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I have to determine the Jacobson radical of this matrix ring: $\begin{pmatrix} \mathbb{Z}_{63} & \mathbb{Z}_{63}\\ 0& \mathbb{Z}_{63} \end{pmatrix}$

I have done the following $(a,b,c,r,s,t \in \mathbb{Z}_{63})$:

$$\begin{pmatrix} 1 & 0\\ 0& 1 \end{pmatrix}- \begin{pmatrix} r & s\\ 0& t \end{pmatrix}\begin{pmatrix} a & b\\ 0& c \end{pmatrix}= \begin{pmatrix} 1-ra & -(rb+sc)\\ 0& 1-tc \end{pmatrix}$$

Then I have: $$\begin{pmatrix} 1-ra & -(rb+sc)\\ 0& 1-tc \end{pmatrix}^{\!\!-1}= \frac{1}{1-tc-ra+ratc}\begin{pmatrix} 1-tc & rb + sc\\ 0 & 1-ra \end{pmatrix}.$$

What should be my next step?

user26857
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Dazzler95
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1 Answers1

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It's quite a bit easier to deduce the elements in another way.

First of all, notice that $I=\begin{bmatrix}0&\mathbb {Z}_{63}\\0&0\end{bmatrix}$ is a nilpotent ideal, so it is contained in the Jacobson radical. Finding the maximal ideals of the ring is therefore equivalent to finding the maximal ideals of $R/I\cong \mathbb Z_{63}\times \mathbb Z_{63}$.

But this is easy, right? The maximal ideals in $\mathbb Z_{63}$ are $(3)$ and $(7)$, so the maximal ideals of the product are $(3)\times \mathbb Z_{63}$, $(7)\times \mathbb Z_{63}$ and $\mathbb Z_{63}\times (3)$ and $\mathbb Z_{63}\times (7)$. This makes the radical of the product ring $\mathbb (21)\times (21)$.

Lifting this back up to the original ring, you have

$$ \begin{bmatrix} (21)&\mathbb Z_{63}\\0&(21)\end{bmatrix} $$

You can find similar descriptions of what's going on here:

https://math.stackexchange.com/a/1050265/29335

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rschwieb
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  • Are nilpotent ideals automatically contained in the Jacobson radical? I can see how I is nilpotent and I can also see how no other ideal would contain it. I don't see why you involve the factor ring. I understand that the maximal ideals of $\mathbb{Z}$ are the prime ideals and that the prime factors of $63$ are $3^{2}\times 7$ and that $21$ would be the lowest common multiple. Does that mean that $21\mathbb{Z}$ is then the intersection of all the maximal ideals? I don't understand that other question because my course on Ring Theory didn't mention modules at all. – Dazzler95 Nov 23 '16 at 20:31
  • @Dazzler95 Nilpotent ideals are contained in every prime ideal, and hence in every maximal ideal. So they would have to lie in the intersection of all maximal ideals (because they lie in the intersection of all prime ideals). It's just an easy consequence of $I^n={0}\subseteq P\implies I\subseteq P$. – rschwieb Nov 23 '16 at 21:00
  • @Dazzler95 The maximal ideals of $R/I$ correspond to the maximal ideals of $R$ containing $I$: but what I'm saying is that all of the maximal ideals of $R$ contain $I$. – rschwieb Nov 23 '16 at 21:01
  • @Dazzler95 Yes, in the integers the intersection of ideals corresponds with the ideal generated by the product of the generators. – rschwieb Nov 23 '16 at 21:02
  • OK, thanks for that. So the Jacobson radical of the ring $R$ would simply be \begin{pmatrix} 21\mathbb{Z} & \mathbb{Z}_{63}\ 0 & 21\mathbb{Z} \end{pmatrix} – Dazzler95 Nov 23 '16 at 21:07
  • @Dazzler95 That's right, just different notation. – rschwieb Nov 23 '16 at 21:23