Skip to content

Adjunction of roots in a ring

May 1, 2012

In the post titled a motivation for the universal property of polynomial rings, I showed how the polynomial ring R[x] can be viewed as the smallest ring containing R which possess a new element x not in R having no relations with elements of R \subset R[x], except for the defining axioms for a ring. But suppose we want our new element to satisfy some relation with other elements of R, how should we proceed?

First, suppose our ring R is contained in another ring R' and we want the smallest ring S containing R and an element \alpha \in R'. Since S is a ring containing \alpha, it must contain all positive powers of \alpha. It must also contain all products of these powers by elements of R, all sums and differences of such expressions. In other words, the ring S must contain all elements of R' of the form

r_0 + r_1 \alpha + r_2 \alpha ^2 + \cdots + r_n \alpha ^n

where r_i \in R for i = 0, \ldots, n, for all n \in \textbf{N}. But since these elements together form a ring and S is wanted to be minimal, we conclude that S must be exactly the set of elements of this form. In other words, S is the image of R[x] under the ring homomorphism ev_{\alpha} : R[x] \to R' defined by ev_{\alpha} (\sum r_i x ^i) = \sum r_i \alpha ^i. We will denote this ring naturally by R[\alpha].

But what if our ring is sitting in the void and is not seen as a subring of another ring? We could still want the smallest ring (with respect to injective morphisms) containing our ring R and some element \alpha satisfying some polynomial relation with the other elements of R. Say you want the smallest ring S containing R and an element \alpha \in S such that

f(\alpha) = r_0 + r_1 \alpha + r_2 \alpha ^2 + \cdots + r_n \alpha ^n = 0

where r_i \in R for i = 0, \ldots, n. In other words, we want to adjoin a root of f(x) to the ring R. Let’s suppose also that f(x) is an irreducible monic polynomial in R[x], (monic means that the coefficient of the term of highest order in f(x) is 1). Now we can’t just hit R[x] with an evaluation morphism to obtain our ring S since we have nothing to evaluate it with, we have to create \alpha abstractly. But since R[x] can be constructed abstractly from R and satisfies a somewhat more general property than what we seek for S, we can in fact obtain S by quotienting R[x] with the right ideal:

Theorem: Let R be a commutative ring and f(x) a non-constant irreducible monic polynomial in R[x]. Then R[x] / (f) is a ring containing an isomorphic image of R satisfying

  1. There is an element \alpha \in R[x]/(f) such that f(\alpha) = 0.
  2. For all ring R' containing an isomorphic image of R, for all \beta \in R' such that f(\beta) = 0, there exists a unique injective ring homomorphism \phi : R[x]/(f) \to R' such that \phi(R) = R and \phi(\alpha) = \beta.

Note that there are some abuse of language in this proposition. When I say f(\alpha) = 0, I identify f(x) = \sum r_ix^i with \sum \iota(r_i)x^i (\text{mod }f) where \iota is the inclusion map from R to R[x]/(f). Also, when I write \phi(R) = R I mean that \phi takes the isomorphic image of R in R[x]/(f) to the isomorphic image of R in R'. ie this diagram commutes

Proof: Recall that R[x]/(f) means the quotient of R[x] by the principal ideal generated by f. Since clearly no non-zero element of R is a multiple of f(x) in R[x], the canonical epimorphism \pi : R[x] \to R[x]/(f) is injective when restricted to R. Moreover, taking \alpha to be the equivalence class \overline{x} of x in R[x]/(f), we have

f(\alpha) = \displaystyle\sum_{i=0}^n \overline{\alpha_i}\overline{x}^i = \overline{f(x)} = \overline{0}

so property 1 is satisfied. To prove 2, suppose that R' is another ring containing R having an element \beta such that f(\beta) = 0. The universal property for polynomial rings assures us that there exists a unique ring homomorphism \varphi : R[x] \to R' such that \varphi(R) = R and \varphi(x) = \beta. Since f(\beta) = 0 \in R', and f is irreducible in R, \varphi(g(x)) = 0 iff g(\beta) =0 iff g(x) = f(x)h(x) for some h(x) \in R[x]. So \ker \varphi = (f) and by the first isomorphism theorem (more exactly the universal property of quotient rings), we have a unique isomorphism

R[x]/(f) \simeq \text{Im}(\varphi) \subset R'

providing us with a unique injective ring homomorphism \phi : R[x]/(f) \to R', as wanted.

