Translating between rings
Is \(1234 \times 5678 = 7{,}006{,}652\)? Here's a quick check. Replace each number by its remainder modulo 9 (the digit sum, from Unit 2): \(1234 \to 1\), \(5678 \to 8\), so the product should be \(1 \times 8 = 8\). And \(7{,}006{,}652\) has digit sum 26, which gives 8. The answer passes. A wrong answer like \(7{,}006{,}752\) (digit sum 27, so 0) would be caught.
Why is this check legitimate? Because "take the remainder mod 9" turns products into products: the remainder of a product is the product of the remainders. The map from \(\mathbb{Z}\) to \(\mathbb{Z}/9\) respects multiplication (and addition). A map like that is called a homomorphism, from Greek for "same shape".
The definition
Definition
A ring homomorphism is a map \(\varphi : R \to S\) between rings such that, for all \(a, b \in R\),
\[ \varphi(a + b) = \varphi(a) + \varphi(b), \qquad \varphi(ab) = \varphi(a)\,\varphi(b), \qquad \varphi(1) = 1. \]In words: it doesn't matter whether you calculate first and then translate, or translate first and then calculate. The condition \(\varphi(1) = 1\) rules out silly maps like "send everything to 0". A few more facts come for free:
- \(\varphi(0) = 0\), since \(\varphi(0) = \varphi(0 + 0) = \varphi(0) + \varphi(0)\), then subtract.
- \(\varphi(-a) = -\varphi(a)\), since \(\varphi(a) + \varphi(-a) = \varphi(0) = 0\).
- If \(u\) is a unit, so is \(\varphi(u)\): \(\varphi(u)\varphi(u^{-1}) = \varphi(1) = 1\).
- \(\varphi\) respects powers and therefore polynomial expressions: \(\varphi(a^3 - 2ab) = \varphi(a)^3 - 2\varphi(a)\varphi(b)\).
A gallery of homomorphisms
| Map | What it does |
|---|---|
| \(\mathbb{Z} \to \mathbb{Z}/n\) | Take the remainder mod \(n\). |
| \(\mathbb{Z}/12 \to \mathbb{Z}/4\) | Remainder mod 4. Allowed because 4 divides 12: a number's remainder mod 12 already determines its remainder mod 4. |
| \(R[x] \to R\), \(f \mapsto f(a)\) | Evaluation at \(a\). Plugging in respects sums and products: \((fg)(a) = f(a)g(a)\). |
| \(\mathbb{R}[x] \to \mathbb{C}\), \(f \mapsto f(i)\) | Evaluate at a complex number: \(x^2 + 3x \mapsto -1 + 3i\). |
| \(\mathbb{C} \to \mathbb{C}\), \(z \mapsto \bar z\) | Complex conjugation: \(\overline{zw} = \bar z\,\bar w\), \(\overline{z + w} = \bar z + \bar w\). |
| \(\mathbb{Z}[x] \to (\mathbb{Z}/p)[x]\) | Reduce every coefficient mod \(p\) (used in Unit 4's irreducibility test). |
| \(F \to F\), \(a \mapsto a^p\) | The Frobenius map in characteristic \(p\) (Unit 3). |
And some maps that look plausible but fail:
- \(a \mapsto 2a\) on \(\mathbb{Z}\): respects addition, but \(2(ab) \neq (2a)(2b)\), and \(1 \mapsto 2\).
- \(a \mapsto a^2\) on \(\mathbb{R}\): respects multiplication, but \((a + b)^2 \neq a^2 + b^2\).
- Remainder mod 4 from \(\mathbb{Z}/6\) to \(\mathbb{Z}/4\): not even well defined, since \(0 \equiv 6 \pmod 6\) but \(0 \not\equiv 6 \pmod 4\).
- Any map \(\mathbb{Z}/3 \to \mathbb{Z}\): it would need \(\varphi(1 + 1 + 1) = 1 + 1 + 1 = 3\), but \(1 + 1 + 1 = 0\) in \(\mathbb{Z}/3\), which must go to 0.
