Prime Ideals

In algebraic number theory, prime ideals of rings of integers generalize prime numbers and are the objects that factor uniquely, restoring the unique factorization that individual algebraic integers can lack. In algebraic geometry, prime ideals of a polynomial ring correspond to irreducible varieties, and the set of all prime ideals of a ring — its spectrum — is the foundation of scheme theory. In commutative algebra broadly, localizing a ring at a prime ideal produces local rings that are the basic local-to-global tool for studying singularities and dimension. In cryptography, the structure of prime ideals in number fields underlies algorithms in computational number theory used for factoring and for lattice-based and ideal-lattice cryptosystems. In coding theory and combinatorics, quotients by prime ideals of polynomial rings appear when analyzing varieties over finite fields, connecting to Gröbner basis computations. Even in statistics and applied algebra, prime ideals of polynomial rings arise in algebraic statistics when describing the zero sets ("varieties") of constraints defining statistical models.

:::quiz[In $\mathbb{Q}[x,y]/(y-x^2) \cong \mathbb{Q}[t]$, what property of $\mathbb{Q}[t]$ shows $(y-x^2)$ is prime?]

• $\mathbb{Q}[t]$ is a field
• $\mathbb{Q}[t]$ is an integral domain because degrees add under multiplication ✓
• $\mathbb{Q}[t]$ has finitely many elements
• $\mathbb{Q}[t]$ is isomorphic to $\mathbb{Q}[x,y]$

Ask the tutor: How does the theorem "$P$ prime iff $R/P$ is a domain" change when we instead ask when $R/P$ is a field?

Think of an ideal $P$ as a "sieve": elements of $R$ either fall through the sieve (land in $P$) or do not. The prime property says the sieve cannot be fooled by a product — if the product $ab$ falls through, at least one of $a$ or $b$ must already have fallen through on its own. Contrast this with a "leaky" ideal like $(6) \subset \mathbb{Z}$: the product $2 \cdot 3 = 6$ falls into $(6)$, yet neither $2$ nor $3$ individually belongs to $(6)$. The sieve $(6)$ is leaky precisely because $6 = 2\cdot 3$ is composite. The reformulation in terms of quotients makes this vivid. Passing to $R/P$ collapses everything in $P$ to zero. The prime property "if $ab \in P$ then $a \in P$ or $b \in P$" becomes, after collapsing, "if $\bar a \bar b = 0$ in $R/P$ then $\bar a = 0$ or $\bar b = 0$" — which is exactly the defining property of an integral domain. So checking primeness of $P$ is the same computation as checking that $R/P$ has no zero divisors; the definition and the theorem are two descriptions of one phenomenon, viewed before and after quotienting.

Example 1: Prime ideals of $\mathbb{Z}$, and a leaky non-example. Let $R = \mathbb{Z}$. We determine which ideals $(n)$, $n \geq 0$, are prime, using the theorem rather than checking the definition by hand. Every ideal of $\mathbb{Z}$ has the form $(n)$ for a unique $n \geq 0$ (a fact from the classification of ideals of $\mathbb{Z}$ as a principal ideal domain). We compute $\mathbb{Z}/(n)$ in each case:

• If $n = 0$: $\mathbb{Z}/(0) \cong \mathbb{Z}$, which is an integral domain. So $(0)$ is prime.
• If $n = 1$: $\mathbb{Z}/(1)$ is the zero ring, excluded by definition since $1 \notin \mathbb{Z}$ would be false ($P = R$ here). So $(1)$ is not prime.
• If $n \geq 2$ and $n$ is prime (in the elementary-number-theory sense): $\mathbb{Z}/(n) = \mathbb{Z}_n$ is a field (every nonzero residue is invertible by unique factorization applied to $\gcd(a,n)=1$), and every field is trivially an integral domain. So $(n)$ is prime.
• If $n \geq 2$ and $n$ is composite, say $n = rs$ with $1 < r, s < n$: then in $\mathbb{Z}/(n)$, the classes $\bar r, \bar s$ are both nonzero (since $0 < r, s < n$), yet $\bar r \bar s = \overline{rs} = \bar n = \bar 0$. So $\mathbb{Z}/(n)$ has zero divisors and is not an integral domain, hence $(n)$ is not prime.

