is a proper ideal of $R$ (why?)". Can you please confirm the argument given is OK.

---------------------------------------------------------------------------

we have that (d) is not proper, that is (d) = R if and only if (d) contains a unit.

Assume (d0 is not proper. Then (d) contains a unit.

Elements of (d) have the form rd = dr for $$\displaystyle r \in R$$ so then if an element of (d) is a unit then there is some $$\displaystyle r_1 \in R$$ such that:

$$\displaystyle (dr)r_1 = 1$$

$$\displaystyle \Longrightarrow d(rr_1) = 1$$

$$\displaystyle \Longrightarrow dr_2 = 1$$ for some $$\displaystyle r_2 \in R$$

BUT d is not a unit in R so this equation cannot hold

Therefore d does not contain a unit ...

Therefore (d) does not contain the unit 1 and so $$\displaystyle (d) \ne R$$, that is (d) is a proper ideal.'

Can you confirm this is a sound argument?

Peter

#### Deveno

##### Well-known member
MHB Math Scholar
Yes, an ideal is proper iff it does not contain a unit (or equivalently is the entire ring iff it does contain a unit). You have a couple of typos but your reasoning is sound.

So the group of units of a ring is important, because it helps us classify the ideals.

A word about motivation:

Rings can be very complicated structures, and the presence of a lot of non-units can make solving equations from them daunting. Zero divisors also pose similar challenges to isolating epxressions we want to evaluate. So the hope is we can find an ideal containing all the "bad" elements, and mod it out, giving a "nicer" ring in which calculation is more straight-forward.

And...if we are fortunate, this nicer ring will allow us to "lift" information gleaned from it, to our original ring. For example, equations involving integers are often solved by passing to the integers mod n (if n = p, a prime, this is a field and we can divide...something we CAN'T do in the original integers), and then going back to the integers. Any argument involving even and odd integer cases is actually using this "trick" (although many people employing this do not even realize this).

Maximal ideals are important, because for commutative rings with unity, we get a field as our quotient ring. Fields are the very pinnacle of "niceness" for rings, algebra works just like our intuition tells us it should. Conversely, the size of a maximal ideal, gives us a rough idea of how much of the ring is going to be hard to work with.

In answer to your earlier question about "how" i inverted the power series:

$a - xf$ it really is just a variation of inverting:

$1 - x$ by noting that:

$1(1 + x + x^2 + x^3 + \cdots) - x(1 + x + x^2 + x^3 + \cdots)$

$= 1 + (x - x) + (x^2 - x^2) + (x^3 - x^3) + \cdots$

$= 1 + 0 + 0 + 0 + \cdots = 1$

Now if $a$ is a unit, so that we have some $b$ with $ab = 1$, we have:

$1 = 1 + 0 + 0 + 0 + \cdots$

$= ab + 0 + 0 + 0 + \cdots$

$= ab + (bxf - bxf) + ((bxf)^2 - (bxf)^2) + ((bxf)^3 - (bxf)^3) + \cdots$

$= ab + (ab)(bxf - bxf) + (ab)((bxf)^2 - (bxf)^2) + (ab)((bxf)^3 - (bxf)^3) + \cdots$

$= ab + ab^2xf - ab^2xf + ab^3x^2f^2 - ab^3x^2f^2 + ab^4x^3f^3 - ab^4x^3f^3 + \cdots$

$= a(b + b^2xf + b^3x^2f^2 + b^4x^3f^3 + \cdots) - xf(b + b^2xf + b^3x^2f^2 + b^4x^3f^3 + \cdots)$

$= (a - xf)(b + (b^2xf + b^3x^2f^2 + b^4x^3f^3 + \cdots))$

$= (a - xf)(b - x(-b)(bf + b^2xf^2 + b^3x^2f^3 + \cdots))$

so that we may take:

$g = (-b^2)(f + bxf^2 + b^2x^2f^3 + \cdots)$

(it appears that I have a typo in my original post. Oh dear. *Sigh*. I try to check these things, but my mind wanders...)

