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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 $f(x),g(x)\in D[x]$.
- Prove that $\deg(f(x)\cdot g(x)) = \deg(f(x))+\deg(g(x))$.
- 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