Determine if a quotient ring meets various definitions

Question

Consider the ring $R=\mathbb{Z}/14\mathbb{Z}$. Give justified answers to the following:

1. List the zero-divisors of $R$.
2. List the elements of $R$ that have a multiplicative inverse.
3. List the subrings of $R$.
4. Is $R$ an integral domain?
5. Is $I=\{0,2,4,6,8,10,12\}$ an ideal of $R$?
6. Does $I$ have a multiplicative identity element?
7. Find and ideal of $R$ having only two elements.

Solution

1. An element $a$ of a ring $R$ if $a\ne0$ there exist $b\in R$, $b\ne0$ such that either $ab=0$ or $ba=0$. Our zero-divisors are $2,4,6,7,8,10,12$. As

2. The elements of $R$ that have a multiplicative inverse are $\{1,3,5,9,11,13\}$. An inverse is defined as $x\times x^{-1}=e$, where $e$ is the identity for multiplication, which is $1$.

3. If S is a subring of $R$ it contains e and hence also contains $e,2e,3e,...$ However these elements are $1+14Z,2+14Z,3+14Z....$ In other words $S$ contains all the elements of $R$ and so $S=R$.

4. Our ring is not an integral domain as we have zero-divisors.

5. The number of elements of an ideal must divide 14. So our ideal must have 2 or 7 elements. I has 7 elements and can be generated by 2, making it a cyclic group. I is an ideal of R.

6. No it does not, the multiplicative identity of $R$ is $1$.

7. The subring of order $2$ is ${0,7}$.

What mistakes have I made for all parts?

Answer 1

1. Correct. Do you have a proof as well?
2. Incorrect; there are more units. In particular, the multiplicative identity is always a unit.

A nice general fact about finite rings, which you may be able to prove for yourself, is the following:

In a finite unital ring, every nonzero element is either a zero divisor or a unit.

3. Big hint: A subring must contain $1$ by definition, and hence also $1+1$, and $1+1+1$, etc.
4. Correct.
5. Your reasoning fails since you have updated your answer to part 1; reconsider your answer.
6. Correct.
7. Incorrect; this set is not closed under addition, and not closed under multiplication by elements of $R$. Big hint: For each element $x$ of the ideal, the set $$xR=\{0,x,2x,\ldots,13x\},$$ is a subset of the ideal, and hence consists of at most two elements.