Local rings

Posted: August 2, 2022 in Local & Semilocal Rings, Noncommutative Ring Theory Notes
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All rings in this post are assumed to have 1 and they may or may not be commutative. As always, for a ring R, we denote by J(R), U(R) the Jacobson radical and the group of units of R, respectively. Also, M_n(R) is the ring of n \times n matrices with entries from R.

Definition. A ring R is said to be local if R has only one maximal left ideal.

So division rings are obviously local and a commutative ring is local if and only if it has only one maximal ideal. We have seen commutative local rings in this blog several times. It is easy to find examples of commutative local rings. For example, if k is a field, then the ring of formal power series k[[x]] is a local ring with the unique maximal ideal \langle x \rangle. We can also use localization to generate local rings; pick any commutative ring R and any prime ideal P of R; then the localization S^{-1}R, where S:=R \setminus P, is a commutative local ring with the unique maximal ideal S^{-1}P. But this straightforward construction of local rings does not work for noncommutative rings.

Let’s begin with proving a few statements equivalent to the definition of local rings.

Theorem 1. For a ring R, let I:=R \setminus U(R). The following statements are equivalent.

i) R is local.

ii) R/J(R) is a division ring.

iii) I=J(R)

iv) I is an ideal of R.

Proof. i) \implies ii). Let \mathfrak{m} be the unique maximal left ideal of R. Then, by definition, J(R)=\mathfrak{m}. Let x +\mathfrak{m} be a non-zero element of R/\mathfrak{m}. So x \notin \mathfrak{m} and hence Rx+\mathfrak{m}=R, which gives 1-rx \in \mathfrak{m} for some r \in R. Thus r+\mathfrak{m} is the (left) inverse of x+\mathfrak{m} and hence R/\mathfrak{m} is a division ring.

ii) \implies iii). Clarealy J(R) \subseteq I. Now let x \notin J(R). Then x+J(R) is a non-zero element of R/J(R) and thus, since R/J(R) is a division ring, x+J(R) is invertible. So x \in U(R), by Problem 1, i), in this post.

iii) \implies iv). Trivial.

iv) \implies i). Let \mathfrak{m} be a maximal left ideal of R. It is clear that \mathfrak{m} \subseteq I. Since I is a proper ideal, it is also a proper left ideal and so \mathfrak{m}=I. Thus I is the unique maximal left ideal of R. \ \Box

Remark. If R \setminus U(R) \subseteq I for some proper ideal of R, then R \setminus U(R)=I and so, by Theorem 1, iv), R is local.

Example. Let k be a field of characteristic \ne 2. By Theorem 1 and part ix) of the Theorem in this post, the exterior algebra \Lambda_n(k) is a local ring.

Note. The next theorem is Corollary 19.19 in T.Y. Lam’s book A First Course in Noncommutative Rings. However, my proof is completely different from the proof given in that book.

Recall that an element x in a ring is called an idempotent if x^2=x. An idempotent x is called non-trivial if x \ne 0, 1.

Theorem 2. i) If R is a local ring, then R has no non-trivial idempotent.

ii) A left Artinian ring R is local if and only if R has no non-trivial idempotent.

Proof (Y. Sharifi). i) Let x \in R be an idempotent. So x(1-x)=0 hence if x \in U(R), then x=1 and if 1-x \in U(R), then x=0. If x \notin U(R) and 1-x \notin U(R), then, by Theorem 1, iv), 1=x+1-x \notin U(R), which is nonsense. So x= 0,1 are the only idempotents of R.

ii) One side is clear by i). Conversely, suppose that R has no non-trivial idempotent. By the Example in this post, every idempotent of R/J(R) can be lifted and hence R/J(R) has no non-trivial idempotent. On the other hand, by the Artin-Wedderburn theorem,

\displaystyle R/J(R) \cong \prod_{i=1}^kM_{n_i}(D_i), \ \ \ \ \ \ \ \ \ (*)

for some positive integers k, n_i, \ 1 \le i \le k, and some division rings D_i. It is clear that \prod_{i=1}^k D_i, \ k \ge 2, has non-trivial idempotents. Also each M_{n_i}(D_i), \ n_i \ge 2, has non-trivial idempotents, e.g. the matrix with the (1,1)-entry 1 and all the other entries 0. Thus, since R/J(R) has no non-trivial idempotent, (*) gives k=n_1=1 and hence R/J(R) \cong D_1 is a division ring. Thus, by Theorem 1, ii), R is local. \ \Box

Exercise 1. Show that every local ring is Dedekind-finite.
Hint. Use Theorem 1, ii), and Example 8 in this post.

Exercise 2. Show that a ring is local if and only if it has only one maximal right ideal.

Exercise 3. Give an example of a ring R which has only one maximal ideal but R is not local.
Hint. Try R=M_n(\mathbb{C}), \ n \ge 2.

Exercise 4. Show that a ring R is local if and only if the following condition is satisfied:

\forall x,y \in R: \ x+y \in U(R) \implies x \in U(R) \ \text{or} \ y \in U(R).

Exercise 5. Let R be a ring such that |R| \ge 3 and |U(R)| =1. Show that R is not local. Conclude that if R is a Boolean ring with identity, then R is local if and only if R \cong \mathbb{Z}_2.
Hint. If x \in R \setminus \{0,1\}, then x, 1-x \notin U(R). Now use Theorem 1, iv).

Exercise 6. Let R be a ring. Show that

i) if I is a nil left ideal of R, then I \subseteq J(R),

ii) if every element of R is either nilpotent or a unit, then R is local.

iii) the converse of ii) is not true.
Hint. i) If x \in I, \ r \in R, then rx \in I and so rx is nilpotent. Hence 1-rx \in U(R) giving x \in J(R).
ii) By the Remark following Theorem 1, we only need to show that R \setminus U(R) \subseteq J(R). Let x \in R \setminus U(R). So x is nilpotent and hence rx \notin U(R) for all r \in R. Thus Rx is a nil left ideal of R. Now use i).
iii) The ring of formal power series k[[x]] over a field k is a counter-example.

Exercise 7. Show that a von Neumann regular ring R is local if and only if R is a division ring.
Hint. Suppose that R is local, and let 0 \ne a \in R. By definition, a=axa, for some x \in R, and therefore (ax)^2=ax. Now use Theorem 2, i), to finish the proof.

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