Concretely take $n = 91 = 7 \times 13$, an asymmetric composite (not a small "textbook" number like $4$ or $6$). In $\mathbb{Z}/(91)$, $\overline{7} \neq \bar 0$ and $\overline{13} \neq \bar 0$, but $\overline{7}\cdot\overline{13} = \overline{91} = \bar 0$. Directly in $\mathbb{Z}$: $7 \cdot 13 = 91 \in (91)$, but $7 \notin (91)$ and $13 \notin (91)$ (since $91 \nmid 7$ and $91 \nmid 13$). So $(91)$ fails the prime property, confirming via the theorem that $\mathbb{Z}/(91)$ is not an integral domain — matching the direct check above. Conclusion: the complete list of prime ideals of $\mathbb{Z}$ is $(0)$ together with $(p)$ for each prime number $p$.

Example 2: A prime ideal in a polynomial ring, checked two ways. Let $R = \mathbb{Q}[x,y]$ and let $P = (y - x^2)$, the ideal generated by the single polynomial $y - x^2$. Quotient computation. Consider the evaluation-type homomorphism

sending $x \mapsto t$ and $y \mapsto t^2$ (well-defined by the universal property of polynomial rings: a ring homomorphism out of $\mathbb{Q}[x,y]$ is determined by where $x,y$ go, subject to $\mathbb{Q}$-coefficients being fixed). Surjectivity: $\varphi(x) = t$, so $t$, and hence every polynomial in $t$, is in the image. Kernel: We claim $\ker \varphi = P = (y-x^2)$. $(\supseteq)$ Since $\varphi(y - x^2) = t^2 - t^2 = 0$, the generator $y - x^2$ lies in $\ker\varphi$, and since $\ker\varphi$ is an ideal, $(y - x^2) \subseteq \ker\varphi$. $(\subseteq)$ Let $f(x,y) \in \mathbb{Q}[x,y]$. Perform polynomial division of $f$ by $y - x^2$, treating $f$ as a polynomial in $y$ with coefficients in $\mathbb{Q}[x]$: since $y - x^2$ is monic of degree $1$ in $y$, division gives

for some $q \in \mathbb{Q}[x,y]$ and remainder $r \in \mathbb{Q}[x]$ (degree $0$ in $y$). Applying $\varphi$:

If $f \in \ker\varphi$, then $r(t) = 0$ as a polynomial in $t$, forcing $r = 0$ in $\mathbb{Q}[x]$ (a polynomial in one variable over a field vanishes identically only if all its coefficients vanish). Hence $f = q\cdot(y-x^2) \in (y-x^2)$. So $\ker\varphi = (y-x^2) = P$. By the First Isomorphism Theorem for rings (from The Isomorphism Theorems for Rings),

Since $\mathbb{Q}[t]$ is an integral domain (a polynomial ring over a field has no zero divisors: the degree of a product of nonzero polynomials equals the sum of the degrees, which cannot be $-\infty$ unless a factor is zero), $R/P$ is an integral domain. By the theorem, $P = (y - x^2)$ is a prime ideal. Direct check, as a sanity confirmation. Suppose $f(x,y) g(x,y) \in (y - x^2)$ in $\mathbb{Q}[x,y]$. Substituting $y = t^2, x = t$ (i.e. applying $\varphi$) must send the product to $0$ in $\mathbb{Q}[t]$, since $\varphi$ kills $(y-x^2)$. So $\varphi(f)\varphi(g) = 0$ in $\mathbb{Q}[t]$, an integral domain, so $\varphi(f) = 0$ or $\varphi(g) = 0$, i.e. $f \in \ker\varphi = P$ or $g \in \ker\varphi = P$ — exactly the prime property, matching the quotient argument. Geometrically, $P$ consists of all polynomials vanishing on the parabola $y = x^2$, and the fact that $P$ is prime reflects that the parabola is an irreducible curve — it cannot be split into two smaller algebraic pieces, foreshadowing the dictionary between prime ideals and irreducible varieties used throughout Unique Factorization Domains and algebraic geometry.