1. Dec 31, 2016

### Math Amateur

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am focused on Section 6.1 The Jacobson Radical ... ...

I need help with the proof of Corollary 6.1.3 ...

Corollary 6.1.3 (including the preceding Proposition) reads as follows:

My questions are as follows:

Question 1

In the proof of Corollary 6.1.3 above we read:

"... ... Since $R$ is generated by $1, J(R) \neq R$. ... ...

My question is as follows: why, given that $R$ is generated by $1$, is it true that $J(R) \neq R$ ... ... ?

Question 2

Bland seems to argue that if we accept that $J(R) \neq R$, then the Corollary is proved ... ... that is that

$J(R) \neq R \Longrightarrow \text{ Rad}(M) \neq M$ ... ...

But ... why would this be true ...?

Hope someone can help ... ...

Peter

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In order to give forum readers the notations, definitions and context of the above post, I am providing the first two pages of Chapter 6 of Bland ... ... as follows ... ... :

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2. Jan 1, 2017

### Staff: Mentor

Hi, Peter,
I don't get it either. In my understanding a maximal ideal is always a proper ideal and thus $J(R) \subsetneq R$ is also a proper ideal. The $1$ simply illustrates it, because $1 \notin J(R)$ and so is $J(R) \subsetneq R$. To me this only excludes the case $J(R)=R=\{0\}$. Maybe $J(R)=R$ means there are no maximal ideals at all, but in my understanding this would only mean $J(R)=\{0\}$ rather than $J(R)=R$. Anyway, since we have a $1$ we need not to bother.
I don't think this is part of the proof. I think the proof is missing, because it is obvious by the proposition above. The convention $Rad(M)=M$ means $M$ has no (proper) maximal submodules, so the corollary is simply another formulation of the proposition itself: "there is at least one maximal submodule" (which has to be proper) is the same as "according to our convention $Rad(M) \neq M \,.$

The remark $J(R)\neq R$ (as soon there is a $1$ or there is at least one maximal ideal) stands for itself (see my confusion under question 1).

3. Jan 1, 2017

### Math Amateur

Hi fresh_42 ...

Thanks for the answer to Question 2 ...

I received the following answer to question 1 from Euge on the MHB Forum:

Since maximal right ideals of $R$ don't contain $1$ by definition, then $J(R)$, the intersection of all maximal right deals, cannot contain $1$. That's why $J(R) \neq R$.

Hope Euge's answer helps ... ...

Regarding the conditions $J(R) = R$ and $J(R) = 0$ ... ... ... ... I am still a little confused ...

Regarding $\text{ Rad}(M)$ ... Bland is clear when it comes to $\text{ Rad}(M) = M$ ... indeed Bland's definition of $\text{ Rad}(M)$ is as follows:

"... ... If $M$ is an $R$-module, then the radical of $M$, denoted by $\text{ Rad}(M)$, is the intersection of the maximal submodules of $M$. If $M$ fails to have maximal submodules, then we set $\text{ Rad}(M) = M$. ... ...

"Presumably $\text{ Rad}(M) = \{ 0 \}$ when there exist maximal submodules in $M$, but their intersection is $\{ 0 \}$

Now, with $J(R)$ Bland does not explicitly give a condition for which $J(R) = R$ ... indeed his definition of the Jacobson radical is as follows:

"... ... The Jacobson radical of $R$, denoted $J(R)$, is the intersection of maximal right ideals of $R$. If $J(R) = 0$, then $R$ is said to be a Jacobson semisimple ring ... ..."

Is it correct to assume that if there are no maximal right ideals in $R$, then $J(R) = R$ ... ... ?

... ... and $J(R) = 0$ if there do exist maximum right ideals but their intersection is $\{ 0 \}$ ... ... ?

What do you think?

