Pathologies in "rng"

Solution 1:

Note- In the following list, $R$ or $R_i$ will denote a "rng" and $S$ or $S_i $ will denote its "subrng"

• Every Ideal is a subrng.
• $S$ may not have unity even if $R$ has.

Example- $R=\Bbb{Z}$ and $S=2\Bbb{Z}$

• $S$ has a unity where $R$ does not.

Example- $R= \Bbb{Z} \times \Bbb{2Z}$ and $S=\Bbb{Z}\times 0$

• $S$ and $R$ may both have unity but they may not be same.

Example- $R$ = $\Bbb{Z_6}$ and $S= \{\bar{0}, \bar{2}, \bar{4}\}$ where $1_R=\bar{1}\ \text{and }\ 1_S=\bar{4}$.

• If $R$ and $S$ have different unities, then $a\in S$ may be a unit in $S$ but not in $R$

Example- $R$ = $\Bbb{Z_6}$ and $S= \{\bar{0}, \bar{2}, \bar{4}\}$ then $a=\bar{2}$ is a unit in $S$ but not in $R$

• $\text{Char}{(S)}< \text{Char}{(R)}$

Example- The above example works as $char(S)=3$ and $char(R)=6$

• $a\in S $ may be a zero-divisor in $R$ but not in $S$.

Example- $R=\Bbb{Z}$ and $S=\{0\}$

• Direct Product of "rngs" $R_1$ and $R_2$, i.e. $D=R_1 \times R_2$ is a ring iff both $R_1$ and $R_2$ are rings.
  In this case $1_D=(1_{R_1},1_{R_2}) $

• Let $M=M_n(R)$, then there is no one-one correspondence between $2$- sided ideals of $R$ and $M$.

Example- Let $R=2\Bbb{Z}$, then $M=M_2{(2\Bbb{Z})}$. Then ideal $$J=\{\begin {pmatrix} a & b \\ c & d \end {pmatrix}\ |\ a,b,c,d \in 2\Bbb{Z} \text{with}\ a \in 4\Bbb{Z} \}$$.
Now it is easy to see that $J$ is an ideal strictly between $M_2{(4\Bbb{Z})}$ and $M_2(\Bbb{2Z})$ but there is no ideal between $\Bbb{4Z}$ and $2\Bbb{Z}$ in $\Bbb{Z}$.

• If $a\in R$ is a zero-divisor in $R$ then $a+I$ need not be a zero-divisor in $R/I$.

Example- $\bar{3}$ is a zero divisor in $\Bbb{Z_6}$ but not in $\Bbb{Z_6}/I$ where $I=\{\bar{0}, \bar{2}, \bar{4}\}$

• If $f$ is a "rng" homomorphism, then $f(1_R)$ need not equal $f(1_{R'})$

Example- Inclusion homomorphism $\{\bar{0}, \bar{2}, \bar{4}\} \hookrightarrow \Bbb{Z}_6$.

• Jacobson radical for a "rng" is defined as, $$\text{rad}(R)= \{a \in R :\ Ra\ \text{is left quasi-regular}\}$$

• Given an abelian group $(G,+)$ we can always make it into a "rng" by defining trivial multiplication i.e. $a.b=0$ for all $a,b\in G$, but that is not always true for non-finitely generated abelian groups, as if your group has the property that every element has finite order, but there is no upper bound on the orders of the elements, then it cannot be the additive abelian group of a ring with identity. The reason is that if there were such a ring structure with an identity $1$, then $1$ would have finite additive order $k$, and then for all $a$ in your group, $k\cdot a=(k\cdot1)a=0a=0$, which forces $a$ to have order at most $k$.

Example- Prüfer $p$-group $\mathbb Z(p^\infty)$ or the quotient group $\mathbb Q/\mathbb Z$.

Courtesy- Jonas Meyer's answer

Edit on 19/08/15-
For a commutative "rng", it is not necessary that every maximal ideal is prime, as $4\Bbb{Z}$ is maximal in $2\Bbb{Z}$ but not prime as $2.2\in 4\Bbb{Z}$ but $2\notin 4\Bbb{Z}$.

Solution 2:

I will add a little pathological thing to your list. Suppose $R$ is a commutative ring which may or may not have a unital element. Take $a\in R$. Then one may ask what is the ideal $(a)$?

A first answer is the following :

$$(a)=aR:=\{ar\mid r\in R\} $$

But if one does this, it may happen that $a\notin (a)$. For instance take the unital ring $A:=\mathbb{Z}[X]$ and define $R:=(X)=\{P\in A|P(0)=0\}$. Then $R$ is a ring without unity. Now take $a:=2X$ then :

$$(a)=\{2XP(X)|P\in R\} $$

I claim that $2X\notin (2X)$ (the valuation of any non-zero element of $(2X)$ is $>1$).

Hence one sees that $aR$ is not the smallest ideal containing $a$.

A second answer is the following :

$$(a)=\langle a,ar\mid r\in R\rangle $$

That is the additive group generated by $a$ and all elements $ar$ where $r\in R$. I claim that this is the smallest ideal containing $a$ but in this case $(a)\neq aR$ which is (IMHO) very disturbing...

Solution 3:

It seems like no one has mentioned the pathology that first comes to my mind for rings without identity:

A ring without identity may not have maximal (left/right/two-sided) ideals.

The fact that a ring with identity must have maximal ideals is a sort of very weak finiteness condition. Of course, it is not a spectacular finiteness condition since there are rings without identity that have maximal ideals anyway (even finite rings without identity) but still it is a special condition.

Example- $(\Bbb{Q},+,\circ)$ where $\circ$ is trivial multiplication has no maximal ideals as ideals in this rng are same as subgroups of $(\Bbb{Q},+)$ and we can prove that $(\Bbb{Q},+)$ has no maximal subgroups.