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The subsets a ring can be quotiented by: absorption rather than mere closure, why every ideal of the integers is the multiples of one number, and the quotient ring that results.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to tell an ideal from a subring by testing absorption, find the single generator of an ideal of the integers and connect it to Bézout's identity, build the quotient ring and count its elements, explain why absorption is what makes coset multiplication well defined, and prove that the kernel of a ring homomorphism is always an ideal.
Unit 4 asked which subgroups a group can be quotiented by, and the answer was the normal ones — the subgroups stable under conjugation, which was exactly what made the cosets multiply. The same question for rings has the same shape of answer, and the condition turns out to be absorption rather than stability.
A subring is a subset closed under subtraction and multiplication. An ideal $I$ is an additive subgroup with $ra \in I$ and $ar \in I$ for every $a \in I$ and every $r$ in the ring. An ideal is principal when it is the multiples of a single element, written $(a)$. A principal ideal domain is an integral domain in which every ideal is principal. The quotient ring $R/I$ has the cosets $a + I$ as its elements.
An ideal $I$ of a ring $R$ is an additive subgroup that absorbs multiplication:
$$a \in I, \ r \in R \implies ra \in I \text{ and } ar \in I.$$
That is strictly stronger than being a subring. The integers sit inside the rationals as a subring — closed under subtraction and multiplication — and are not an ideal, because $\tfrac{1}{2} \times 2$ escapes.
Why the stronger condition. The cosets $a + I$ are to be multiplied by $(a + I)(b + I) = ab + I$, and the check that this does not depend on the representatives needs exactly absorption: if $a' = a + i$ and $b' = b + j$ then
$$a'b' = ab + (aj + ib + ij),$$
and the bracket is in $I$ only because $I$ absorbs multiplication by $a$ and by $b$. So ideals are to rings what normal subgroups are to groups, and for the same reason.
In the integers, every ideal is principal. Take a non-zero ideal $I$, let $d$ be its least positive element, and divide any $a \in I$ by $d$: the remainder $a - qd$ lies in $I$ and is smaller than $d$, so it is $0$. Hence $I = (d)$. It is the division-algorithm argument from the lesson on cyclic groups, word for word with ideal in place of subgroup.
Bézout is the same fact. The ideal generated by $a$ and $b$ is $\{ax + by\}$, which is principal, generated by $\gcd(a, b)$. So the highest common factor is a combination of the two numbers — not by a separate theorem but by the same one.
Two ideals are always there: the zero ideal $\{0\}$ and the whole ring. An ideal containing a unit is the whole ring, since it then contains $u^{-1}u = 1$ and absorbs everything. In particular, a field has only those two ideals.
Another way: picture
Think of an ideal as a black hole inside the ring. Anything in the ring, multiplied by something in the ideal, falls into the ideal and cannot get out. A subring is merely a self-contained smaller ring — things inside stay inside when combined with each other, but multiplying by something outside can lift you out. Only a black hole can be collapsed to a single point without tearing the multiplication.
Another way: steps
To decide whether a subset is an ideal: 1. Is it closed under subtraction, and does it contain $0$? 2. Take a member $a$ and an arbitrary ring element $r$: is $ra$ still inside? And $ar$? 3. If it contains a unit, it is the whole ring. 4. In the integers, find the least positive member: the ideal is its multiples. 5. For two generators, the single generator is their highest common factor.
Given an ideal $I$, the cosets $a + I$ form a ring under
$$(a + I) + (b + I) = (a + b) + I, \qquad (a + I)(b + I) = ab + I.$$
The additive part is the quotient group of unit 4, since $(R, +)$ is abelian and every subgroup of it is normal. The multiplicative part is what needs absorption, and it is why ideal is a stronger condition than additive subgroup.
The projection $\pi : R \to R/I$, $\pi(a) = a + I$, is a surjective ring homomorphism with kernel $I$. Together with this lesson's last item — every kernel is an ideal — that gives the same correspondence as before: the ideals of $R$ are exactly the kernels of ring homomorphisms out of $R$.