QED

This shows that the first candidate we considered (when R was thought of as a subring of another ring R' and the new element belonged to R'), namely R[\alpha], obtained with the evaluation morphism, is isomorphic to the quotient R[x]/(f) where f is the irreducible polynomial for \alpha in R. For example \textbf{R}[x]/(x^2+1) \simeq \textbf{C} and \textbf{Q}[x]/(t^2-2) \simeq \textbf{Q}(\sqrt2). Note that these isomorphisms might not be unique! Indeed, we deduce from the proposition that there are exactly n distinct embedding of such a quotient in the larger field where n is the number of distinct roots of the polynomial in the larger field. For example, there are two distinct embeddings of \textbf{Q}[x]/(t^2-2) in \textbf{R} since \textbf{R} possess two distinct roots of (t^2-2), namely \sqrt2 and -\sqrt2.

I will conclude this post with a very useful observation. Note that if we restrict our attention to the abelian group structure of R_{d-1}[x], the polynomials over R of degree at most (d-1), it can be identified with R^{\oplus_d} with the group isomorphism

\psi (r_0, r_1, \ldots, r_{d-1}) = r_0 + r_1 x + \cdots + r^{d-1}.

Consider also, for f(x) \in R[x] of degree d, the function \varphi_f : R[x] \to R_{d-1}[x] taking a monic polynomial g(x) to its remainder when divided by f(x).

Lemma: The function \varphi_f is well defined. ie if f(x) is a monic polynomial and

f(x)q_1(x) + r_1(x) = f(x)q_2(x) + r_2(x)

with \deg (r_1) and \deg(r_2) both smaller than \deg(f), then r_1(x) = r_2(x).

Proof: We have

f(x)(q_1(x) - q_2(x)) =r_2(x) - r_1(x).

So if r_1(x) \neq r_2(x), (assuring also that q_1(x) \neq q_2(x)), then \deg(r_2(x) - r_1(x)) < \deg f but \deg (f(x)(q_1(x)-q_2(x))) \geq \deg f, a contradiction.

QED

Proposition: The function \varphi_f is a surjective homomorphism of abelian groups from R[x] to R_{d-1}[x].

Proof: The function is clearly surjective. To see that it is a group homomorphism, suppose that

g_1(x) = f(x)q_1(x) + r_1(x) and g_2(x) = f(x)q_2(x) + r_2(x)

with \deg r_1 and \deg r_2 smaller than d. Then

g_1(x) + g_2(x) = f(x)((q_1(x) + q_2(x)) + (r_1(x) + r_2(x))

with \deg (r_1 + r_2) < d. So

\varphi_f(g_1(x) + g_2(x)) = r_1(x) + r_2(x) = \varphi_f(g_1(x)) + \varphi_f(g_2(x)).

QED

But as in the proof of the theorem, \varphi_f(g(x)) = 0 iff g(x) = f(x)h(x) for some h(x) \in R[x]. So \ker \varphi_f = (f) and by the first isomorphism theorem and with the identification given by \psi, we have

R[x]/(f) \simeq R^{\oplus_d}

as abelian groups, where d is the degree of f. So for every monic polynomial f of degree d, this construction can be used to give a different ring structure on the abelian group R^{\oplus d}.

Writing R[\alpha] for R[x]/(f) where \alpha is a root of f in the new ring, we see that the set \{1, \alpha, \alpha^2, \ldots, \alpha^{d-1} \} is a basis for R[\alpha] over R.

References: M. Artin, Algebra, 2nd edition;

E. Artin, Galois Theory;

P. Aluffi, Algebra : Chapter 0.

Advertisement

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: