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Consider the following condition on a ring:

For every nonzero $a$, there is a nonunit $b$ with the property that $a+b$ is a unit.

Observe that if $a$ is already a unit, then $b=0$ will do just fine. So the content of this condition is what it says about the nonzero nonunits of the ring.

Some rings satisfying the condition (I will write $\mathbf{1}$ for the multiplicative identity element of each).

  • For any $n \geq 1$ and field $F$, the polynomial ring $F[x_1, \dots, x_n]$.

  • For any $n \geq 1$ and field $F$, the ring $M_n(F)$ of $n \times n$ matrices over $F$.

    Sketch of proof: Fix a nonzero $a \in M_n(F)$ and regard it as a linear transformation $T: F^n \to F^n$ in the obvious way. It is possible to find new bases $B$ and $B'$ of $F^n$ so that the matrix of $T$ with respect to these bases is a nonzero diagonal $\{0,1\}$ matrix $D$. (Extend a basis for $\ker T$ to a basis $B$ for $F^n$ by adding a list of vectors $B_0$; then note that $\{T(v): v \in B_0\}$ is a linearly independent subset of $F^n$, which you can extend to a basis $B'$ of $F^n$.) This implies the existence of invertible $U$ and $V$ in $M_n(F)$ with $a = UDV$. Take $b = U(\mathbf{1} - D)V$.

  • Any $C^*$-algebra $A$.

    Proof: fix a nonzero $a \in A$. If the spectrum $\sigma(a)$ of $a$ contains a nonzero element $\lambda$ then we can take $b = \lambda \mathbf{1} - a$, so assume $\sigma(a) = \{0\}$. As $a$ is nonzero, one of $a + a^*$ and $a - a^*$ is nonzero; by replacing $a$ with $ia$ if necessary we may assume $a + a^*$ is nonzero. Since the spectral radius of a self-adjoint element coincides with its norm, there is a nonzero number $\lambda$ in $\sigma(a + a^*)$. Take $b = \lambda \mathbf{1} - (a + a^*)$. (The element $a + b = \lambda \mathbf{1} - a^*$ is invertible because $\lambda$ is nonzero and $\sigma(a^*) = \{\overline{z}: z \in \sigma(a)\} = \{0\}$.)

Some rings that do not satisfy the condition:

  • The ring of formal power series $F[[x_1, \dots, x_n]]$ (where $F$ is a field and $n \geq 1$).

  • More generally, any local ring.

  • For any $n > 1$ and field $F$, the ring of upper-triangular elements of $M_n(F)$.

  • Any commutative subalgebra of $\mathcal{B}(H)$ (the bounded operators on a Hilbert space $H$) containing a nonzero operator $a$ with $\sigma(a) = \{0\}$. (Due, e.g., to the nontrivial fact that if $b$ commutes with $a$ and $\sigma(a) = \{0\}$ then $\sigma(a + b) = \sigma(b)$.)

I don't know much ring theory. Is there a specific name for rings satisfying this condition?

If the answer to that is straightforward, here is a vaguer question: what does this condition "say" about a ring? Are there more familiar ring-theoretic properties it is related to? (For example: if you knew a lot of ring theory, and someone gave you a ring, how would you check whether or not it satisfied this condition?)

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Isn't it the case that the ring can have no Jacobson radical (assuming the usual chain conditions on ideals). If $a \in J(R)$ and $ a+b = u$ for some non-unit $b$ and unit $u,$ then $b = -a + u.$ However, if $uv = vu = 1,$ then $vb = 1-va$ which is unit as $va$ is nilpotent. So te property seems closely related to semisimplicity, especialy in view of your examples. –  Geoff Robinson Feb 29 '12 at 0:11
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$\rm b$ nonunit $\rm \Rightarrow b\in M\ max$. So $\rm\: 0\ne a\in J(R)\subset M\Rightarrow a+b\in M$ so nonunit $\Rightarrow\Leftarrow\:$ So $\rm\: J(R)=0$. $\quad $ –  Bill Dubuque Feb 29 '12 at 1:12
    
@GeoffRobinson and BillDubuque, thanks for the comments. (The more I read about the structure of abstract rings, the more it seems that the rings I am professionally most familiar with have almost no structure at all...) –  leslie townes Mar 2 '12 at 23:15
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@BillDubuque What is $\mathrm M$ in your comment? And what does $\mathrm M\; \max$ mean? (I must admit that I sometimes struggle to understand your notation.) –  user23211 May 11 '12 at 19:54
    
@ymar $\:$ Nonunit $\rm\:b\:\Rightarrow\:(b)\ne (1)\:\Rightarrow\:(b)\subset M\subsetneq (1)\:$ for some maximal ideal $\rm\:M.$ See also the theorem here on the Jacobson radical. –  Bill Dubuque May 11 '12 at 20:28

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