@Manish: Does the Wedderburn theorem really apply here? It says that every semisimple Artinian ring is a product of matrix rings, but you're asking about the converse.
@Manish: I mean the problem is just that the darn thing isn't simple, unless I'm missing some really silly distinction (which I might be): it has a nontrivial left-submodule where the only nonzero elements are in the first column.
@EricStucky:(in reply to your 1st comment) In TY LAM its stated that Mn(D) is simple, left semisimple etc. This is used to prove Wedderburn theorem. I included Weddernburn theorem to give you a context about the question, I didn't mean to say it comes from that.
Then what's the exact difference between $1~\text{mol}^2$ (of dimension $\mathsf{N}^2$) and $(1~\text{mol}^2)\cdot N_{\text{A}}$ (of dimension $\mathsf{N}$)?
@Manish: Hm, I think I said that wrong, but the point is: you were looking at the ideal generated by an element r... but in order for you to conclude that that ideal is Mn(D), you have to agree that when you say 'ideal' you mean 'two-sided ideal'
@ EricStucky Also If you scroll down a bit on that page you would discover that Mn(D) is also left semisimple (with one component) and hence should not have any left submodule. The example you suggested is not submodule because of matrix multiplication (am i correct on this)
so, yeah, again, when you apply Thm 3.3 to say that the 'ideal' is Mn(D), you have to mean 'two-sided ideal'. Of course you can consider the left-ideal, but then Thm 3.3 gives no guarantee that this thing is Mn(D). [And, as arctic and I mentioned; it is not true.]
[in fact, one can go further: if R is a ring and M is left Mn(R)-module, it is of the form M=N^n where N is a left R-module. to get the summands of M, multiply by diagonal matrices e_ii for i=1,..,n.]
@riker oh ok. That site is really confusing. That makes sense I suppose. I asked the guy who I presume drew it for their game, but they never responded.
@EricStucky Since you agree that Mn(D) is left semisimple, by wedderburn it also implies that its right semisimple. So r_R = U1 circle+ U2 = R_r, (where R_r and r_R is right and left semisimple ring) then it implies U1 and U2 are simple ideal which implies R is not simple-a contradiction, hence r_R has only one component. (Can you find a flaw in my argument if you believe its wrong)
Strange. Someone flagged my joke in the jee chat today and I got suspended for 30 mins. I'm pretty sure that most of the regular people didn't flag it. Some mod maybe :P.
Well, if you're flagged someone will check it out, and then I believe mods may have discretionary authority. 5x flag is something I'd expect is when it automatically happens
Some help with this simple linear algebra problem please: give an example where the span of an intersection of subsets is not equal to the intersection of the spans of the individual subsets.
@arctictern If that's true then would u find a flaw in my argument mentioned; (which i repeared previously) Since you agree that Mn(D) is left semisimple, by wedderburn it also implies that its right semisimple. So r_R = U1 circle+ U2 = R_r, (where R_r and r_R is right and left semisimple ring) then it implies U1 and U2 are simple ideal which implies R is not simple-a contradiction, hence r_R has only one component. (Can you find a flaw in my argument if you believe its wrong)
@arctictern Suppose Mn(D) has two simple components U1 and U2 ie Mn(D)= U1 direct sum U2. Now Mn(D) is both left and right semisimple, this implies U1 and U2 are ideals which should be simple. This Forces Mn(D) not to be simple which is a contradiction. hence it should have only 1 component. (Now could you find a flaw in this)
contradiciton because Mn(D) is simple because D is division ring
well then how are you concluding Mn(D) has two simple components in the first place?
left semisimple only implies it has a decomposition into left submodules, and right semisimple only implies it has a decomposition into right submodules. this doesn't give you a decomposition into two-sided ideals.
yeah, but just because there exists decomposition into left submodules and there also exists decomposition into right submodules doesn't mean there is any two-sided ideals
left submodules may not be right submodules and vice versa
Let C_i be the subset of Mn(D) comprised of matrices with all zero entries except possibly in the ith column. Let R_i be the subset of Mn(D) comprised of matrices with all zero entries except in the ith row. Then all the C_i's are left submodules, and Mn(D) is a direct sum of the C_i's, and all the R_i's are right submodules, and Mn(D) is a direct sum of the R_is, but none of the C_i's are right modules and none of the R_i's are left modules.
so, Mn(D) is not a simple left module over itself, is not a simple right module over itself, but is both left semisimple and right semisimple. at no point can you use any of these facts to deduce the existence of a (proper nontrivial) two-sided ideal of Mn(D).
@arctictern I understand your point, even I had my doubts about having same structure (and hence the question) but then I couldn't make sense of lemma 3.8. I wish you could read that (I think you should recheck the link ; pg 33 and 34 are not in preview but pg 35 is atleat in my country)
@ManishKumarSingh in your edition, is lemma 3.8 saying that if R=B1+...+Br=C1+...+Cs are direct sums of indecomposable ideals then r=s and {B1,...Br}={C1,...,Cs}?
@arctictern I know that I must express some vector x as a linear combination of vectors s in S, and a linear combination of vectors t in T, but all s and t's must not be in the intersection of S and T
we can decompose it into simple left submodules, and we can decompose it into simple right submodules, but not into (proper nontrivial) two-sided ideals
@ManishKumarSingh yes, "if it did exist it'd be unique" is a good way to state lemma 3.8
@MGA Say your sets are A and B with intersection C. You already have span(C) contained in span(A) and span(B). in order for something to be in both span(A) and span(B) but not in span(C), you want something expressible with things from A and also expressible with things from B but not with things from C.
I have a dumb question. If I have a function $f:\mathbb{R}\to\mathbb{R}$, the first two derivatives exist everywhere and $f''(x)\not=0\forall x$, then is the concavity of $f$ the same everywhere?
@arctictern So far, so good. I'm with you. I'm just failing to see how such an example can be constructed. I'm thinking about the intersection of 2 planes in 3D space. You can't have a vector expressible on both planes that does not lie on their intersection line.
