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Definition: For a ring with unit $R$, Jacobson Radical is defined as the ideal $J(R) = \{r \in R: rM = 0 \}$ where $M$ is simple.

  1. Let $R$ be a ring with $1_R$, $u = a + x $ is a unit and $x \in J(R)$, prove $a$ is a unit.
  1. So I wrote that if $M$ is simple, let $m \in M$. Then $um = (a + x)m = am + xm = am$. So therefore $um = am \iff (u-a)m = 0 \iff u - a \in J(R).$ But this is where I am stuck. To show $a$ is a unit, I have to show $a \not\in J(R)$ at least. Now I think here I have to assume $J(R) \neq R$, then this gives me $u \not\in J(R)$ since $J(R)$ is an ideal of $R$, so therefore $a \not\in J(R)$. But I am still stuck. I tried also using $u$ being a unit to generate $1$, but I can't get any leads.

EDIT: I just realized $J(R)$ can be defined as the intersection of right maximal ideals in $R$ and I can turn the problem into showing $x\not\in J(R) $, but I was only able to show $1-sx$ is non-unit. where $s\in R.$

rschwieb
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Lemon
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2 Answers2

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EDIT: I just realized $J(R)$ can be defined as the intersection of right maximal ideals in $R$ and I can turn the problem into showing $x\not\in J(R) $, but I was only able to show $1-sx$ is non-unit. where $s\in R.$

If you believe the bolded characterization above, AND that its left-hand analogue is the same set, then there is a fast solution (not along the lines you were pursuing.)

Suppose $u=a+x$ as in your hypotheses, and that $a$ is not a unit. Then $a$ is contained in some maximal left ideal or some maximal right ideal, call it $A$. In that case, $x\in A$ also, by the characterizations mentioned above.

But since $A$ is additively closed, that would mean $u\in A$ as well, but when an ideal contains a unit, it is the entire ring. This would contradict the fact that $A$ is proper.

Therefore $a$ must be a unit.

rschwieb
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  • When I wrote the bolded statement, I was thinking about a commutative case. When I tried to justify it for the general case, it falls short. Let $X$ be a simple left module, since $J(R) = \cap \operatorname{ann}{X}$

    then $\operatorname{ann}(X) = {r \in R: rX = 0} = {r : rR/m=0} = {r :r(s + m) = m, s\in R } = {r: rs \in m }$

    But if we aren't in a commutative ring, how do we know $r \in m$?

     – Lemon Apr 11 '21 at 02:29
  • @Hawk what exactly are you trying to justify now? Symmetry of the radical? – rschwieb Apr 11 '21 at 02:51
  • The two equivalent definitions of the radical. – Lemon Apr 11 '21 at 02:52
  • That is, for a $R$ with $1$ and $M$ is a simple right module, $J(R) = {x: Mx = 0 } = \cap m$ where m is some maximal right ideal in $R.$ – Lemon Apr 11 '21 at 02:57
  • @Hawk Usially one resorts to other characterizations to show the symmetry. For example it is explained here – rschwieb Apr 11 '21 at 02:59
  • Why does he have $m =1$? I thought $M$ is a module, what is the $1$ element in a module? – Lemon Apr 11 '21 at 03:41
  • I actually have a very similar active question relating to this, but I think I figured he meant $r = 1$ – Lemon Apr 11 '21 at 03:43
  • @Hawk you’re not seeing that in the case he’s talking about, $M=R/I$ and he really should have written the coset $1+I=m$ – rschwieb Apr 11 '21 at 03:51
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If $a$ is not a unit then $aR$ is a proper right ideal of $R$, so there is a maximal right ideal $m$ with $aR \subset m$. But then $a \in m$, and $x \in J(R) \subset m$ since $J(R)$ is the intersection of all the maximal right ideals, so $u \in m$, but $u$ is a unit, which contradicts $m$ being a maximal ideal.

Flounderer
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  • "If $a$ is not a unit then $aR$ is a proper right ideal of $R$". This is not necessarily true unless you assume commutativity. There are rings in which there are elements $a,b$ such that $ab=1$ and $ba\neq 1$, and in such rings $a$ is a nonunit and is not contained in any proper right ideal but it is contained in a proper left ideal. – rschwieb Apr 09 '21 at 13:26