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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