Anyway, that's how you derive the formula for $g$.

#### Peter

##### Well-known member
MHB Site Helper
If it is *given* that $R$ is quasi-local, then if $d$ is any non-unit of $R$, we have that:

$(d)$ is a proper ideal of $R$ (why?), so $(d)$ is contained in a maximal ideal, so (because there is only one maximal ideal) $(d) \subseteq M$, so $d \in M$.

Thus $M$ contains every non-unit, and none of the units.

On the other hand, if an ideal contains every non-unit, and is not the entire ring, it must be maximal, and no other maximal ideal can possibly exist (since any proper ideal only has non-units in it, and is thus contained in our ideal of every non-unit).

The discussion up to now indicates we should look at the constant terms of elements of $R[[x_1]]$. So this is what you need to do:

A) Prove that the set of power series whose constant term is not a unit forms an ideal. By our discussion above, this will be a unique maximal ideal, showing that $R[[x_1]]$ is quasi-local. If you are perceptive enough, you will see that this boils down to a similar question about $R$, where the matter is already settled.

B) Next, consider the case $R[[x_1,x_2]]$. Once you have this settled, you can use an induction argument on the number of indeterminates.
Thanks so much for the help Deveno.

You write that the first thing we must do is as follows:

"Prove that the set of power series whose constant term is not a unit forms an ideal."

Consider f to be a power series, that is of the form:

$$\displaystyle f = a_0 + a_1x + a_2x^2 + ... \ ...$$

then define the set M* defined as follows:

M* = $$\displaystyle \{ f \ | \ f \in R[[x_1]]$$ and $$\displaystyle a_0$$ is a non-unit $$\displaystyle \}$$

We wish to show that M* is an ideal under the operations of addition and multiplication of power series. That is, specifically, we wish to show:

(i) $$\displaystyle M^* \ne \emptyset$$
(ii) whenever $$\displaystyle f,g \in M^*$$ , then $$\displaystyle f+g \in M^*$$ and
(iii) whenever $$\displaystyle f \in M^*$$ and $$\displaystyle h \in R[[x_1]]$$ then $$\displaystyle hf \in M^*$$

To show $$\displaystyle M^* \ne \emptyset$$

The zero element of $$\displaystyle R[[x_1]]$$ (viz. $$\displaystyle 0 = 0 + 0.x_1 + 0.x^2 + ... \ ...$$ ) belongs to M* since 0 is a non-unit, Therefore M* is not the empty set.

To show that whenever $$\displaystyle f,g \in M^*$$, then $$\displaystyle f+g \in M^*$$

Consider $$\displaystyle f, g \in M^*$$

where $$\displaystyle f = a_0 + a_1x + a_2x^2 + ... \ ...$$

and $$\displaystyle g = b_0 + b_1x + b_2x^2 + ... \ ...$$

Then $$\displaystyle f + g = (a_0 + b_0) + ... \ ...$$

Now $$\displaystyle a_0, b_0 \in M$$ where M is the (unique) maximal ideal in R containing all non-units of R

We have $$\displaystyle a_0, b_0 \in M \Longrightarrow a_0 + b_0 \in M$$

$$\displaystyle \Longrightarrow a_0 + b_0$$ is a non-unit in R

$$\displaystyle \Longrightarrow f + g \in M^*$$

Finally ... ...

To show that whenever $$\displaystyle f \in M^*$$ and $$\displaystyle h \in R[[x_1]]$$ then $$\displaystyle hf \in M^*$$

Consider $$\displaystyle f \in M^*$$ and $$\displaystyle h \in R[[x_1]]$$

where $$\displaystyle f = a_0 + a_1x + a_2x^2 + ... \ ...$$

and $$\displaystyle h = h_0 + h_1x + h_2x^2 + ... \ ...$$

Then $$\displaystyle hf = h_0a_0 + ... \ ...$$

But $$\displaystyle h_0a_0 \in M$$

$$\displaystyle \Longrightarrow h_0a_0$$ is a non-unit

$$\displaystyle \Longrightarrow h_0a_0 \in M^*$$

Thus M* is an ideal

Can you confirm that the above proof is sound?

