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Problem 35 (When Is A Polynomial Factor Ring An Integral Domain)

Let $R=\mathbb{Z}[x]$. Suppose that $A=\left<p(x)\right>$ is a principle ideal generated by a single polynomial.

1. Find a polynomial $p(x)$ of degree 2 such that $R/A$ is not an integral domain. Then find one of degree 3 and degree 4.
2. Find a polynomial $p(x)$ of degree 3 such that $R/A$ is an integral domain.
3. If you know that $R/A$ is not an integral domain, what do you know about $p(x)$?
4. State a simple condition on $p(x)$ that is equivalent to $R/A$ being an integral domain.

Problem 37 (Properties Of Ring Homomorphisms)

Let $\phi:R\to S$ be a ring homomorphism. Prove the following properties.

1. For any $r\in R$ and any positive integer $n$, $\phi(n\cdot r) = n\phi(r)$ and $\phi(r^n) = (\phi(r))^n$.
2. If $A$ is a subring of $R$, then $\phi (A) = \{\phi(a)|a\in A\}$ is a subring of $S$.
3. If $A$ is an ideal, then $\phi(A)$ is an ideal of $\phi(R)$.
4. If $B$ is an ideal of $S$, then $\phi^{-1}(B) = \{r\in R|\phi(r)\in B\}$ is an ideal of $R$.
5. If $R$ is commutative, then $\phi(R)$ is commutative.
6. If $R$ has a unity 1, if $S\neq \{0\}$, and if $\phi$ is onto, then $\phi(1)$ is the unity of $S$.
7. We know $\phi$ is an isomorphism if and only if $\phi$ is onto and $\ker\phi = \{r\in R|\phi(r)=0\} = \{0\}$.
8. If $\phi$ is an isomorphism from $R$ onto $S$, then $\phi^{-1}$ is an isomorphism from $S$ onto $R$.

Theorem (First Isomorphism Theorem For Rings)

Let $\phi$ be a ring homomorphism from $R$ to $S$. The mapping from $R/\ker\phi$ to $\phi(R)$, given by $r+\ker\phi \to \phi(r)$ is an isomorphism. In symbols, we have $R/\ker\phi\cong \phi(R).$

Problem 38 (First Isomorphism Theorem For Rings Proof)

Prove the first isomorphism theorem for rings. You'll need to first prove that the map defined in the statement of the theorem is well-defined, then you can use the properties of ring homomorphisms to show that the map is an isomorphism.

Problem 39 (Some Polynomial Factor Rings)

Complete the following:

1. Use the first isomorphism theorem to prove that $\mathbb{Z}[x]/\left<x^2+1\right>$ is isomorphic to $\mathbb{Z}[i]$. From this we know that $\mathbb{Z}[x]/\left<x^2+1\right>$ is an integral domain and not a field.
2. Is $\mathbb{Q}[x]/\left<x^2+1\right>$ an integral domain? a field?
3. Is $\mathbb{R}[x]/\left<x^2+1\right>$ an integral domain? a field?
4. Is $\mathbb{C}[x]/\left<x^2+1\right>$ an integral domain? a field?

The next few definitions are things you already know, I am just including them here for completeness. Make sure you read the definition of the degree of a polynomial.

Definition (Polynomial Ring With Coefficients In R)

Let $R$ be commutative ring. The set of formal symbols $$R[x] = \{a_nx^n+\cdots+a_1x+a_0|a_i\in R, n\in Z^+\}$$ is called the ring of polynomials over $R$ in the indeterminate $x$. Two elements are considered equal if and only if the have the same coefficients. Addition is defined component wise, and multiplication is defined using regular polynomial distribution.

Definition (Degree Of A Polynomial)

If $f(x) = a_nx^n+\cdots+a_1x+a_0$ with $a_n\neq 0$ then we say the degree of $f(x)$ is $n$. The polynomial $f(x)=0$ has no degree (it is not degree zero).

Let's now prove that given any polynomials $f(x)$ and $g(x)$, we can always perform long division and write $f(x)=q(x)g(x)+r(x)$ where the degree of $r$ is less than the degree of $g$, or $r=0$. To figure out a proof, just pretend like you are going to do long division. What would your first step be? You should be able to reduce the degree of $f(x)$, and then use induction on the degree of $f(x)$.

