I have to know the Jacobson radicals of the rings $$ R=\left\{\left [\begin{array}\ \mathbb x & \mathbb y\\ 0 & \mathbb x \end{array} \right ] \mid x,y\in \mathbb Z_2\right\}$$ and $$S= \left\{\left [\begin{array}\ \mathbb x & \mathbb y\\ 0 & \mathbb x \end{array} \right ]\mid x\in \mathbb Z_4, y\in \mathbb Z_4 \oplus \mathbb Z_4\right\}.$$ I know that the Jacobson radical of $\left [\begin{array}\ \mathbb Z_2 & \mathbb Z_2\\ 0 & \mathbb Z_2 \end{array} \right ]$ is $\left [\begin{array}\ \mathbb 0 & \mathbb Z_2\\ 0 & \mathbb 0 \end{array} \right ]$, and this contains $J(R)$. Similarly, $J(S)$ is contained in $\left [\begin{array}\ 2\mathbb Z_4 & \mathbb Z_4\oplus\mathbb Z_4\\ 0 & 2\mathbb Z_4 \end{array} \right ]$. Any suggestion?

  • $\begingroup$ Btw, all of the links I mentioned to your earlier question explain this. Please consider reading them sometime! $\endgroup$ – rschwieb Mar 3 '17 at 14:31
  • $\begingroup$ @rschwieb: What I tried in this question was based on the "links" you mentioned. Thank You! $\endgroup$ – karparvar Mar 3 '17 at 16:51
  • $\begingroup$ Hm, well I thought that in those links I mention the technique of confirming that your set contains the radical by showing the quotient is semiprimitive. (This is the same technique Marc Bogaerts is suggesting.) Yes, this indeed happens at least one place: math.stackexchange.com/a/1050265/29335 in step 4. $\endgroup$ – rschwieb Mar 3 '17 at 17:27
  • $\begingroup$ It's OK don't take me too seriously. I myself miss stuff on the first three times around, at least. $\endgroup$ – rschwieb Mar 3 '17 at 17:30
  • $\begingroup$ @rschwieb: But, what I deduced concerning the Jacobson radicals of the two rings was an immediate consequence of the fact that $J(A)$ is contained in $J(B)$ whenever $A$ is a subring of $B$. And I wanted to know $J(A)$ in the two cases! $\endgroup$ – karparvar Mar 3 '17 at 18:04

For both rings, if you take the quotient by the ideals you already found you obtain a ring of scalar matrices, thus a field. The intersectin of all the maximal subspaces is in both cases the zero ideal, so the inverse images of the zero ideal under the projections are the intersections of the maximal ideals, i.e. the ideals you already found.

  • $\begingroup$ Well said! (Autocorrect borked my earlier comment, I just now noticed.) $\endgroup$ – rschwieb Mar 3 '17 at 14:30
  • $\begingroup$ I understood something in the sense of enough said for one week :-] $\endgroup$ – Marc Bogaerts Mar 3 '17 at 14:53

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