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6.7 Prime ideals

Another way to write generators of groups of ideals is to use prime ideals. An ideal P of R is prime if for every a and b in R so that ab lies in P at least one of a and b lies in P; an equivalent assertion is that R/P is a domain--non-zero elements give non-zero products. It is clear that the restriction P $ \cap$ $ \mathbb {Z}$ satisfies the same conditions for integers a and b. In other words P $ \cap$ $ \mathbb {Z}$ is generated by a prime number p. Thus P determined by the ideal P/pR in R/pR. Now, if R = $ \mathbb {Z}$ . w1 + ... + $ \mathbb {Z}$ . wn, then R/pR is a vector space of rank n over the finite field $ \mathbb {Z}$/p$ \mathbb {Z}$. Instead of solving this specific problem, we can ask for a structure theorem for commutative rings with identity with underlying additive group a vector space of rank n over $ \mathbb {Z}$/p$ \mathbb {Z}$.

Lemma 15   Any ring with underlying additive group a finite dimensional vector space over $ \mathbb {Z}$/p$ \mathbb {Z}$ is a direct sum of rings of the form ($ \mathbb {Z}$/p$ \mathbb {Z}$)[T]/(P(T)m), where P(T) is an irreducible polynomial over the field $ \mathbb {Z}$/p$ \mathbb {Z}$.

This result follows from the Chinese Remainder theorem and Euclidean division applied to the polynomial ring in one variable ($ \mathbb {Z}$/p$ \mathbb {Z}$)[T]. If we use the symbol $ \mathbb {F}$q to denote the field with q elements (where q is a prime power), then this result can be refined further as follows

Lemma 16   The ring ($ \mathbb {Z}$/p$ \mathbb {Z}$)[T]/(P(T)m), where P(T) is an irreducible polynomial of degree d, is isomorphic to the ring $ \mathbb {F}$pd[h]/(hn).

This result follows by constructing (using Newton's method of successive approximations) a polynomial $ \tilde{T}$ of the form T + P(T)Q1(T) + ... + P(T)n - 1Qn - 1(T) so that P($ \tilde{T}$) is divisible by P(T)n. Then h = $ \tilde{T}$ - T. Combining these results, we see that R/pR has the form

% latex2html id marker 16521
$\displaystyle {\frac{{\mathbb F}_{p^{f_1}}[h_1]}{(h_1^{e_1})}}$ $\displaystyle \oplus$ ... $\displaystyle \oplus$ % latex2html id marker 16524
$\displaystyle {\frac{{\mathbb F}_{p^{f_g}}[h_g]}{(h_g^{e_g})}}$

Corresponding to each factor we get a surjective ring homomorphism R/pR$ \to$$ \mathbb {F}$pfi. The kernel of this has the form Pi/pR for a prime ideal Pi whose restriction is p. The number fi is called the residual degree (or more simply the degree) of Pi over p, while the number ei is called the ramification degree (or order of ramification) of Pi over p.

Now, let I be any ideal. Suppose first that the restriction i of I is of the form i1i2, with i1 and i2 co-prime. We can apply the Chinese Remainder theorem to write R/I as a direct sum of R/I1 and R/I2, where I1 = I + i1R and I2 = I + i2R. Thus I = I1 $ \cap$ I2; if I is invertible one easily shows that I = I1 . I2. Thus we can reduce our study of groups of ideals to the study of ideals Q so that the restriction q is the power of a prime p. Now, R/(Q + pR) is a quotient of the ring studied above and is not zero. Thus there is a prime Pi as above so that Q is contained in Pi. By successively removing such Pi (if Q is invertible) we can write Q as a product of various powers of the Pi considered above. Thus we have written any invertible ideal as a product of (invertible) prime ideals. It is not difficult to show that any such product expression is unique. Thus, we obtain the unique factorisation of ideals in terms of prime ideals.

next up previous
Next: 7 Quadratic fields Up: 6 Algebraic Number Fields Previous: 6.6 Minkowski's Geometry of
Kapil Hari Paranjape 2002-10-20