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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.

Problem 44 (Not Every Subring Is An Ideal)

Give an example of a ring $R$ and a subring $S$ so that $S$ is not an ideal of $R$. Make sure you prove that $S$ is a subring, but not an ideal.

Problem 45 (A Homomorphism From Z To A Ring With Unity)

Let $R$ be a ring with unity 1. After you have finished the first step below, you should be able to give answers to the remaining parts by referring to the first part and using the properties of ring homomorphisms.

1. Show that the mapping $\phi:\mathbb{Z}\to R$ given by $\phi(n)=n\cdot 1$ is a ring homomorphism.
2. Show that if $R$ has characteristic $n>0$, then $R$ contains a subring isomorphic to $Z_n$.
3. Show that if $R$ has characteristic $n=0$, then $R$ contains a subring that is isomorphic to $Z$.
4. Show that for any positive integer $m$, the mapping of $\phi:\mathbb{Z}\to \mathbb{Z}_m$ given by $x\to x$ mod $m$ is a ring homomorphism.

Problem 46 (Every Field Contains A Subfield Isomorphic To $\mathbb{Z}_p$ or $\mathbb{Q}$)

Suppose that $F$ is a field.

1. If $F$ has prime characteristic $p$, show that $F$ contains a subfield isomorphic to $Z_p$.
2. If $F$ has characteristic 0, show that $F$ contains a subfield isomorphic to $\mathbb{Q}$.

Problem 47 (The Remainder Theorem)

Let $F$ be a field, $a\in F$, and $f(x)\in F[x]$. Prove that $f(a)$ is the remainder in division of $f(x)$ by $x-a$.

Problem 48 (The Factor Theorem)

Let $F$ be a field, $a\in F$, and $f(x)\in F[x]$. Prove that $a$ is a zero of $f(x)$ if and only if $x-a$ is a factor of $f(x)$.

Problem 49 (Polynomials of degree $n$ have at most $n$ zeros)

Show that a polynomial of degree $n$ over a field has at most $n$ zeros, counting multiplicity.

Problem 50 (We have $I=\left<g(x)\right>$ if and only if $g(x)$ is a polynomial of minimal degree in $I$)

Let $F$ be field and $I$ a nonzero ideal in $F[x]$, and $g(x)$ an element of $F[x]$. Show that $I=\langle g(x)\rangle$ if and only if $g(x)$ is a nonzero polynomial of minimal degree in $I$.

In the previous problem, we showed that any ideal in $F[x]$ is generated by a single polynomial. That's pretty remarkable, in that no matter what polynomials we use to generate an ideal, we can always find a single polynomial that generates the whole ideal. This property turns out to be extremely useful, and as such we'll give any ring with this property a special name.

Definition (Principal Ideal Domain PID)

A principal ideal domain (PID) is an integral domain in which every ideal has the form $\left<a\right> = \{ra|r\in R\}$ for some $a$ in $R$.

We know that $F[x]$ is a principle ideal domain provided $F$ is a field. Are there other principle ideal domains?

Problem 51 (Polynomial Rings Over PIDs need not be PIDs)

Show that $\mathbb{Z}$ is a principle ideal domain. Then show that $\mathbb{Z}[x]$ is not a principle ideal domain.

Problem 52 (The Degree Of A Product Of Polynomials)

Suppose that $D$ is an integral domain, and suppose that $f(x),g(x)\in D[x]$.

1. Prove that $\deg(f(x)\cdot g(x)) = \deg(f(x))+\deg(g(x))$.
2. Then give an example of a commutative ring $R$ and two polynomials so that $\deg(f(x)\cdot g(x)) < \deg(f(x))+\deg(g(x))$.

For more problems, see AllProblems