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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 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 $\deg(f(x)\cdot g(x)) < \deg(f(x))+\deg(g(x))$.

We'd like to know when we obtain polynomials $p(x)\in F[x]$ such that $F[x]/\left<p(x)\right>$ is a field. We already know this means that $I=\left<p(x)\right>$ must be a maximal ideal in $F[x]$. We've also seen that if we can factor $p(x)$ as the product of two polynomials (that don't have multiplicative inverses), then $I$ is not maximal. Let's make a definition to make this precise.

Definition (Irreducible Polynomial Reducible Polynomial)

Let $D$ be an integral domain. A polynomial $f(x)$ from $D[x]$ that is neither the zero polynomial nor a unit in $D[x]$ is said to be irreducible over $D$ if, whenever $f(x)$ is expressed as a product $f(x)=g(x)h(x)$, with $g(x)$ and $h(x)$ from $D[x]$, then $g(x)$ or $h(x)$ is a unit in $D[x]$. A nonzero, nonunit element of $D[x]$ that is not irreducible over $D$ is called reducible over $D$.

In general, it's a hard problem to determine when a polynomial is reducible or irreducible. However, there are a few cases where it's easy.

Problem 53 (Reducibility Test For Degrees 2 And 3)

Let $F$ be a field. If $f(x)\in F[x]$ and deg $f$ is 2 or 3, then prove that $f(x)$ is reducible over $F$ if and only if $f(x)$ has a zero in $F$.

Before we look at any other reducibility tests, let's prove a key theorem, namely that $p(x)$ is irreducible if and only if $\left<p(x)\right>$ is maximal. We really want to know when we can guarantee that $F[x]/\left<p(x)\right>$ is a field.

Problem 54 (We Know $\left<p(x) \right>$ Is Maximal Iff $p(x)$ Is Irreducible)

Let $F$ be a field and let $p(x)\in F[x]$. Prove that $\left<p(x) \right>$ is a maximal ideal if and only if $p(x)$ is irreducible over $F$.

This next definition just gets rid of a complication from some polynomials, by removing from the polynomial any common factors.

Definition (Content Of A Polynomial Primitive Polynomial)

The content of a polynomial in $\mathbb{Z}[x]$ is the greatest common divisor of the coefficients. A primitive polynomial has content 1.

Problem 54.5 (The Product Of Primitives Is Primitive)

Prove that the product of two primitive polynomials is primitive.

As you work through the problems in the next few weeks, you'll want to pay close attention to the assumptions. In some problems we assume that we are working in a field. In some problems, we assume that we are working in an integral domain. The next problem shows that if you can show something is reducible over the field $\mathbb{Q}$, then it is reducible over $\mathbb{Z}$.

As you work through the problems in the next few weeks, you'll want to pay close attention to the assumptions. In some problems we assume that we are working in a field. In some problems, we assume that we are working in an integral domain. The next problem shows that if you can show something is reducible over the field $\mathbb{Q}$, then it is reducible over $\mathbb{Z}$.

Problem 55 (Reducibility Over Q Implies Reducibility Over Z)

Let $f(x)\in \mathbb{Z}[x]$. Prove that if $f(x)$ is reducible over $\mathbb{Q}$, then $f(x)$ is reducible over $\mathbb{Z}$.

The contrapositive to the previous problem is extremely powerful, namely if a polynomial with integer coefficients is not reducible over $\mathbb{Z}$, then it is not reducible over $\mathbb{Q}$. For this reason, we'll now study irreducibility tests over $\mathbb{Z}$.

Problem 56 (Mod P Irreducibility Test)

Let $p$ be a prime and suppose that $f(x)\in \mathbb{Z}[x]$. Let $\bar f (x)$ be the polynomial in $\mathbb{Z}_p[x]$ obtained by reducing the coefficients of $f(x)$ modulo $p$. Prove that if if $\bar f (x)$ is irreducible over $\mathbb{Z}_p$ and $\text{deg }\bar f(x) = \text{deg }f(x)$, then $f(x)$ is irreducible over $\mathbb{Q}$.

Problem 57 (Eisenstein's Criterion)

Let $f(x)=a_nx^n + \cdots +a_1x +a_0$. Prove that if there is a prime $p$ such that $p$ divides every coefficient but $a_n$ and $p^2$ does not divide $a_0$, then $f(x)$ is irreducible over $\mathbb{Q}$.

Problem 58 (Rational Root Test)

Suppose that $$f(x) = a_nx^n+\cdots +a_1x+a_0\in \mathbb{Z}[x],$$ with $a_n\neq 0$. Prove that if $r$ and $s$ are relatively prime and $f(r/s)=0$, then we must have $r\mid a_0$ and $s\mid a_n$.

Problem 59 (Irreducibles Behave Like Prime Numbers)

Let $F$ be a field and suppose that $p(x)\in F[x]$ is irreducible over $F$. Suppose also that $p(x)$ divides the product $a_1(x)a_2(x)\cdots a_n(x)$ where $a_i(x)\in F[x]$ for each $i$. Prove that $p(x)$ must divide $a_k(x)$ for some $k$.

For more problems, see AllProblems