# Adjunction of roots in a ring

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

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

where for for all But since these elements together form a ring and is wanted to be minimal, we conclude that must be exactly the set of elements of this form. In other words, is the image of under the ring homomorphism defined by We will denote this ring naturally by

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 and some element satisfying some polynomial relation with the other elements of Say you want the smallest ring containing and an element such that

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

Theorem:Let be a commutative ring and a non-constant irreducible monic polynomial in Then is a ring containing an isomorphic image of satisfying

- There is an element such that .
- For all ring containing an isomorphic image of for all such that , there exists a unique injective ring homomorphism such that and

Note that there are some abuse of language in this proposition. When I say I identify with where is the inclusion map from to Also, when I write I mean that takes the isomorphic image of in to the isomorphic image of in ie this diagram commutes

**Proof:** Recall that means the quotient of by the principal ideal generated by Since clearly no non-zero element of is a multiple of in the canonical epimorphism is injective when restricted to Moreover, taking to be the equivalence class of in we have

so property 1 is satisfied. To prove 2, suppose that is another ring containing having an element such that The universal property for polynomial rings assures us that there exists a unique ring homomorphism such that and Since and is irreducible in iff iff for some So and by the first isomorphism theorem (more exactly the universal property of quotient rings), we have a unique isomorphism

providing us with a unique injective ring homomorphism as wanted.

QED

This shows that the first candidate we considered (when was thought of as a subring of another ring and the new element belonged to ), namely obtained with the evaluation morphism, is isomorphic to the quotient where is the irreducible polynomial for in For example and Note that these isomorphisms might not be unique! Indeed, we deduce from the proposition that there are exactly distinct embedding of such a quotient in the larger field where is the number of distinct roots of the polynomial in the larger field. For example, there are two distinct embeddings of in since possess two distinct roots of namely and

I will conclude this post with a very useful observation. Note that if we restrict our attention to the abelian group structure of the polynomials over of degree at most it can be identified with with the group isomorphism

Consider also, for of degree d, the function taking a monic polynomial to its remainder when divided by

Lemma:The function is well defined. ie if is a monic polynomial andwith and both smaller than then

**Proof: **We have

So if (assuring also that ), then but a contradiction.

QED

Proposition:The function is a surjective homomorphism of abelian groups from to .

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

and

with and smaller than d. Then

with So

QED

But as in the proof of the theorem, iff for some So and by the first isomorphism theorem and with the identification given by we have

as abelian groups, where is the degree of So for every monic polynomial of degree this construction can be used to give a different ring structure on the abelian group

Writing for where is a root of in the new ring, we see that the set is a basis for over

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

E. Artin, Galois Theory;

P. Aluffi, Algebra : Chapter 0.

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