The first isomorphism theorem transfers unchanged: $R/\ker\phi \cong \operatorname{im}\phi$, with the same four checks. The most familiar instance is the one every schoolchild meets: reduction modulo $n$ is a surjective ring homomorphism $\mathbb{Z} \to \mathbb{Z}_n$ with kernel $(n)$, so
$$\mathbb{Z}/(n) \cong \mathbb{Z}_n.$$
Modular arithmetic is a quotient ring, and always was.
One caution about ideals in non-commutative rings: absorption on the left and on the right are different conditions, giving left ideals and right ideals, and only a two-sided ideal produces a quotient ring. Every ring in this course is commutative, where the distinction collapses — but the words exist for a reason.
Checking closure under the subset's own multiplication. That is the subring condition. An ideal must survive multiplication by everything in the ring.
Forgetting that an ideal containing $1$ is everything. More generally, an ideal containing any unit is the whole ring — which is why a field has only the two trivial ideals and why fields have no interesting quotients.
Expecting an ideal to contain $1$. Most do not. The multiples of $5$ form an ideal of the integers and contain no unit at all, so an ideal is usually not a ring with unity in its own right.
Thinking every ring has only principal ideals. The integers and $F[x]$ do; $\mathbb{Z}[x]$ does not, since the ideal generated by $2$ and $x$ needs both generators.
Quotienting by a subring. There is no ring there. The cosets can be added but not multiplied.
In the integers, take the ideal generated by $12$ and $18$. It contains every $12x + 18y$.
All combinations.
The least positive such combination is $\gcd(12, 18) = 6$, reached by $12 \times (-1) + 18 \times 1$.
Bézout.
So the ideal is the multiples of $6$, and the two generators collapse to one. The quotient has six elements.
Principal after all.
Inside $\mathbb{Q}[x]$, take the constant polynomials. Closed under subtraction and under multiplication by each other, so a subring.
A subring.
But $x \times 3 = 3x$, which is not constant.
Absorption fails.
So it is not an ideal, and there is no quotient ring by it. By contrast the polynomials with zero constant term are an ideal, and the quotient is $\mathbb{Q}$.
One subset each way.
Closed under subtraction: a difference of two multiples of $x^{2} + 1$ is a multiple of it.
The additive part.
Absorption: any polynomial times a multiple of $x^{2} + 1$ is again a multiple of it.
The multiplicative part.
So it is an ideal — the principal ideal generated by $x^{2} + 1$. The quotient turns out to be the complex numbers, which is the last unit's construction two lessons early.
Select every subset that is an ideal of the ring it sits in.
This task has no paper form; do it on a device.
In the integers, the ideal generated by $7$ and $17$ is in fact generated by a single number. Which?
Answer:
Quotient the integers by each of these ideals. Say how many elements the quotient ring has.
| Elements in the quotient | |
|---|---|
| the multiples of 8 | |
| the multiples of 16 | |
| the multiples of 1 |
Match each ideal of the integers to the quotient ring it produces.
| the integers modulo $2$ | the integers modulo $5$ | the integers modulo $6$ | the integers | the zero ring | |
|---|---|---|---|---|---|
| the multiples of $2$ | |||||
| the multiples of $5$ | |||||
| the multiples of $6$ | |||||
| the zero ideal |
Is the integers, inside the rationals an ideal, or only a subring?
Build the proof that the kernel of a ring homomorphism is an ideal.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Quotient the integers by each of these ideals. Say how many elements the quotient ring has.
| Elements in the quotient | |
|---|---|
| the multiples of 3 | |
| the multiples of 6 | |
| the multiples of 1 |
You can decide whether a subset is an ideal and describe the quotient ring it gives. Say in your own words why being a subring is not enough, and give an example of a subring that is not an ideal. Next: which ideals give a domain and which give a field.
9. Your turn: is the set of polynomials divisible by $x^{2} + 1$ an ideal of the polynomials with rational coefficients?, step 3