Will now move on to thinking about Deveno's suggested second step ... namely

"Next, consider the case $R[[x_1,x_2]]$. Once you have this settled, you can use an induction argument on the number of indeterminates"

Peter

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#### Deveno

##### Well-known member
MHB Math Scholar
That's the idea exactly: reduce the problem of $R[[x_1]]$ where we are a bit uncertain of how things work, to the same problem of $R$, where we already know the answer.

Now, for the next step:

Is there an isomorphism between $R[[x_1,x_2]]$ and $(R[[x_1]])[[x_2]]$? If it existed, what would it have to look like?

(I'll give you a hint: use the analogous isomorphism that takes $R[x][y] \to R[x,y]$ that is, map the terms "polynomially").

#### Peter

##### Well-known member
MHB Site Helper
That's the idea exactly: reduce the problem of $R[[x_1]]$ where we are a bit uncertain of how things work, to the same problem of $R$, where we already know the answer.

Now, for the next step:

Is there an isomorphism between $R[[x_1,x_2]]$ and $(R[[x_1]])[[x_2]]$? If it existed, what would it have to look like?

(I'll give you a hint: use the analogous isomorphism that takes $R[x][y] \to R[x,y]$ that is, map the terms "polynomially").
Again, thank you for the guidance and help ...

You write:

"Now, for the next step:

Is there an isomorphism between [FONT=MathJax_Math]R[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]][/FONT][FONT=MathJax_Main]][/FONT] and [FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]R[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]][/FONT][FONT=MathJax_Main]][/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]][/FONT][FONT=MathJax_Main]][/FONT]? If it existed, what would it have to look like?

(I'll give you a hint: use the analogous isomorphism that takes [FONT=MathJax_Math]R[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]][/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]][/FONT][FONT=MathJax_Main]→[/FONT][FONT=MathJax_Math]R[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]][/FONT] that is, map the terms "polynomially"). "

I checked a number of textbooks on the construction of the two rings $$\displaystyle R[x][y]$$ and $$\displaystyle R[x,y]$$ including the isomorphism between them, and found Papantonopoulou: Algebra: Pure and Applied most helpful.

To understand the isomorphism between $$\displaystyle R[x][y]$$ and $$\displaystyle R[x,y]$$ we must first understand the nature of the two rings.

Drawing on Papantonopoulou we have the following:

The elements of $$\displaystyle R[x][y]$$ are polynomials in y whose coefficients are taken from the ring of polynomials in x, namely R[x].

Thus for example, one element would be

$$\displaystyle f_0 + f_1y + f_2y^2$$

$$\displaystyle = (x^2 +1) + (x^3 + 2x)y + (x+2)y^2$$ ... ... ... ... (1)

However, we can also define a ring R[x,y] whose elements are finite sums $$\displaystyle \sum a_{ij}x^iy^j$$

An element of R[x,y] for example would be

$$\displaystyle x^2 + 1 + x^3y + 2xy + xy^2 + 2y^2$$ ... ... ... ... (2)

where (2) for illustrative purposes is (1) multiplied out

---------------------------------------------------------------------------

Now to focus on the isomorphism between R[x][y] and R[x,y] ... ...

Let $$\displaystyle \phi$$ map an element of R[x][y] to an element of R[x,y]

obtained by "multiplying out" in the way (1) is turned in (1) above is turned into (2).

So, consider an element h of R[x][y]:

$$\displaystyle h = f_n(x)y^n + ... \ ... f_1(x)y + f_0$$

where $$\displaystyle f_i(x) = {\sum}_{j=1}^{m_i} a_{ij}x^j$$

Then we have

$$\displaystyle \phi (h) = {\sum}_{i=0}^{n} {\sum}_{j=0}^{m_i} a_{ij} x^j y^i$$

Papantonopoulou claims that $$\displaystyle \phi$$ is an isomorphism ...

If we accept this, then where to now ... can you help?

