Exploration 3: Polynomials

• WTS the product of two nonzero polynomials $f,g$ in $R[x]$ is always nonzero.
• Let $a_n$ and $b_m$ be the leading coefficients of $f,g$ respectively. Since $f,g$ are nonzero, $a_n$ and $b_m$ are nonzero.
• Then the leading coefficient of $fg$ is $a_nb_m$. Since $R$ is an integral domain, this product is nonzero.
• Then since the leading coefficient of $fg$ is nonzero, $fg$ is nonzero, and we are done.

To see this, towards contradiction assume we have any two distinct solutions $q,r$ and $q',r'$ where $f=gq+r=gq'+r$ implies $g(q-q')=r'-r$.

• We know that $\deg r<\deg g$ and therefore the RHS $r'-r$ has degree $<\deg g$.
• Then the LHS $g(q-q')$ has degree $\ge\deg g+\deg(q-q')$.
• Since the LHS and RHS should have the same degree, this is a contradiction.

You can only divide by nonzero polynomials in $R[x]$ whose leading coefficient is a unit in $R$, but every leading coefficient is a unit when $R$ is a field. Thus you can divide by any nonzero polynomial in $R[x]$, making $R[x]$ a Euclidean domain.

Following from the division algorithm, you can only divide by polynomials whose leading coefficient is a unit, and $1$ is always a unit. Therefore you can always divide by monic polynomials.

Since every nonzero element in a field is a unit, you can already divide by every element in a field: $a=qb$. Therefore all fields have a division algorithm and are Euclidean domains.

The Euclidean condition norm requires that for every $a,b\in R$, we have either $a=qb+r$ or where either $r=0$ or $\phi(r)<\phi(b)$. This is exactly the required division.

• If $f$ and $g$ divide each other, we have $f=gq$ and $g=fq'$.
• $\deg f=\deg gq$ implies $\deg f\ge\deg g$.
• $\deg g=\deg fq'$ implies $\deg g\ge\deg f$.
• Therefore $\deg f=\deg g$, which implies that $q,q'$ are constant polynomials that don’t affect the degree of the product.
• The fact that both $f,g$ are monic shows that the leading coefficient is $1$ before and after multiplying by $q,q$’, which implies that $q,q'$ are both $1$.
• Therefore $f=g$.

• $f$ and $g$ being nonzero in an integral domain implies that neither of them are zero divisors.
• If $f$ and $g$ generate each other, we have $f=gq$ and $g=fq'$, and therefore $f=fqq'$.
• Then $0=f(qq'-1)$ where $f$ is not a zero divisor, implying $qq'-1=0$.
• This means $qq'=1$, i.e. $q,q'$ are units.
• Thus $f=gq$ means $f$ differs from $g$ by a unit.
• The converse is trivial — if $f,g$ differ by a unit $u$, then they generate each other via the unit: $f=ug$ and $g=u^{-1}f$.

Since there is a constant polynomial $r\in R[x]$ for every $r\in R$, and constant polynomials remain unchanged by the evaluation map, every element of $R$ gets mapped to.

• The evaluation map $\varphi_0:R[x]\to R$, which evaluates a polynomial at $0$, has kernel $(x)$. To see this, notice that for every $f\in R[x]$, $\varphi_0(f)=f(0)=0$ implies the constant coefficient is $0$, and therefore $$\begin{aligned} f(x)&=a_nx^n+\ldots+a_2x^2+a_1x+0a_0\\ f(x)&=a_nx^n+\ldots+a_2x^2+a_1x\\ f(x)&=x(a_nx^{n-1}+\ldots+a_2x+a_1)\\ f(x)&\in (x) \end{aligned}$$ so the kernel of $\varphi_0$ is $(x)$.
• The first ring isomorphism theorem says $R[x]/\ker\varphi_0\iso\im\varphi_0$. We just showed $\ker\varphi=(x)$, and since $\varphi_0$ is surjective, $\im\varphi_0=R$, therefore the above becomes $R[x]/(x)\iso R$.

Since $x-a$ is monic, the division algorithm implies some unique $q,r$ such that $f=(x-a)q+r$. Evaluating at $a$ gives $f(a)=(a-a)q+r=r$. Therefore, the remainder $r$ is $f(a)$.

The evaluation $f(a)$ replaces $x$ with $a$. If $f=(x-a)q$, then $f(a)=(a-a)q=0q=0$. Conversely, if $f(a)=0$, then since $x-a$ is monic, by the remainder theorem $f$ divided by $x-a$ results in $f=(x-a)q+f(a)$. But $f(a)=0$, so this becomes $f=(x-a)q$.

This is easily proved by induction by applying the factor theorem $n$ times.

• Let $c/d$ be a rational root of the polynomial $f=a_0+a_1x+a_2x^2+\ldots+a_nx^n\in\ZZ[x]$.
• Assume $c,d$ are coprime integers – if they are not, divide both $c$ and $d$ by their GCD to make them coprime.
• Since $c/d$ is a root of $f$, we have $f(c/d)=0$: $$a_0+a_1(c/d)+a_2(c/d)^2+\ldots+a_n(c/d)^n=0$$
• Multiply both sides by $d^n$: $$a_0d^n+a_1d^{n-1}c+a_2d^{n-2}c^2+\ldots+a_nc^n=0$$
• From here, we can go in two directions. First, isolate the $a_0d^n$ term and factor out $c$ from the other terms: $$c(a_1d^{n-1}+a_2d^{n-2}c+\ldots+a_nc^{n-1})=-a_0d^n$$ This means $c$ is a factor of $-a_0d^n$. Since $c,d$ are coprime integers, $c$ must divide $a_0$.
• Second, isolate the $a_nc^n$ term instead and factor out $d$ from the other terms: $$d(a_0d^{n-1}+a_1d^{n-2}c+a_2d^{n-3}c^2+\ldots+a_{n-1}c^{n-1})=-a_nc^n$$ Similarly, this means $d$ is a factor of $-a_nc^n$. Since $c,d$ are coprime integers, $d$ must divide $a_n$.

• Try to interpret $\sqrt{m}$ as a rational root ($\in\QQ$) of some polynomial $\in\ZZ[x]$.
• If $\sqrt{m}\in\QQ$, it would be a rational root $c/d$ of the polynomial $x^2-m$. We know $\sqrt{m}$ is one such root.
• By Rational Roots Theorem, $c\mid m$ and $d\mid 1$, therefore $d=\pm 1$. Then $c/d=\pm c$ so the root $\sqrt{m}$ must be some integer $\pm c$. Therefore $m$ is the square of some integer.

• (Same proof as above)
• If $\sqrt[n]{m}\in\QQ$, it would be a rational root $c/d$ of the polynomial $x^n-m$. We know $\sqrt[n]{m}$ is one such root.
• By Rational Roots Theorem, $c\mid m$ and $d\mid 1$, therefore $d=\pm 1$. Then $c/d=\pm c$ so the root $\sqrt[n]{m}$ must be some integer $\pm c$. Therefore $m$ is the $n$th power of some integer.

The proof often requires complex analysis or topology. Here is a proof due to Artin, trying not to use many ideas outside of ring theory:

• It is a theorem that if you treat evaluation of the polynomial $f\in\CC[x]$ as a map $\CC\to\CC$, then $f$ is continuous.
• Evaluate $f$ at each point on a circle $C_r$ of radius $r$ around the origin of the complex plane. Since $f$ is continuous, the images $f(C_r)$ represent some loop on the complex plane. Each point on the circle $C_r$ can be represented in polar coordinates as $z=re^{i\theta}$ for some angle $\theta$. Let its corresponding point on the loop $f(C_r)$ be $f(z)$.
• There are two loops we want to consider:
  • First, with $r$ approaching $0$, we know the images $f(C_r)$ are all close to the constant coefficient $c_0$ of $f$. So $f(C_r)$ is a tiny loop around $c_0$. We assume $c_0$ is nonzero – if it’s zero then $f(0)=0$ implies $0$ is a root and we are done.
  • Second, with $r$ approaching $\infty$, we know the images $f(C_r)$ are very large but also represent some loop. Since for large values the leading term $c_nz^n$ of $f(z)$ dominates the other terms, which we can write as $f(z)-c_nz^n$, we know $|f(z)-c_nz^n|<|c_nz^n|$ and (because $|c_nz^n|=c_n|r^ne^{in\theta}|=c_nr^n$) therefore $|f(z)-c_nz^n|<c_nr^n$. This implies the distance between $f(z)$ and the leading term $c_nz^n$ is always less than $c_nr^n$, no matter what $\theta$ is. If you imagine $\theta$ increasing, then as $c_nz^n=c_nr^ne^{in\theta}$ walks a circle of radius $c_nr^n$ around the origin, while $f(z)$ (being continuous) is following a distance less than $c_nr^n$ behind and therefore the loop it traces must also enclose the origin.
• Importantly, the first loop does not enclose the origin, while the second one does. Since $f$ is continuous, varying $r$ within $(0,\infty)$ will continously vary the corresponding loop between one that doesn’t enclose the origin and one that does. So at some $r$ between $0$ and $\infty$, the resulting loop crosses the origin, and therefore we get a root $f(re^{i\theta})=0$ for some $\theta$.

