Ideals $\neq0$ in $F[x]$ & Euclidean domains are generated by elements $\neq0$ of minimal degree (Euclidean size), so Euclidean $\Rightarrow$ PID

Question

Consider the following

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

Now it gets completely foggy what is meant by "minimum degree". I constantly come across the "minimal polynomial". This seems to be linked, but it seems strange that there is a division here.

Could someone give a pointer to what the second line actually means. Maybe some example or diagram (if possible) would help a lot here.

Answer 1

Consider $I=\{p(x) \in \Bbb{R}[x] \, | \, p(0)=0 \}$. It is easy to see that this is an ideal of $\Bbb{R}[x]$. For example, $x^2+x, x^{100}-7x$, they are all members of $I$. But is there a non-zero polynomial with smallest degree that lives in here?

If you think a bit more you will see that $x \in I$ is an example of a non-zero polynomial of least degree that lives in $I$. Furthermore every member of $I$ can be represented as $xk(x)$ for some $k(x) \in \Bbb{R}[x]$. Thus one can reason that $I=\langle x \rangle$.

Answer 2

In this situation, when they say "$g(x)$ is a nonzero polynomial of minimum degree in $I$" they simply mean that of all of the polynomials in $I$, $g(x)$ has the smallest degree. Recall that for polynomials in one variable that the degree is the maximal power of $x$ appearing in $g(x)$ that has a nonzero coefficient. In other words, if $f(x) = f_nx^n+f_{n-1}x^{n-1}+...+f_1x+f_0$ and $f_n\neq 0$ then the degree of $f(x)$ is $n$.

Answer 3

Theorem $ $ An ideal $I\neq 0\,$ in a Euclidean domain (i.e. having division with smaller remainder) is generated by any element $\,0\neq \color{0a0}{g\in I}\,$ with min Euclidean size (= least polynomial degree in OP).

Proof $\ $ The key idea is ideals are closed under remainder (mod), so a minimal size $\,\color{#0a0}{g\in I}\,$ must $ \rm\color{#c00}{divide}$ every $\,f\in I\,$ (else the remainder $\,0\neq f\bmod g = f-q\,g\in I\,$ and is smaller size than $\,g,\,$ contra minimality of $\,g).\,$ Thus $\,\color{#c00}{(g)\supseteq I}\color{#0a0}{\supseteq (g)}\,$ so $\,I = (g).\ \ \small\bf QED$

The descent in this proof can be interpreted constructively as computing a generator of $\,I\,$ by computing the gcd of its elements (by taking repeated remainders as in the Euclidean algorithm). Thus "minimal" is with respect to the size measure used for remainders in (Euclidean) division with smaller remainder.

Remark $ $ The minimal polynomial of an element $\:\!\alpha\:\!$ algebraic over a field $F$ is indeed a special case, since it is the (monic) minimal degree element of the ideal of polynomials $\,f\in F[x]\,$ with $\,f(\alpha) = 0$. As above, it can be viewed as the gcd of all such polynomials having $\alpha$ as a root, and the gcd yields $\rm\color{#c00}{descent\!:}$ $\,f_1(\alpha)=0=f_2(\alpha)\,\Rightarrow\, g(\alpha) = 0,\,$ $\ g := \gcd(f_1,f_2) = af+bg,\,$ and the gcd will have $\rm\color{#c00}{smaller}$ degree than $f_1,f_2\,$ if they are incomparable (i.e. if neither divides the other). If we disassemble the gcds into assembly language = iterated remainders (mods) then this is the same descent by remainder that is used in the above proof that $I$ is principal.

Generalization the idea extends to PIDs: (Dedekind-Hasse criterion) a domain $\,D\,$ is a PID iff for $\:0\neq a, b \in D,\:$ either $\:a\:|\:b\:$ or some $D$-linear combination $\:ad+bc\:$ is "smaller" than $\,a.\,$ In a PID we can choose as a "size" measure the number of prime factors (counting multiplicity).