Theorem (Division Algorithm For Polynomials)

Let $F$ be a field and let $f(x)$ and $g(x)\in F[x]$ with $g(x)\neq 0$. Then there exist unique polynomials $q(x)$ and $r(x)$ such that $f(x)=g(x)q(x)+r(x)$ and either $r(x)=0$ or $\deg r(x) < \deg g(x)$.

Problem 40 (Division Algorithm For Polynomials Proof)

Prove Theorem Division Algorithm For Polynomials.

A lot of our work has been focused on determining when a ring must be a field. If it's a finite ring, there's a really easy way to immediately throw out the possibility that the ring is a field, by just counting elements. If the order of the ring is not a power of a prime, then it can't be a field. The next problem has you show this.

Problem 41 (The Order Of A Finite Field)

Suppose that $F$ is a finite field of characteristic $p$. Show that the order of $F$ must be $p^n$ for some integer $n$.

Suppose we know that $D$ is an integral domain. If $D$ is finite, we've already shown it must be a field. However, if $D$ is not finite, can we somehow obtain a field from $D$? As an example, we already know that we can obtain the rationals $\mathbb{Q}$ from $\mathbb{Z}$ by defining division, in that we say $a/b=c/d$ if and only $ad=bc$ where $b\neq 0$ and $d\neq 0$. We'll show that this approach allows us to take any integral domain and from it create a field (called the field of quotients).

To obtain this field of fractions, let's first review what an equivalence relation is, as we use equivalence relations to define division.

Definition (Equivalence Relation)

Let $S$ be set. Let $\cong$ be a relation on $S$, meaning $\cong$ is a collection $\mathscr{C}$ of ordered pairs of $S$. We way that $A\cong B$ if and only if the ordered pair $(A,B)$ is an element of $\mathscr{C}$. We say that $\cong$ is an equivalence relation if and only if

1. (Reflexive) For every $A\in S$, we know that $A\cong A$ (so the ordered pair $(A,A)$ is always in $\mathscr{C}$).
2. (Symmetric) If $A\cong B$, then $B\cong A.$
3. (Transitive) If $A\cong B$ and $B\cong C$, then $A\cong C$.

Problem 42 (The Rational Are Obtained From The Integers From An Equivalence Relation)

Consider the set of ordered pairs $S=\{(a,b)\mid a,b\in\mathbb{Z},b\neq0\}$. Define a relation on $S$ by saying that $(a,b)\cong(c,d)$ if and only if $ad=bc$. We'll generally write $(a/b)=(c/d)$ to mean that $(a,b)\cong(c,d)$.

1. Prove that the relation above is an equivalence relation.
2. If we replace $\mathbb{Z}$ with any integral domain $D$, prove that we still obtain an equivalence relation on $S$. (If your work on part 1 didn't refer to the integers specifically, then this part should automatically follow.)

Problem 43 (The Field Of Quotients Of An Integral Domain)

Consider again the set of ordered pairs $S=\{(a,b)\mid a,b\in\mathbb{Z},b\neq0\}$, together with the equivalence relation $(a,b)\cong(c,d)$ if and only if $ad=bc$. Let $F$ be the set of equivalence classes of $S$ under the relation $\cong$. We can write an element in $F$ as $ [a/b] $ where the brackets remind us that there are infinitely many ways to represent elements in $F$ (for example we know $ [2/4]=[3/6] $ because $2\cdot 6=3\cdot 4$). On the set $F$, define the operations $$ [a/b]+[c/d]=[(ad+bc)/(bd)]\quad \text{and}\quad [a/b]\cdot[c/d]=[ac/bd]. $$

1. Prove $+$ and $\cdot$ are binary operations on $F$. This will require you show that if $ [a/b] \cong [a'/b']$ and $ [c/d] \cong [c'/d']$ then we must have $ [(ad+bc)/(bd)] \cong [(a'd'+b'c')/(b'd')] $, and a similar fact for multiplication.
2. Prove that $(F,+,\cdot)$ is a field.
3. If we replace $\mathbb{Z}$ with any integral domain $D$, prove that $(F,+,\cdot)$ is a field.

For more problems, see AllProblems