Try it. For which \(m\) is "remainder mod \(m\)" a homomorphism \(\mathbb{Z}/n \to \mathbb{Z}/m\)?
Show answer
Exactly when \(m\) divides \(n\). The map must send \(n \cdot 1 = 0\) to \(n \cdot 1 = 0\) in \(\mathbb{Z}/m\), so \(m \mid n\). Conversely if \(m \mid n\), numbers that agree mod \(n\) also agree mod \(m\), so the map is well defined.
Proving things impossible
Homomorphisms are a powerful way to prove that equations have no solutions. If an equation had a solution in \(R\), then applying \(\varphi\) would give a solution in \(S\). So if there's no solution in a small, easy ring, there's none in the big one.
Example: \(x^2 - 3y^2 = 2\) has no integer solutions
Apply reduction mod 3. A solution would give \(x^2 \equiv 2 \pmod 3\). But squares mod 3 are only \(0^2 = 0\), \(1^2 = 1\), \(2^2 = 4 \equiv 1\). No square is 2, so there's no solution, in \(\mathbb{Z}\) or anywhere that maps to \(\mathbb{Z}/3\).
Example: \(x^3 + x + 1 = 0\) has no integer root
Apply reduction mod 2. Plug in \(0\): \(1\). Plug in \(1\): \(1 + 1 + 1 = 3 \equiv 1\). The polynomial is never 0 mod 2, so it can't be 0 at any integer.
The trick works in one direction only. Having a solution mod \(n\) doesn't guarantee an integer solution: the homomorphism forgets information, and a solution may only exist in the forgetful version.
Kernel and image
Two sets measure what a homomorphism does. The image is what gets hit. The kernel is what gets crushed to zero.
Definition
For a homomorphism \(\varphi : R \to S\),
\[ \ker \varphi = \{a \in R : \varphi(a) = 0\}, \qquad \operatorname{im} \varphi = \{\varphi(a) : a \in R\}. \]| Homomorphism | Kernel |
|---|---|
| \(\mathbb{Z} \to \mathbb{Z}/n\) | the multiples of \(n\), written \(n\mathbb{Z}\) |
| evaluation \(F[x] \to F\) at \(a\) | polynomials with \(a\) as a root: the multiples of \(x - a\) |
| \(\mathbb{R}[x] \to \mathbb{C}\), \(f \mapsto f(i)\) | the multiples of \(x^2 + 1\) |
| complex conjugation | just \(\{0\}\) |
The kernel tells you how much information is lost. The key fact: a homomorphism is one-to-one exactly when its kernel is \(\{0\}\). If \(\varphi(a) = \varphi(b)\), then \(\varphi(a - b) = 0\), so \(a - b\) is in the kernel. When the kernel is only \(0\), \(a = b\).
Notice a pattern in the table: every kernel is "all multiples of something". That's no accident. Kernels always have two special properties:
- Closed under addition: if \(\varphi(a) = \varphi(b) = 0\), then \(\varphi(a + b) = 0\).
- Absorbs multiplication: if \(\varphi(a) = 0\) and \(r\) is any element of \(R\), then \(\varphi(ra) = \varphi(r) \cdot 0 = 0\).
The second property is striking. Multiply something in the kernel by anything at all and you stay in the kernel, like a black hole. Subsets with these properties are called ideals, and they're the subject of the next unit. There you'll see that every ideal is a kernel, and that the image of \(\varphi\) is a perfect copy of \(R\) with its kernel "set to zero".
The image, on the other hand, is a subring of \(S\): it contains 1 and is closed under \(+\), \(-\) and \(\times\). For example, the image of evaluation \(\mathbb{Z}[x] \to \mathbb{C}\) at \(x = i\) is \(\mathbb{Z}[i]\), the Gaussian integers.
Isomorphisms: the same ring in disguise
Definition
An isomorphism is a homomorphism that is a bijection (one-to-one and onto). If one exists, \(R\) and \(S\) are isomorphic, written \(R \cong S\).
