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39

Let $A$ be a finite commutative ring (not assumed to contain an identity). Suppose that $a \in A$ is not a zero-divisor. Then multiplication by $a$ induces an injection from $A$ to itself, which is necessarily a bijection, since $A$ is finite. Thus multiplication by $a$ is a permutation of the finite set $A$, and hence multiplication by some power of $a$ (...

39

Let $R$ be a ring, not necessarily with identity, not necessarily commutative. An ideal $\mathfrak{P}$ of $R$ is said to be a prime ideal if and only if $\mathfrak{P}\neq R$, and whenever $\mathfrak{A}$ and $\mathfrak{B}$ are ideals of $R$, then $\mathfrak{AB}\subseteq \mathfrak{P}$ implies $\mathfrak{A}\subseteq \mathfrak{P}$ or $\mathfrak{B}\subseteq \... 29 Many authors take the existence of$1$as part of the definition of a ring. In fact, I would disagree with Alessandro's comment and claim that most authors take the existence of$1$to be part of the definition of a ring. There is another object, often called a rng (pronounced "rung"), which is defined by taking all the axioms that define a ring except you ... 20 The "right" notion of a substructure of an algebraic gadget is the kernel of a homomorphism. For abelian groups, and more generally modules, these are subgroups, respectively submodules. For groups, we need normal subgroups. For rings, we need ideals. 19 Any ring$R$is a module over itself, in the obvious manner:$R$is an abelian group, and we define$a\cdot b$for$a\in R$and$b\in R$to be$ab$. A submodule of$R$-as-an-$R$-module is precisely an ideal of$R$(work out the relevant definitions to see this). Thus, the definition of a Noetherian ring is really saying that it is a Noetherian module over ... 16 One general source of such examples is functional analysis. One of the easiest examples to describe is the space$C_{0}{(X)}$of functions vanishing at infinity, where$X$is locally compact, with pointwise addition and multiplication as operations. This ring is commutative and it is unital if and only if$X$is compact. Another class of examples is ... 16 Let$a$be an element of$A$. Then$a$can be expressed as product of two elements, each of which can be expressed as a product of two elements, and so on forever. By the finiteness of$A$, among these expressions for$a$, some$y\in A$appears to arbitrarily high powers. Again by finiteness, we have$y^i=y^j$for some positive integers$i<j$. Take a ... 16 As Thomas points out,$2\mathbb Z$is not a "ring", since it does not contain any identity element$1.$It is true that every maximal ideal of a commutative ring with identity is prime. 15 You don't need an identity. Let$P$be a prime ideal and let$I$be an ideal which strictly contains$P$. I will show that$I$is the whole ring. Let$i \in I - P$. For any element$r$in the ring, $$i(r-ri) = ri-ri^2 = ri - ri = 0 \in P.$$ Since$P$is a prime ideal, this implies that$r-ri$is in$P$. Since$ri \in I$, this shows that$r = (r-ri)+ri \...

15

Non-unital rings are employed heavily in the general study of radical theories for rings. Perhaps you will find the following remarks of interest, excerpted from the preface of Gardner and Wiegandt: Radical Theory of Rings, 2004. Some authors deal exclusively with rings with unity element. This assumption is all right and not restrictive, if the ring is ...

14

You won't find any examples of maximal, non-prime ideals other than those given in Arturo Magidin's lovely answer. I won't even assume commutativity. And I freely admit that this is basically the same argument as in Arturo's answer! Claim: If $R$ is a rng with a maximal ideal $M$ that is not prime, then $R^2 \subseteq M$. Proof: Let $M$ be such an ideal, ...

13

Any ideal in a ring is itself a ring (but generally without unity, unless it's the full ring). So, there are plenty of examples.

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Take $R=2\mathbb{Z}$, the ring of even integers. The ideal $4\mathbb{Z}\subset R$ is maximal (the only larger ideal is $R$ itself), but not prime, as $2\cdot 2\in 4\mathbb{Z}$, but $2\not\in 4\mathbb{Z}$.

12

Assume that $n=nk$ for some $k\in R$. Because $R$ has no unity, there exists an element $r\in R$ such that $kr\neq r$. In other words $kr-r\neq0$. But by standard applications of rng axioms $$n(kr-r)=n(kr)-nr=(nk)r-nr=nr-nr=0.$$ Therefore $n$ is a zero-divisor.

