Homomorphisms and Polynomial Rings

Section 8.4 Homomorphisms and Polynomial Rings

Subsection Homomorphisms

“When you have a map, you know where to go.”
―Shakira

Definition 8.46. Ring Homomorphism.

If \(R\) and \(S\) are rings, a ring homomorphism from \(R\) to \(S\) is a function \(\varphi: R \to S\) that satisfies:

\(\varphi(x + y) = \varphi(x) + \varphi(y)\) for all \(x,y \in R\text{,}\)
\(\varphi(x \cdot y) = \varphi(x) \cdot f(y)\) for all \(x,y \in R\text{.}\)

Ring homomorphisms are often referred to as ring maps.

So basically the same as a group homomorphism, but with an extra operation that we need to preserve.

Definition 8.47. Ring Isomorphism.

A ring homomorphism \(f: R \to S\) that is bijective is called a ring isomorphism. Two rings \(R\) and \(S\) are isomorphic, written \(R \cong S\text{,}\) if there is an isomorphism from \(R\) to \(S\text{.}\)

Ring isomorphisms are not usually referred to as super ring maps, but it would be cool if they were.

Proposition 8.48. Isomorphism Invariants.

The following are ring isomorphism invariants:

all group isomorphism invariants of the additive group, including the isomorphism class (i.e., if \(R\cong S\) then \((R,+)\cong(S,+)\)).
being unital, being commutative, division ring, field, integral domain
the number of zerodivisors.
if \(R\) is unital, all group isomorphism invariants of the group of units, including the isomorphism class (i.e., if \(R\cong S\) then \((R^\times,\cdot)\cong(S^\times,\cdot)\)).
the isomorphism type of the center (i.e., if \(R\cong S\) then \(Z(R)\cong Z(S)\)).

Example 8.49. Examples of Ring Maps.

The identity map is a ring isomorphism.
The zero map is a ring map.
Let \(S\) be a subring of a ring \(R\text{.}\) The inclusion mapping of \(S\) into \(R\) is a ring homomorphism.
Projection maps are ring homomorphisms.

Lemma 8.50. Properties of Ring Maps.

If \(\varphi: R \to S\) is a ring homomorphism, then

\(\varphi(0_R) = 0_S\) and \(\varphi(-x)=-f(x)\text{.}\)
if \(R\text{,}\) \(S\) are unital then \(\varphi(1_R)\) can be either \(0_S, 1_S\) or a zerodivisor.
If \(\varphi(1_R)=1_S\) and \(u\in R^\times\) then \(\varphi(u^{-1})=\varphi(u)^{-1}\text{.}\)
If \(\varphi: R \to S\) and \(\phi:S\to T\) are ring homomorphisms (or isomorphisms, respectively), then \(\phi\circ \varphi:R\to T\) is a ring homomorphism (or isomorphism).

Proof.

Since \(1_R1_R=1_R\) we have \(\varphi(1_R)\varphi(1_R)=\varphi(1_R)\text{,}\) thus
\begin{equation*} \varphi(1_R)(\varphi(1_R)-1_S)=0_S. \end{equation*}
Now either \(\varphi(1_R)=0_S\) or \(\varphi(1_R)-1_S=0_S\) (which yields \(\varphi(1_R)=1_S\)) or both of these are nonzero and then they are complementary zerodivisors (in particular, \(\varphi(1_R)\) is a zerodivisor).

Exercise 8.51. Equivalent Field Characterizations.

Let \(R\) be a nontrivial ring. Then \(R\) is a field if and only if every homomorphism of \(R\) into a nonzero ring \(S\) is injective.

Exercise 8.52. Isomorphisms and Idempotents.

Suppose \(R\) is commutative and \(e\) an idempotent. Let \(Re:=\{re\mid r\in R\}\text{.}\) Prove that the map \(\phi:R\to Re \times R(1-e)\) given by \(\phi(r)=(re, r(1-e))\) is a ring isomorphism.

Exercise 8.53. More Isomorphisms.

Let \(a, b\) be squarefree integers and set \(R=\Z[\sqrt{a}]\) and \(S=\Z[\sqrt{b}]\text{.}\)

There is a group isomorphism \((R,+) \cong (S,+)\text{.}\)
There is a ring isomorphism \(R \cong S\) if and only if \(a = b\text{.}\)

Subsection Polynomial Rings

“I had a polynomial once. My doctor removed it.”
―Michael Grant

Definition 8.54. Polynomial Ring.

For any commutative ring \(R\text{,}\) the polynomial ring in the variable \(x\text{,}\) written \(R[x]\text{,}\) is the set

\begin{equation*} R[x]=\{a_nx^n+\ldots+a_0\mid n\in \Z, n\geq 0, a_i\in R\} \end{equation*}

with addition defined by

\begin{equation*} (a_nx^n+\ldots+a_0)+(b_nx^n+\ldots+b_0)=(a_n+b_n)x^n+\ldots++(a_0+b_0) \end{equation*}

and multiplication defined by

\begin{equation*} (a_nx^n+\ldots+a_0)(b_mx^n+\ldots+b_0)=\sum_{k=0}^{n+m}\left(\sum_{i=0}^k a_ib_{k-i}\right)x^i. \end{equation*}

For any commutative ring \(R\text{,}\) the polynomial ring in \(x_1, \dots, x_n\text{,}\) written \(R[x_1, \dots, x_n]\text{,}\) is defined inductively as \(R[x_1, \dots, x_n]=R[x_1, \dots, x_{n-1}][x_n]\text{,}\) but more easily thought of as the set consisting of (finite) sums of the form

\begin{equation*} R[x_1, \dots, x_n]=\left\{\sum_{e_1, \dots, e_n \in \Z_{\geq 0}} r_{e_1, \dots, e_n} x_1^{e_1} x_2^{e_2} \cdots x_n^{e_n}\right\} \end{equation*}

with addition and multiplication defined by rules similar to the ones seen above.

Remark 8.55.

One often views \(R\) as the subring of \(R[x_1, \dots, x_n]\) consisting of the constant polynomials.

Let’s remind ourselves of some classic notions of polynomials.

Definition 8.56. Degree.

Let \(f(x)\in R[x]\text{,}\) \(f(x)\neq0\text{.}\) Say \(f(x)=a_nx^n+\cdots+a_1x+a_0\text{,}\) where \(a_n\neq0\text{.}\) Then \(n\) is the degreeof \(f\text{.}\)

Definition 8.57. Monic.

A polynomial \(f(x)=a_nx^n+\cdots+a_1x+a_0\) is monic if \(a_n=1\text{.}\)

Please do not confuse this with the notion of a yonic polynomial, which does not yet exist and should probably stay that way. Algebraists are already on very thin ice for some of the things they’ve named over the years.

Proposition 8.58. Polynomials, Domains, Degrees, Units.

If \(R\) is an integral domain, then

\(R[x]\) is an integral domain
for any nonzero polynomials \(f,g\in R[x]\text{,}\) \(\deg(fg)=\deg(f)+\deg(g)\)
the units of \(R[x]\) are the units of \(R\) (\(R[x]^\times=R^\times\))

Its been far too long since we’ve had ourselves a universal mapping property, I think we’ve earned one. As a treat.

Theorem 8.59. UMP for Polynomial Rings.

Let \(R\) and \(S\) be commutative rings, \(\phi: R \to S\) is a ring homomorphism and \(s_1, \dots, s_n\) arbitrary elements of \(S\text{.}\) Then there exists a unique ring homomorphism

\begin{equation*} \tilde{\phi}: R[x_1, \dots, x_n] \to S \end{equation*}

such that \(\tilde{\phi}|_R = \phi\) and \(\tilde{\phi}(x_i) = s_i\) for all \(i\text{,}\) namely

\begin{equation*} \tilde{\phi}\left(\sum_{e_1, \dots, e_n \geq 0} r_{e_1, \dots, e_n} x_1^{e_1} \cdots x_n^{e_n}\right) =\sum_{e_1, \dots, e_n \geq 0} \phi(r_{e_1, \dots, e_n}) s_1^{e_1} \cdots s_n^{e_n} \end{equation*}

Proof.

Let’s observe first that if such a map exists it is unique. For if \(\tilde\phi: R[x_1, \dots, x_n] \to S\) is a ring map extending \(\phi\) and sending \(x_i\) to \(s_i\text{.}\) Then

\begin{equation*} \begin{aligned*} \tilde\phi(\sum_{e_1, \dots, e_n \geq 0} r_{e_1, \dots, e_n} x_1^{e_1} \cdots x_n^{e_n}) &= \sum_{e_1, \dots, e_n \geq 0} \phi(r_{e_1, \dots, e_n}) \phi(x_1)^{e_1} \cdots \phi(x_n)^{e_n} \\ &= \sum_{e_1, \dots, e_n \geq 0} \phi(r_{e_1, \dots, e_n}) s_1^{e_1} \cdots s_n^{e_n}, \end{aligned*} \end{equation*}

using that \(\tilde\phi\) preserves \(+\) and \(\cdot\text{.}\)

For existence, let’s assume \(n = 1\) at first. Given \(\phi: R \to S\) and \(s \in S\text{,}\) define

\begin{equation*} \tilde{\phi}: R[x] \to S \end{equation*}

\begin{equation*} \tilde{\phi}(\sum_i r_i x^i) = \sum_i \phi(r_i) s^i. \end{equation*}

It is elementary (but tedious) to check \(\tilde{\phi}\) really is a ring homomorphism. The fact that it restricts to \(\phi\) is clear, however.

For the general case, we proceed by induction on the number of variables \(n\text{.}\) The induction hypothesis shows that there is a ring homomorphism

\begin{equation*} \psi: R[x_1, \dots, x_{n-1}] \to S \end{equation*}

such that \(\psi|_R = \phi\) and \(\psi(x_i) = s_i\text{,}\) \(i = 1, \dots, n-1\text{.}\) Applying the \(n=1\) case to \(\psi\) gives

\begin{equation*} \tilde{\psi}: (R[x_1, \dots, x_{n-1}])[x_n] \to S \end{equation*}

with \(\tilde\psi|_{R[x_1, \dots, x_{n-1}]} = \psi\) and \(\tilde{\psi}(x_n) = s_n\text{.}\) Setting \(\tilde\phi=\tilde\psi\) gives a map \(\tilde{\phi}\) with the needed properties.

Exercise 8.60. Evalutation Homomorphism.

If \(R\subseteq S\) are commutative rings with \(1\neq 0\) and \(a\in S\text{,}\) then the evaluation at \(a\) function \(\phi:R[x]\to S\) given by \(\phi(f(x))=f(a)\) is a ring homomorphism.

Example 8.61. Ring Maps and Coefficients.

Given a ring map \(\phi: R \to S\) between commutative rings, we may apply Theorem 8.59 to the composition \(R \xra{\phi} S \hookrightarrow S[x]\) using the element \(s = x\) of \(S[x]\) to get an induced ring map

\begin{equation*} \tilde{\phi}: R[x] \to S[x] \end{equation*}

that sends \(\sum_i r_i x^i\) to \(\sum_i \phi(r_i) x^i\text{.}\) That is, the map \(\tilde{\phi}\) applies \(\phi\) to the coefficients of a polynomial. This can be generalized to more than one variable in the obvious way.

Exercise 8.62. Reduction Homomorphism.

Continuing with Example 8.61, we could have \(S = R/I\) for an ideal \(I\) of \(R\) and \(\phi\) could be the quotient map. Then

\begin{equation*} \tilde{\phi}: R[x_1, \dots, x_n] \to (R/I)[x_1, \dots, x_n] \end{equation*}

takes a polynomial and “reduces” its coefficients modulo \(I\text{.}\) We will usually denote the image of \(f(x)\) through this reduction homomorphism by \(\ov{f}(x)\text{.}\)

Luckily for us, it turns out that the Division Algorithm holds in polynomial rings as well.

Theorem 8.63. Polynomial Division Algorithm.

Let \(R\) be a commutative ring with identity and \(f,g\in R[x]\text{.}\) Assume the leading coefficient of \(g\) is a unit in \(R\text{.}\) Then there exists a unique \(q,r\in R[x]\) such that \(f=gq+r\text{,}\) and \(\operatorname{deg}r<\operatorname{deg}g=n\text{.}\)

Theorem 8.64. Factor Theorem.

Let \(R\) be a commutative ring with identity, \(f(x)\in R[x]\) and \(a\in R\text{.}\) Then \(f(a)=0\) if and only if \((x-a)|f(x)\text{.}\)

Summary

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