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Jul
20
comment Question about Jacobson's proof of structure theorem for semi-simple Artinian rings
On the module side there's no issue with identities. All that aside, in the end none of this matters in your current context since all products/sums are finite. :)
Jul
20
comment Question about Jacobson's proof of structure theorem for semi-simple Artinian rings
I see what you mean about sums/products now... The direct sum / product, for rings $R_\alpha$, $\alpha \in I$ is defined as follows: The direct product is the cartesian product of $R_\alpha$'s given the natural coordinatewise operations. This is the product ring. If you take the subring where all but finitely many coordinates are 0 (this is a subring in the "rings without unity" sense it is not a subring in Jacobson's sense) you get an external direct sum of $R_\alpha$'s. So this isn't in the category of rings with 1 (unless we are only dealing with a finite collection of $R_\alpha$'s).
Jul
20
comment Question about Jacobson's proof of structure theorem for semi-simple Artinian rings
It's not an isomorphism of rings really. We have that $R$ is decomposed as a $M(R)$-module internally. So $R = I_1 \oplus I_2 \oplus \cdots$. This says that every element of $R$ can be written uniquely as a sum $r_1+r_2+\dots$ where $r_j \in I_j$. [This is what the lemma gives us.] Next, we have (from previous arguments or my answer above) that each $I_j$ is in fact an ideal of $R$. Therefore, $R = I_1 \oplus I_2 \oplus \cdots$ is a direct sum of ideals. Finally, irreducible = simple in this context. Done.
Jul
19
answered Question about Jacobson's proof of structure theorem for semi-simple Artinian rings
Jul
16
answered Determining remainders when dividing polynomials.
Jul
10
comment Number of automorphisms of $\mathbb Z _{91}$ of order 3
What about $(b,1)$, $(1,a)$, $(b,a)$, $(a,a)$, and $(b,b)$? $\mathbb{Z}_{13}^*$ and $\mathbb{Z}_7^*$ are indeed cyclic. Both have orders ($13-1=12$ and $7-1=6$) divisible by 3, so they both have 2 elements of order 3 plus 1 of order 1 (identity). There are $3 \cdot 3 =9$ ways to combine these elements as ordered pairs. Toss out the identity and get $9-1=8$ elements of order 3.
Jul
8
comment An infinite subset of a countable set is countable
It's finite since $\{0,1,2,\dots,q\}$ has exactly $q+1$ elements (a finite number of elements).
Jun
24
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Jun
20
reviewed Approve Fredholm equation
Jun
17
answered Is $\max(0, x)$ a differentiable function?
Jun
17
revised Why are there infinitely many orthonormal vectors?
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16
revised Why are there infinitely many orthonormal vectors?
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Jun
16
answered Why are there infinitely many orthonormal vectors?
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14
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Jun
7
answered Determine the degree of the splitting field of the polynomial $x^4-2$
Jun
6
comment What is the difference between tensors and tensor products?
Yes.............
Jun
3
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3
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