What can you set to zero?
Suppose you want a new ring in which some element \(a\) becomes zero. What else is forced to become zero along with it?
- Anything times zero is zero, so every multiple \(ra\) must become zero too, for every \(r\) in the ring.
- Zero plus zero is zero, so sums of such things must become zero as well.
So you can never set just one element to zero. You must set a whole collection to zero, and that collection must be closed under addition and must swallow every multiple of its members. In \(\mathbb{Z}\), declaring \(5 = 0\) forces every multiple of 5 to be zero: \(\dots, -10, -5, 0, 5, 10, \dots\). That set is exactly what disappears in \(\mathbb{Z}/5\).
Ideals
Definition
An ideal of a commutative ring \(R\) is a subset \(I\) such that
- \(0 \in I\), and if \(a, b \in I\) then \(a + b \in I\) and \(-a \in I\);
- if \(a \in I\) and \(r \in R\), then \(ra \in I\). (Absorption.)
You met ideals already in Unit 5: every kernel of a homomorphism is an ideal. The name comes from 19th-century number theory, where Kummer and Dedekind introduced "ideal numbers" to repair failures of unique factorisation (Unit 8).
| Ring | Ideal? | Why |
|---|---|---|
| Even numbers in \(\mathbb{Z}\) | Yes | Even + even is even; anything × even is even. |
| Multiples of \(x^2 + 1\) in \(\mathbb{R}[x]\) | Yes | The kernel of \(f \mapsto f(i)\). |
| Polynomials with constant term 0 in \(\mathbb{Z}[x]\) | Yes | The kernel of \(f \mapsto f(0)\); it's all multiples of \(x\). |
| \(\{0\}\) and \(R\) itself | Yes | The two "trivial" ideals in every ring. |
| Odd numbers in \(\mathbb{Z}\) | No | \(1 + 1 = 2\) isn't odd, and 0 isn't odd. |
| \(\mathbb{Z}\) inside \(\mathbb{Q}\) | No | A subring, but \(\tfrac12 \cdot 1 \notin \mathbb{Z}\): it doesn't absorb. |
The last example shows the difference between a subring and an ideal. A subring is closed under multiplying its own elements. An ideal must absorb multiplication by everything in the big ring.
Theorem: units blow ideals up
If an ideal \(I\) contains a unit, then \(I = R\). Consequently, a field has only two ideals: \(\{0\}\) and the whole field.
Show proof
If \(u \in I\) is a unit, then \(1 = u^{-1}u \in I\) by absorption, and then every \(r = r \cdot 1 \in I\). In a field, every non-zero element is a unit, so any ideal other than \(\{0\}\) contains a unit and is the whole field. ∎
In terms of "setting things to zero": if you declare a unit to be zero, then \(1 = u^{-1}u\) becomes zero, and everything collapses to the zero ring. That's why fields can't be partially collapsed, as we saw in Unit 5.
Generators and divisibility
The simplest ideals consist of all multiples of a single element:
Definition
The principal ideal generated by \(a\) is \((a) = \{ra : r \in R\}\). More generally, \((a_1, \dots, a_k) = \{r_1a_1 + \dots + r_ka_k\}\) is the smallest ideal containing \(a_1, \dots, a_k\).
In \(\mathbb{Z}\), what is \((4, 6)\), the set of all \(4x + 6y\)? It contains \(6 - 4 = 2\), and every \(4x + 6y\) is even. So \((4, 6) = (2)\). This is Bézout's identity from Unit 2 in a new language: \((a, b) = (\gcd(a, b))\). In fact every ideal of \(\mathbb{Z}\) is principal: it's \((d)\) where \(d\) is its smallest positive element (Unit 8 proves this).
Containment of principal ideals is divisibility turned upside down:
Here are all the ideals of \(\mathbb{Z}/12\), arranged with bigger ideals higher up. There's one for each divisor of 12:
Quotient rings
Now the machine. Given an ideal \(I\) of \(R\), call two elements congruent modulo \(I\) if their difference lies in \(I\):
\[ a \equiv b \pmod I \iff a - b \in I. \]For \(R = \mathbb{Z}\) and \(I = (n)\), this is ordinary congruence mod \(n\). The set of everything congruent to \(a\) is the coset \(a + I = \{a + i : i \in I\}\): think of it as "\(a\), give or take something in \(I\)". The cosets are the elements of the new ring.
Definition and theorem: the quotient ring
The cosets of \(I\) form a commutative ring \(R/I\) (read "\(R\) mod \(I\)"), with
\[ (a + I) + (b + I) = (a + b) + I, \qquad (a + I)(b + I) = ab + I. \]Its zero is \(0 + I = I\) and its one is \(1 + I\). The map \(R \to R/I\), \(a \mapsto a + I\), is a homomorphism with kernel \(I\).
Show proof that the operations are well defined
This is the same check as for \(\mathbb{Z}/n\) in Unit 2, and it's exactly where absorption is needed. Suppose \(a' = a + i\) and \(b' = b + j\) with \(i, j \in I\). Then \(a' + b' = (a + b) + (i + j)\), and \(i + j \in I\). For products,
\[ a'b' = ab + (aj + ib + ij). \]Each of \(aj\), \(ib\), \(ij\) is something in \(I\) times something in \(R\), so it's in \(I\) by absorption, and so is their sum. So the answer doesn't depend on the representatives. The ring axioms are inherited from \(R\). ∎
Notice how the proof would fail for \(\mathbb{Z}\) inside \(\mathbb{Q}\): multiplying representatives by \(\tfrac12\) could take you out of the "zero" set. Ideals are precisely the subsets for which the quotient construction works.
How to compute in \(R/I\): calculate in \(R\) as usual, but whenever something in \(I\) shows up, you may replace it by 0. In \(\mathbb{Z}/(5)\), replace 5 by 0. In \(\mathbb{R}[x]/(x^2 + 1)\), replace \(x^2 + 1\) by 0, that is, replace \(x^2\) by \(-1\).
Building \(\mathbb{C}\) from polynomials
Take real polynomials and declare \(x^2 + 1 = 0\). What ring do we get?
By the division algorithm, every polynomial \(f\) can be written as \(f = q \cdot (x^2 + 1) + (a + bx)\). The first part is in the ideal, so it vanishes: every element of \(\mathbb{R}[x]/(x^2 + 1)\) has a unique representative \(a + bx\). Adding is obvious. Multiplying:
\[ (a + bx)(c + dx) = ac + (ad + bc)x + bd\,x^2 = (ac - bd) + (ad + bc)x, \]using \(x^2 = -1\). That's exactly the rule for multiplying complex numbers \((a + bi)(c + di)\). The class of \(x\) behaves exactly like \(i\).
The same trick builds many other rings:
| Quotient | Rule | What it is |
|---|---|---|
| \(\mathbb{Q}[x]/(x^2 - 2)\) | \(x^2 = 2\) | \(\{a + b\sqrt2 : a, b \in \mathbb{Q}\}\), a field |
| \(\mathbb{Z}[x]/(x^2 + 1)\) | \(x^2 = -1\) | the Gaussian integers \(\mathbb{Z}[i]\) |
| \(\mathbb{Z}[x]/(x)\) | \(x = 0\) | \(\mathbb{Z}\) itself: setting \(x = 0\) leaves just the constant term |
| \((\mathbb{Z}/2)[x]/(x^2 + x + 1)\) | \(x^2 = x + 1\) | a field with 4 elements (Unit 10) |
| \(\mathbb{R}[x]/(x^2 - 1)\) | \(x^2 = 1\) | not a domain: \((x - 1)(x + 1) = 0\) |
| \(\mathbb{R}[x]/(x^2)\) | \(x^2 = 0\) | the "dual numbers", with a nilpotent \(\varepsilon = x\) |
Going deeper: dual numbers compute derivatives
In \(\mathbb{R}[\varepsilon]/(\varepsilon^2)\), expand \((a + \varepsilon)^n\) with the binomial theorem. Every term with \(\varepsilon^2\) or higher vanishes, leaving \(a^n + na^{n-1}\varepsilon\). The coefficient of \(\varepsilon\) is the derivative of \(x^n\) at \(a\)! For any polynomial \(f\),
\[ f(a + \varepsilon) = f(a) + f'(a)\,\varepsilon. \]For example \((3 + \varepsilon)^3 = 27 + 27\varepsilon\), and indeed \(\frac{d}{dx}x^3 = 3x^2 = 27\) at \(x = 3\). This is the idea behind forward-mode automatic differentiation, used by software that trains neural networks: compute with dual numbers and the derivatives come out for free, exactly, with no approximation.
A picture of a quotient
The Gaussian integers \(\mathbb{Z}[i] = \{a + bi\}\) form a square grid of points in the plane. Here is the ideal \((2 + i)\), all multiples of \(2 + i\), drawn in large purple dots. Multiplying by \(2 + i\) rotates and enlarges the grid (as in Unit 5's picture), so the ideal is a tilted square grid with side \(\sqrt5\).
So \(\mathbb{Z}[i]/(2 + i)\) has exactly 5 elements. Which ring is it? In the quotient, \(2 + i = 0\), so \(i = -2 = 3\). Then \(i^2 = 9 = 4 = -1\) mod 5, consistent with \(i^2 = -1\). Sending \(a + bi \mapsto a + 3b \bmod 5\) identifies the quotient with \(\mathbb{Z}/5\): the colour of the dot at \(a + bi\) is exactly \(a + 3b \bmod 5\). The number 5 isn't a coincidence: it's \(|2 + i|^2 = 2^2 + 1^2\), the area of one tilted square. Unit 9 builds on this.
The first isomorphism theorem
Every homomorphism hides a quotient ring inside it. Its kernel is an ideal, and the image is exactly what's left after setting the kernel to zero:
Theorem: the first isomorphism theorem
If \(\varphi : R \to S\) is a ring homomorphism, then
\[ R / \ker\varphi \;\cong\; \operatorname{im}\varphi, \qquad a + \ker\varphi \;\mapsto\; \varphi(a). \]Show proof
Write \(K = \ker\varphi\) and define \(\bar\varphi(a + K) = \varphi(a)\). It's well defined: if \(a + K = b + K\) then \(a - b \in K\), so \(\varphi(a) = \varphi(b)\). It's a homomorphism because \(\varphi\) is. It's onto the image by construction. It's one-to-one: \(\bar\varphi(a + K) = 0\) means \(a \in K\), so \(a + K\) is the zero coset. ∎
This theorem is the standard way to identify a quotient ring: find a homomorphism onto a ring you know, whose kernel is your ideal.
Example: \(\mathbb{R}[x]/(x^2 + 1) \cong \mathbb{C}\), the slick way
Evaluate at \(i\): \(\varphi : \mathbb{R}[x] \to \mathbb{C}\), \(f \mapsto f(i)\). It's onto, since \(a + bx \mapsto a + bi\). Its kernel is the real polynomials with \(i\) as a root. Such a polynomial also has \(-i\) as a root (take conjugates), so it's divisible by \((x - i)(x + i) = x^2 + 1\). So \(\ker\varphi = (x^2 + 1)\), and the theorem does the rest.
Example: \(\mathbb{Z}[x]/(2, x) \cong \mathbb{Z}/2\)
Map \(\mathbb{Z}[x] \to \mathbb{Z}/2\) by "evaluate at 0, then reduce mod 2": \(f \mapsto f(0) \bmod 2\). It's onto, and its kernel is the polynomials with even constant term, which is exactly \((2, x)\).
Conversely, every ideal \(I\) is the kernel of something, namely the quotient map \(R \to R/I\). So "ideals" and "kernels of homomorphisms" are exactly the same thing, seen from two sides.
Ideals of a quotient
The ideals of \(\mathbb{Z}/12\) in the diagram above correspond to the divisors of 12, and the divisors of 12 correspond to the ideals of \(\mathbb{Z}\) that contain \(12\mathbb{Z}\). That's an instance of a general principle:
Theorem: the correspondence theorem
The ideals of \(R/I\) correspond one-to-one with the ideals of \(R\) that contain \(I\), via \(J \mapsto J/I\). The correspondence preserves inclusion.
Intuitively: the ideals of the quotient are what you get by "setting even more to zero", on top of \(I\). In particular, the quotient \(R/I\) has only the two trivial ideals exactly when there's no ideal strictly between \(I\) and \(R\). Unit 7 turns this into the key test for when a quotient is a field.
Practice set
Mixed questions covering the whole unit. Try each one before opening the answer.
1. Which of these are ideals of \(\mathbb{Z}\): the multiples of 7; the non-negative integers; \(\{0\}\); the numbers divisible by 2 or by 3?
Show answer
Multiples of 7: yes. Non-negative integers: no, \(-1 \cdot 1 = -1\) isn't absorbed. \(\{0\}\): yes. Divisible by 2 or 3: no, \(2 + 3 = 5\) is in neither.
2. In \(\mathbb{Z}\), find a single generator for \((12, 18)\), and for the intersection \((4) \cap (6)\).
Show answer
\((12, 18) = (\gcd) = (6)\). \((4) \cap (6)\) is the numbers divisible by both, which is \((\operatorname{lcm}) = (12)\).
3. In \(\mathbb{R}[x]/(x^2 + 1)\), compute \((2 + 3x)(1 - x)\). Check it against complex numbers.
Show answer
\(2 - 2x + 3x - 3x^2 = 2 + x - 3x^2\). Replace \(x^2\) by \(-1\): \(5 + x\). And \((2 + 3i)(1 - i) = 2 - 2i + 3i - 3i^2 = 5 + i\). ✓
4. Show that \(\mathbb{Q}[x]/(x^2 - 1)\) is not a field. Then identify it, using the Chinese remainder idea from Unit 5.
Show answer
\((x - 1)(x + 1) = x^2 - 1 = 0\) in the quotient, with neither factor zero: zero divisors, so not even a domain. The map \(f \mapsto (f(1), f(-1))\) is onto \(\mathbb{Q} \times \mathbb{Q}\) with kernel \((x^2 - 1)\), so \(\mathbb{Q}[x]/(x^2 - 1) \cong \mathbb{Q} \times \mathbb{Q}\).
5. What familiar ring is \(\mathbb{Z}[x]/(x - 3)\)?
Show answer
\(\mathbb{Z}\). Evaluation at 3, \(f \mapsto f(3)\), is onto \(\mathbb{Z}\) (constants hit everything), and its kernel is the polynomials with root 3, which are the multiples of \(x - 3\) (divide by the monic \(x - 3\)). Setting \(x = 3\) just turns every polynomial into a number.
6. In the dual numbers \(\mathbb{R}[\varepsilon]/(\varepsilon^2)\), compute \(f(2 + \varepsilon)\) for \(f(x) = x^4 - 3x\), and read off \(f'(2)\).
Show answer
\((2 + \varepsilon)^4 = 16 + 4 \cdot 8\varepsilon = 16 + 32\varepsilon\), and \(3(2 + \varepsilon) = 6 + 3\varepsilon\). So \(f(2 + \varepsilon) = 10 + 29\varepsilon\). Indeed \(f(2) = 10\) and \(f'(2) = 4 \cdot 8 - 3 = 29\).
Summary and glossary
- An ideal is closed under addition and absorbs multiplication by anything. Ideals are exactly the sets that can be "set to zero".
- An ideal containing a unit is the whole ring; fields have only \(\{0\}\) and themselves.
- \((a) \supseteq (b)\) iff \(a \mid b\). In \(\mathbb{Z}\), \((a, b) = (\gcd(a, b))\).
- The quotient ring \(R/I\): compute as in \(R\), replacing elements of \(I\) by 0. \(\mathbb{R}[x]/(x^2 + 1) \cong \mathbb{C}\).
- First isomorphism theorem: \(R/\ker\varphi \cong \operatorname{im}\varphi\). Ideals and kernels are the same thing.
- Ideals of \(R/I\) correspond to ideals of \(R\) containing \(I\).
- Ideal
- A subset closed under \(+\), containing 0, absorbing multiplication.
- Principal ideal
- \((a)\): all multiples of one element.
- Coset
- \(a + I\): everything congruent to \(a\) modulo \(I\).
- Quotient ring
- \(R/I\): the ring of cosets.
- Quotient map
- \(R \to R/I\), \(a \mapsto a + I\).
- Dual numbers
- \(\mathbb{R}[\varepsilon]/(\varepsilon^2)\).
- First isomorphism theorem
- \(R/\ker\varphi \cong \operatorname{im}\varphi\).
- Correspondence theorem
- Ideals of \(R/I\) ↔ ideals of \(R\) containing \(I\).