Arithmetic mod $n$ is arithmetic in a quotient ring: $\mathbb{Z}_{n} = \mathbb{Z}/(n)$. The same construction with any ideal produces new rings, including $\mathbb{C}$ out of real polynomials and the finite fields at the end of the course. It is the ring version of a quotient group.
Why it works
Computing in $F[x]/(f)$. Two polynomials name the same coset when they differ by a multiple of $f$, so every coset has exactly one representative of degree less than $\deg f$: the remainder after dividing by $f$. In practice, replace high powers of $x$ using $f = 0$. Over $\mathbb{Z}_{p}$, with $\deg f = n$, there are $p^{n}$ remainders, so $\mathbb{Z}_{p}[x]/(f)$ has $p^{n}$ elements.
Worked example: Complex numbers from polynomials
- Expand: $2 - 2x + 3x - 3x^{2} = 2 + x - 3x^{2}$.
- In the quotient $x^{2} + 1 = 0$, so $x^{2} = -1$: $2 + x + 3 = 5 + x$.
- Compare $(2 + 3i)(1 - i) = 5 + i$: the class of $x$ behaves exactly like $i$, and $\mathbb{R}[x]/(x^{2} + 1) \cong \mathbb{C}$.
Answer: $5 + x$.
Worked example: A quotient of $\mathbb{Z}_{2}[x]$
- Remainders of degree less than $2$: $0, 1, x, x + 1$.
- $x(x + 1) = x^{2} + x$, and $x^{2} = x + 1$ here (since $x^{2} + x + 1 = 0$ and $-1 = 1$ mod 2).
- So $x^{2} + x = (x + 1) + x = 2x + 1 = 1$.
Answer: $x$ and $x + 1$ are inverses; this four-element ring is a field.
Worked example: A finite quotient of $\mathbb{Z}$
- $7 \cdot 5 = 35$, and cosets of $(4)$ are classes mod $4$: $35 \equiv 3$.
Answer: $3 + I$; the ring is a copy of $\mathbb{Z}_{4}$.
The first isomorphism theorem holds for rings too: if $\varphi: R \to S$ is a ring homomorphism, then $R/\ker\varphi \cong \operatorname{im}\varphi$. For example, evaluation at $i$ sends $\mathbb{R}[x]$ onto $\mathbb{C}$ with kernel $(x^{2} + 1)$, which is another way to see $\mathbb{R}[x]/(x^{2} + 1) \cong \mathbb{C}$.
Common mistake
Key idea