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I suspect that the Jacobson radical of the ring

$$A = \begin{bmatrix} R & M \\ 0 & S \\ \end{bmatrix}$$

should be

$$\mathfrak{J}(A) = \begin{bmatrix} \mathfrak{J}(R) & M \\ 0 & \mathfrak{J}(S) \\ \end{bmatrix}$$

I know how to classify the left/right/two-sided ideals of this ring (T.Y. Lam, Chapter 1, Page 18) but I don't know how to find the maximal ones!

I'd like to see a proof of my conjecture if it's indeed true.

EDIT: I believe that this is not a duplicate of the linked question because first of all, this question is more general. Secondly, it requests the proof for something that is worth having its own separate question and thirdly, there's no proof in that linked question.

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    @rschwieb: How is it a duplicate of that question? And I don't see the answer to my question on that link either. Just a statement that is true. Where is the proof? – 2017-01-14
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    The proof is outline there quite clearly: the strictly upper triangular matrices are a nilpotent ideal, so it is contained in the radical. Modding out by that, the radical of the remaining ring is obvious. In the end, you get exactly what you wanted to prove above. – 2017-01-14
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    @rschwieb: I can't disagree more with you about this. I think the clarity of a proof is relative to one's personal background and experience in the subject. Probably that's clear for you because you do have a lot of experience with non-commutative algebra but for me who is learning this stuff for the first time and many of the things you consider trivial facts have not been covered in my book that I'm self-studying from even now that you explained it a bit more in detail I didn't quite get it. There are still some vague points that I'd like to be clarified. – 2017-01-14
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    Ask away here, if you like. (also maybe check out the questions linked to that question.) – 2017-01-14
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    @rschwieb: The comment section is too small for you to write a detailed proof, I believe. Moreover, I'd like the question to be opened so maybe others also can add to it. Maybe someone else uses another approach that is different or easier pedagogically. For example, I don't get why you want to mod out the nilpotent ideal and why after that I get what I wanted to prove. Moreover, I'd like to see if the classification of left/right ideals of $A$ can help us in finding its maximal left/right ideals. I'm just new to all this and the more I learn about this stuff, the better it is for me. – 2017-01-14
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    I didn't intend to spell out in excruciating detail the entire proof in a single comment. Giving you one in-and-out per comment would be fairly easy, though. This is one of those problems that's really better to work through one step at a time. – 2017-01-14
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    Okay. Then let's start step by step. I don't get why you want to mod out the nilpotent ideal and why after that I get what I wanted to prove. What's the idea behind that? – 2017-01-14
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    Well, let me begin: do you know the Jacobson radical contains all nilpotent ideals? It is also the smallest ideal $J$ such that $R/J$ has radical $\{0\}$ – 2017-01-14
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    The idea is that as you find ideals that are in the radical, you can mod out by it and keep looking for the radical of the quotient ring. When you get to a ring that has radical zero, the thing you are modding out by is the Jacobson radical. This is related to my comment above talking about how the radical is the smallest $J$ such that $R/J$ has radical zero. – 2017-01-14
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    @rschwieb: I remember on another post you had told me that the Jacobson radical contains other nilpotent ideals and you said something about primitive ideals but I wasn't following because I didn't know what a primitive ideal is. And I don't know why it's the smallest ideal J such that R/J has the Jacobson radical {0}. – 2017-01-14
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    You would really benefit from reading [section 13B of Martin Isaacs algebra book](https://books.google.com/books?id=5tKq0kbHuc4C&printsec=frontcover&dq=martin+isaacs+algebra&hl=en&sa=X&ved=0ahUKEwimpu7dicTRAhUERiYKHRssDgkQ6AEIHDAA#v=snippet&q=%22called%20primitive%22&f=false) which covers that. Let's say you granted the fact for the moment. After modding out by $M$, do you see why the maximal right ideals of the quotient correspond to those of the original ring? – 2017-01-15
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    I also just remembered there is another simple way to see that nilpotent ideals are in the Jacobson radical: do you know the Jacobson radical is the annihilator of simple right $R$ modules? It's very easy to see why a nilpotent ideal annihilates all simple right modules... – 2017-01-16

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