Peter

4. Jan 1, 2017

### Staff: Mentor

That's what I said.
But a maximal ideal isn't the entire ring anyways, isn't it? Simply by definition? It wouldn't make any sense to speak of maximal ideals being the ring itself. Of course $R \trianglelefteq R$ is always an ideal, so the concept of maximal ideals would be useless, if we allowed them to be the ring. In that case the maximal ideals were always simply $R$. That's why I didn't understand the bypass with $1$. I simply don't see the necessity. The intersection of all maximal ideals, $J(R)$, can only be smaller or equal than any maximal ideal. How could it become the entire ring?
Sorry, I wasn't clear on this one.
Let's assume a ring without any maximal ideals. Oops, this cannot be true, since $\{0\}$ is always an ideal. So at least $\{0\}$ would be a maximal ideal. In this case $J(R)=\{0\}$.

Now I tried to understand the condition $J(R)=R$ somehow. I still think it is trivially always false. But let's see, what Blade might have thought. He makes the convention $Rad(M)=M \Longleftrightarrow R \text{ has no maximal submodules }$ for modules and says something about an analogous case to ideals. Therefore I tried to interpret $J(R)=R$ by the same convention and assumed - as being analog - it might mean: no maximal ideals. But as shown above, this would imply $J(R)=\{0\}$ and both together $R=J(R)=\{0\}$, which again is meaningless. Therefore, my thought, even with his convention for modules, the condition $J(R)=R$ doesn't make any sense. E.g. in my book, maximal ideals $I$ are defined as those, which are not $(1)$ plus there is no ideal $J$ properly in between $I \subsetneq J \subsetneq (1)\, .$

At least I don't see any flaws in my argumentation. Maybe I've overlooked some strange cases, but I don't see it. The argument with $1 \in R$ is only a shortcut for the thoughts above. Since we have a $1$, we don't have do think about crude cases.

Personally I don't understand this convention, since with the same arguments as with ideals, $Rad(M)$ should be zero, if there are no maximal modules, because $\{0\}$ is always a submodule. I assume that Blade's definition of "maximal" - ideal or submodule - might exclude both cases $\{0\}$ and $M$ explicitly. Then he can make this conventions to have the notation $Rad(M)=\{0\}$ available for the cases where there are proper maximal submodules that intersect to zero, i.e. the case where actually something happens.
Yes, no objections.
According to his convention with modules, and given a total analogy, it is correct, yes. I wouldn't make it for the reasons I mentioned above, but if he does, then it is correct.

Let's see what an important theorem says to the case. The theorem goes as follows:
$$x \in J(R) \Longleftrightarrow 1-xy \in R \text{ is a unit for all }y \in R$$
Now if $J(R)=R$ then all elements $1-xy$ are units for all $x,y \in R$. If $I \subsetneq R$ is a maximal ideal, then there is an element $x \notin I$ and $xR+I=R$. So we can write $R \ni 1=xy+z$ for some $z \in I$. Thus $I \ni z=1-xy$ is a unit by our assumption $J(R)=R$ and the theorem. But an ideal that contains a unit is the entire ring. Therefore our assumption that $I$ is a maximal ideal cannot be true and $R$ doesn't have any maximal ideals (other than $\{0\}\,$).

So it seems to make sense to make this convention ($J(R)=R \longleftrightarrow \text{ no maximal ideals }$), which tells me that either I haven't seen it before, since most authors don't deal very much on these extreme cases, or I have forgotten it. As long as we are sure, what is meant by this, we can make this convention. Personally I think it's a source of possible errors and one has to be careful by deductions from $J(R)=R$ other than "no maximal ideal, resp. only $\{0\}$ as maximal ideal", but o.k.

Let me renew an advice here. It's o.k. to spend some time thinking about those special cases and it certainly helps the general understanding, but don't forget to keep the basic ideas behind the definitions and propositions on focus.

5. Jan 2, 2017

### micromass

Staff Emeritus
The point is that there might not be any maximal ideal at all. In rings without $1$, there might not be a maximal ideal. In rings with $1$, there always is.