Peter

#### Deveno

##### Well-known member
MHB Math Scholar
Use the same mapping just change the $m$ and $n$ to $\infty$....

#### Peter

##### Well-known member
MHB Site Helper
Use the same mapping just change the $m$ and $n$ to $\infty$....
Sorry Deveno ... I can see that this will give us an isomorphism between power series rings (as distinct from polynomials) but I am not understanding how this can be used establishing what we need to establish in the exercise viz. in the following:

"Let R be a quasi-local commutative ring with maximal ideal M.

Show that the ring $$\displaystyle R[[X_1, ... \ ... , X_n]]$$ of formal power series in indeterminates $$\displaystyle X_1, ... \ ... , X_n$$ with coefficients in R is again a quasi-local ring, and that its maximal ideal is generated by $$\displaystyle M \cup \{ X_1, ... \ ... , X_n \}$$"

Can you help me to see the connection and progress the problem a little ... would be grateful for some help ...

Peter

#### Deveno

##### Well-known member
MHB Math Scholar
Outline of the entire string of beads strung together:

1) Prove that if an ideal of a ring is such that it contains every non-unit, and no units, it is the unique maximal ideal of the ring, and therefore that the ring is quasi-local.

2) Given a quasi-local ring $R$, it possesses a unique maximal ideal $M$.

3) The units of $R[[x_1]]$ are precisely the power series with unit constant terms, that is to say, the constant term is in $R - M$ (the set difference). Hence the non-units of $R[[x_1]]$ are all power series with constant terms in $M$. These non-units form an ideal, which is equal to the ideal generated by $M$ and the set $\{x_1\}$. By (1), this is a unique maximal ideal, so $R[[x_1]]$ is quasi-local.

4) Since $R[[x_1,x_2]] \cong (R[[x_1]])[[x_2]]$, letting $S = R[[x_1]]$, we have from (1) through (3), that $S$ is quasi-local, so $S[[x_2]]$ is quasi-local, with unique maximal ideal generated by:

$(M \cup \{x_1\}) \cup \{x_2\} = M \cup \{x_1,x_2\}$

By induction on $k$, we see that:

$R[[x_1,\dots,x_k]]$ is quasi-local for any positive integer $k$ (mimicking the steps taken going from (3) to (4)).

Taking $k = n$, we are finished.

Credit where credit is due: Turgul's observations are really the key to understanding this...when you understand the essence of what he was saying, my proof is just "formal machinery".

#### Peter

##### Well-known member
MHB Site Helper
Outline of the entire string of beads strung together:

1) Prove that if an ideal of a ring is such that it contains every non-unit, and no units, it is the unique maximal ideal of the ring, and therefore that the ring is quasi-local.

2) Given a quasi-local ring $R$, it possesses a unique maximal ideal $M$.

3) The units of $R[[x_1]]$ are precisely the power series with unit constant terms, that is to say, the constant term is in $R - M$ (the set difference). Hence the non-units of $R[[x_1]]$ are all power series with constant terms in $M$. These non-units form an ideal, which is equal to the ideal generated by $M$ and the set $\{x_1\}$. By (1), this is a unique maximal ideal, so $R[[x_1]]$ is quasi-local.

4) Since $R[[x_1,x_2]] \cong (R[[x_1]])[[x_2]]$, letting $S = R[[x_1]]$, we have from (1) through (3), that $S$ is quasi-local, so $S[[x_2]]$ is quasi-local, with unique maximal ideal generated by:

$(M \cup \{x_1\}) \cup \{x_2\} = M \cup \{x_1,x_2\}$

By induction on $k$, we see that:

$R[[x_1,\dots,x_k]]$ is quasi-local for any positive integer $k$ (mimicking the steps taken going from (3) to (4)).

Taking $k = n$, we are finished.

Credit where credit is due: Turgul's observations are really the key to understanding this...when you understand the essence of what he was saying, my proof is just "formal machinery".

Hi Deveno,

You and Turgul have made this exercise a great learning experience for me

Thanks to you both!

Peter