Isomorphic rings are the same ring with the elements renamed. Every property expressible in terms of \(+\) and \(\times\) transfers: being a field, the number of units, the number of solutions of \(x^2 = 1\), the characteristic.
Example: complex numbers as matrices
Send \(a + bi\) to the matrix \(\begin{pmatrix} a & -b \\ b & a \end{pmatrix}\). This is a homomorphism from \(\mathbb{C}\) into \(2 \times 2\) real matrices. For instance \(i \mapsto \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\), and squaring that matrix gives \(-I\), just as \(i^2 = -1\). It's one-to-one, so \(\mathbb{C}\) is isomorphic to the ring of matrices of this form. Geometrically, multiplying by \(a + bi\) is the linear map given by that matrix: a rotation combined with a scaling.
Here's a more surprising isomorphism. Send each \(k \in \mathbb{Z}/6\) to the pair \((k \bmod 2,\ k \bmod 3)\) in \(\mathbb{Z}/2 \times \mathbb{Z}/3\), where pairs are added and multiplied coordinate by coordinate. Each coordinate is a homomorphism, so the pair map is too. Is it a bijection?
\(\mathbb{Z}/6 \to \mathbb{Z}/2 \times \mathbb{Z}/3\)every cell filled once
\(\mathbb{Z}/4 \to \mathbb{Z}/2 \times \mathbb{Z}/2\)collisions and gaps
So \(\mathbb{Z}/6 \cong \mathbb{Z}/2 \times \mathbb{Z}/3\): arithmetic mod 6 is secretly the same as doing arithmetic mod 2 and mod 3 side by side. It works because 2 and 3 share no factor. This is the Chinese remainder theorem, which returns in Unit 11.
To prove two rings are not isomorphic, find a property that one has and the other lacks. \(\mathbb{Z}/4 \not\cong \mathbb{Z}/2 \times \mathbb{Z}/2\): in the product every element satisfies \(a + a = 0\) (characteristic 2), but in \(\mathbb{Z}/4\), \(1 + 1 = 2 \neq 0\). And \(\mathbb{Q} \not\cong \mathbb{R}\), because \(x^2 = 2\) has a solution in \(\mathbb{R}\) but not in \(\mathbb{Q}\).
Homomorphisms out of fields
Theorem
Every homomorphism from a field to a non-zero ring is one-to-one.
Show proof
Suppose \(\varphi(a) = 0\) with \(a \neq 0\). Then \(a\) has an inverse, and \(1 = \varphi(1) = \varphi(a^{-1}a) = \varphi(a^{-1}) \cdot 0 = 0\) in the target, which is impossible in a non-zero ring. So the kernel is \(\{0\}\). ∎
So a field can't be partially collapsed: a homomorphism either keeps it intact or doesn't exist. Every map out of a field is an embedding, placing a perfect copy of the field inside the target. This is why fields are the "atoms" of ring theory, and it's the reason the prime subfield of Unit 3 is unique.
Going deeper: \(\mathbb{Z}\) maps everywhere, and symmetries of fields
For any ring \(R\) there is exactly one homomorphism \(\mathbb{Z} \to R\): it must send \(1 \mapsto 1\), hence \(n \mapsto 1 + 1 + \dots + 1\). Its kernel is \(n\mathbb{Z}\) where \(n\) is the characteristic of \(R\). That's a cleaner way to define characteristic.
An isomorphism from a ring to itself is an automorphism, a symmetry of the ring. The automorphisms of \(\mathbb{C}\) that fix every real number are exactly the identity and complex conjugation: the only way to rename \(\mathbb{C}\) without changing anything real is to swap \(i\) and \(-i\). The study of field automorphisms like these is Galois theory, which explains why there's no formula for the roots of a general degree-5 polynomial.
Practice set
Mixed questions covering the whole unit. Try each one before opening the answer.