No, that's not the point. Sure, there always is an ideal, but that might not mean there always is a maximal ideal. For example, take the group $\mathbb{Q}$ with addition. This has no maximal subgroup. Define also the trivial multiplication $xy = 0$ for all $x$ and $y$. Then this is an example of a ring (necessary wthout unit) that has no maximal ideal. But $\{0\}$ is always an ideal though.

6. Jan 2, 2017

### Staff: Mentor

Thanks, that's indeed a good example.

7. Jan 4, 2017

### Math Amateur

Thanks for your help fresh_42 ...

Still reflecting on what you have said ...

Thanks again for moving my understanding forward on these issues ...

Peter

8. Jan 4, 2017

### Math Amateur

Thanks for the intriguing example micromass ...

Still reflecting on it ...

Thanks again ...

Peter

9. Jan 4, 2017

### Staff: Mentor

Don't forget about @micromass's example $R=\mathbb{Q}$ with all $s\cdot t = 0$. Additive subgroups here are automatically also ideals or $\mathbb{Q}-$modules. But the ring has no $1$, because all products equal $0$. Why are there no maximal ideals, although there are ideals different from the trivial ideal $\{0\}\,$?

10. Jan 10, 2017

### Math Amateur

Hi fresh_42 ... sorry for late reply ... I have been quite ill with a severe case of the 'flu ...

you write:

"... ... Don't forget about @micromass's example ... ... ... Why are there no maximal ideals ... ... ?

"Hmmm ... not sure of myself here ...

BUT ... thoughts ...

Consider ideal generated by $a_1 \in \mathbb{Q}$ ... it is not maximal because it is included in ideal generated by $a_1$ and $a_2$ .... and that is not maximal because it is included in the ideal generated by $a_1, a_2$ and $a_3$ where $a_i \in \mathbb{Q}$ ... and so on ...

Get infinite chain ...

$(a_1) \subseteq (a_1, a_2) \subseteq (a_1, a_2, a_3) \subseteq (a_1, a_2, a_3, a_4) \subseteq$ ... ...

But ... of course ... that is not a proof that there are no maximal ideals ...

Can you help ... ...

Peter

Peter

11. Jan 10, 2017

### Staff: Mentor

Hi Peter,

I hope you're better now. The good news is: they get rarer with age as you accumulate immunities. The bad news is, you have to gather them.

Your idea is basically correct. Because multiplication is of no help, we only have addition and subtraction. Let us denote our ring $R=(\mathbb{Q}\, , \,s\cdot t =0)$ to avoid confusion with the normal $\mathbb{Q}$. Multiplications now are also only an abbreviation for successive additions, i.e. $n \cdot a = \underbrace{a+ \ldots +a}_{n-\text{times}}$ for integers $n$ and not the ring multiplication, since the latter is trivial, i.e. identically $0$.

That's given. What we want to show is, that for any ideal $I$ there is always another ideal $J$ that properly contains $I$ and isn't $R$ itself. So we need the situation $\{0\} \neq I \subset J \subset R$. In order to find $J$ we only know, that there is an element $a \in R$ with $a \notin I$, so $J := I + \mathbb{Z}\cdot a$ is an ideal that properly contains $I$. What has to be shown is, that $J$ cannot be $R$. Until here, this is your idea.

For the rest let us assume $R=J=I+ \mathbb{Z}\cdot a$ and look for a contradiction. When it comes to $R=\mathbb{Q}$ then it is always an idea to think about primes. So let us write $a=\frac{m}{p_1^{t_1} \cdot \ldots \cdot p_n^{t_n}}$ with primes $p_i$ and some integer nominator $m$.
According to our assumption $J=R$ all elements $r \in R$ can be written as $r=u+z \cdot \frac{m}{p_1^{t_1} \cdot \ldots \cdot p_n^{t_n}}$ with some $u \in I$ and $z \in \mathbb{Z}$. (Remember the product here is only a long addition.)
So it remains to find an $r$ which cannot be written this way. I think $r=\frac{1}{q}$ with a prime $q \notin \{p_1, \ldots , p_n\}$ does it, but it's late here and I may have made a mistake. Can you finish it?

Last edited: Jan 11, 2017