11

No. Consider the ideal generated by $2$ in $\mathbb{Z}/4\mathbb{Z}.$

11

Proceed like this $a0 = a(0+0)= a0 + a0$, property of $0$ and distributivity. Thus $a0+ (-a0) = (a0 + a0) +(-a0)$, using existence of additive inverse. Finally $0 = a0$ by associativity and properties of additive inverse. Your lemma is also true, you can now prove it easily: Just note that $ab +a(-b)= a(b + (-b))= a0= 0$.

10

Well, one interesting fact about the dual numbers of $\mathbb{R}$: consider its polynomial ring, and specifically identify an object $f(x) = \sum_{i=0}^n a_ix^i , a_i \in \mathbb{R}[\epsilon]/\epsilon^2$. Now evaluating $f(a + b\epsilon), a,b \in \mathbb{R}$ will yield $f(a) + bf'(a)\epsilon$ (hint: binomial theorem) which allows for automatic ...

10

If $R$ is a commutative ring such that $R^2=R$ (in particular if R has an identity), then every maximal ideal is prime.

10

Not every non-unital ring (or rng) has a maximal ideal. For example take $(\mathbb{Q},+)$ with trivial multiplication, i.e. $xy=0$ for all $x,y\in \mathbb{Q}$, then a maximal ideal is nothing more than a maximal subgroup. See this question why such a group does not exist.

9

The usual fashion nowadays is to build the existence of a multiplicative identity into the definition of commutative ring. However, the stated result is correct even if one does not. This is not because the existence of a multiplicative identity follows from finiteness. As already posted examples show, if one does not build a "$1$" into the definition of ...

9

Edit: Undeleted and expanded upon as per my comments. The statement you are trying to prove is only necessarily true for commutative rings. In this case, you can argue that $$I\cap J\subseteq (I\cap J)R=(I\cap J)(I+J)=I(I\cap J)+ J(I\cap J)\subseteq IJ+ IJ=IJ$$ But you could use the stronger step $I\cap J= (I\cap J)R$ , when the ring has 1, this doesn't ...

9

Consider the rng of cofinitely zero infinituples over $C$ or $R$. I.e. the set of all infinituples that are entirely zero after a while. This doesnt have a multiplicative identity but every element can be expressed as a product of rng elements. As for the terminology or References, I remain woefully ignorant.

9

Neither do you need to assume that the ring is unital nor the ring is commutative. Definition: Those rings where for each $x \in R$ there is a positive integer $n(x) >1$ (depending on $x$) s.t. $x^{n(x)}=x,$ are called Jacobson-rings or J-rings. General result due to Jacobson: J-rings are commutative. For the proof, you may take a look at Non-...

9

It can be proven with simple steps which are lemmas that come in handy in other situations: Noetherian Boolean rings have identity. Proof: let $\hat{R}$ be the unitization of $R$, so that $R\lhd \hat{R}$. $R$ is finitely generated in $\hat{R}$ and $R^2=R$, so by Nakayama's Lemma, there exists an $x\in R$ such that $(1-x)R=0$. But this means that $x$ is an ...

9

Proposition: If a rng $R$ which does not have nonzero zero divisors, a nonzero idempotent of $R$ must be an identity for the ring. Proof: Let $e$ be a nonzero idempotent. Since $e(er-r)=0=(re-r)e$ for all $r\in R$ and $e$ is nonzero, we conclude $er-r=0=re-r$, and so $e$ is an identity element.

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I've always heard it said "rung" since that is close to a phonetic attempt. But really, in most serious stuff, I think it's preferred to say "in this paper, rings are not assumed to have identity." You might also enjoy knowing that there is a similar thing done for semirings, that is, a structure like a ring, but the requirement for additive inverses is ...

9

One can always find a ring with unity containing a ring (without unity). There are many different ways to do that ; let's look at one of them. Let $R$ denote the ring without unity and consider $A=\mathbb{Z}\times R$ with the canonical addition and the following multiplication $$(m,a)\cdot (n,b)=(mn, na+mb+ab)$$ $A$ with these two operations is a ring and $... 9 I'm currently teaching out of the 4th edition of Stewart's Galois Theory textbook. Stewart defines a ring to be what other authors might call a commutative ring with unity. The reason is simple: in this book, there is not much call for noncommutative rings, nor for rings without unity, and it gets old writing "commutative ring with unity" over and over, when ... 9 In the commutative case (there are probably simple non-commutative counterexamples coming from matrix rings): We say that a commutative rng$R$has property$\mathcal{P}$if, for all nonzero$a\in R$, there is some$b\in R$with$ab\neq 0$. The zero ring has property$\mathcal{P}$, and it has a unit. Let$R\$ be a finite nonzero commutative rng such that ...

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