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

There is an element $\alpha \in R[x]/(f)$ such that $f(\alpha) = 0$ .

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.

From → Algebra, Galois Theory

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Adjunction of roots in a ring

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