1. Is \(\varphi : \mathbb{Z} \to \mathbb{Z}\), \(\varphi(a) = -a\), a ring homomorphism?
Show answer
No. \(\varphi(1) = -1 \neq 1\). Also \(\varphi(2 \cdot 3) = -6\) but \(\varphi(2)\varphi(3) = 6\). In fact the only homomorphism \(\mathbb{Z} \to \mathbb{Z}\) is the identity.
2. Find the kernel of \(\mathbb{Z} \to \mathbb{Z}/4 \times \mathbb{Z}/6\), \(a \mapsto (a \bmod 4, a \bmod 6)\). Is the map onto?
Show answer
The kernel is the integers divisible by both 4 and 6: the multiples of \(\operatorname{lcm}(4, 6) = 12\). The map is not onto: the target has 24 elements, but the image is a copy of \(\mathbb{Z}/12\). For example \((1, 0)\) is never hit, since a number that is \(\equiv 0 \pmod 6\) is even.
3. Use reduction mod 8 to show that no integer of the form \(8k + 7\) is a sum of three squares.
Show answer
Squares mod 8: \(0, 1, 4, 1, 0, 1, 4, 1\), so only \(0, 1, 4\). Sums of three of these give \(0, 1, 2, 3, 4, 5, 6\) (e.g. \(4+1+1 = 6\), \(4 + 4 + 1 = 9 \equiv 1\)) but never 7. Legendre proved the full answer in 1798: a non-negative integer is a sum of three squares exactly when it is not of the form \(4^a(8k + 7)\).
4. How many ring homomorphisms are there from \(\mathbb{Z}[x]\) to \(\mathbb{Z}/5\)?
Show answer
5. A homomorphism must send each integer \(n\) to \(n \bmod 5\), and then it's determined by where \(x\) goes: any of the 5 elements works, giving "reduce mod 5 and evaluate at \(a\)".
5. Is \(\mathbb{Z}/15 \cong \mathbb{Z}/3 \times \mathbb{Z}/5\)? Is \(\mathbb{Z}/9 \cong \mathbb{Z}/3 \times \mathbb{Z}/3\)?
Show answer
Yes for 15: 3 and 5 share no factor, so \(k \mapsto (k \bmod 3, k \bmod 5)\) fills the \(3 \times 5\) grid. No for 9: \(\mathbb{Z}/3 \times \mathbb{Z}/3\) has characteristic 3, while in \(\mathbb{Z}/9\), \(1 + 1 + 1 = 3 \neq 0\).
6. Show there is no homomorphism from \(\mathbb{Q}\) to \(\mathbb{Z}\).
Show answer
A homomorphism would send \(2 \mapsto 2\), and since \(\tfrac12 \cdot 2 = 1\), it would send \(\tfrac12\) to an integer \(m\) with \(2m = 1\). There's no such integer. (Units must go to units, and 2 is not a unit in \(\mathbb{Z}\).)
Summary and glossary
- A homomorphism respects \(+\), \(\times\) and 1. Reduction mod \(n\), evaluation and conjugation are key examples.
- If an equation has no solution after applying a homomorphism, it has no solution before. This proves many impossibility results.
- The kernel is what maps to 0. It absorbs multiplication by anything: it's an ideal. A homomorphism is one-to-one iff its kernel is \(\{0\}\).
- An isomorphism is a bijective homomorphism; isomorphic rings are the same up to renaming. \(\mathbb{Z}/6 \cong \mathbb{Z}/2 \times \mathbb{Z}/3\).
- Homomorphisms out of fields are always one-to-one.
- Homomorphism
- A map preserving addition, multiplication and 1.
- Kernel
- Everything that maps to 0.
- Image
- Everything that gets hit; a subring of the target.
- Isomorphism
- A homomorphism that is also a bijection.
- Embedding
- A one-to-one homomorphism.
- Automorphism
- An isomorphism from a ring to itself.
- Evaluation map
- \(f \mapsto f(a)\), from a polynomial ring.
- Product ring
- \(R \times S\): pairs with coordinate-wise operations.