This is repeated application of the FTA. You apply FTA to find a root $a$, factor it out with the factor theorem to get $f=(x-a)g$, and repeat with $g$ until $g$ is a constant polynomial $k$. Since we’re always factoring out a monic polynomial $x-a$, the leading coefficient never changes, and so the final $k$ is the leading coefficient of $f$.

• Since complex conjugation $\overline{x}$ fixes the real numbers, $f(\overline{a+bi})=\overline{f(a+bi)}$ when $f$ is a real polynomial.
• Therefore, if $f(a+bi)=0$, then $\overline{f(a+bi)}=\overline{0}$ and therefore $\overline{f(\overline{a-bi})}=f(a-bi)=0$ implies $a-bi$ is also a root of $f$.

• Do the same thing you did for complex polynomials, factoring $f$ into a product $f=k\prod_i(x-u_i)$ where the roots $u_i$ are complex.
• Since the coefficients of $f$ are real, by the Conjugate Root Theorem, the complex roots come in conjugate pairs. Then the product of their corresponding factors, $(x-(a+bi))(x-(a-bi))=x^2-2ax+(a^2+b^2)$, is a monic irreducible real quadratic, i.e one of the $q_i$.
• Therefore, after combining all complex factors into monic real quadratic factors, $f$ can be written in the form $k\prod_i(x-u_i)\prod_iq_i$ where $q_i$ are the product of each conjugate pair of complex factors within the original $u_i$.

• Let $R$ be a Euclidean domain with norm $f$. Then for every element $a\in R$, we can divide $a$ by nonzero $b\in R$ to get $a=bq+r$ where either $r=0$ or $f(r)<f(b)$.
• We can show that for every ideal $I$ in $R$, every element $a$ of $I$ is generated by some element $b$ with the smallest norm $f(b)$. That is, $a=bq$ for some $q\in R$.
• This follows immediately from the division algorithm. We have $a=bq+r$, and since $f(b)$ is minimal by definition, there is no $f(r)$ such that $f(r)<f(b)$, therefore $r=0$.

We identify $x^2$ with $0$, so all the degree $2+$ polynomials get sent to $0$, leaving only the degree $1$ and $0$ polynomials (and the zero polynomial).

• Here we identify $x^2+1$ with $0$. We can understand this as $x=\sqrt{-1}=i$ (in $\CC$).
• Since every resulting polynomial is at most degree $1$, they are in the form $c_0+c_1x$.
• In other words, the elements of $\RR[x]/(x^2+1)$ are $c_0+c_1i$.
• This is exactly how we write elements $a+bi$ of $\CC$, so the two rings are isomorphic.

• Here we identify $x^3-2$ with $0$. We can understand this as $x=\sqrt[3]{2}$ (in $\RR$).
• Since every resulting polynomial is at most degree $2$, they are in the form $c_0+c_1x+c_2x^2$.
• In other words, the elements of $\RR[x]/(x^2+1)$ are $c_0+c_1\sqrt[3]{2}+c_2\sqrt[3]{2}^2$.
• But this is the same as adjoining an element $\sqrt[3]{2}$ to $\QQ$.

• Since $F[x]$ is a PID, the result follows directly from this theorem.
• (Note that if $h$ is reducible, then $F[x]/(h)$ sends $h=fg$ to $0$, meaning there are zero divisors $f,g$ so the result fails to be even an integral domain.)

• Since $f$ has $\alpha$ as a root, factor out $f=(x-\alpha)g$. To show that $\alpha$ is a multiple root, we can show that $g$ also has $\alpha$ as a root, i.e. $g(\alpha)=0$.
• By the product rule, $f'=(x-\alpha)g'+g$. Since $f'(\alpha)=0$, we have $0=(\alpha-\alpha)g'(\alpha)+g(\alpha)$, implying $0=g(\alpha)$, implying $\alpha$ is a root of $g$.

Irreducible polynomials $f$ can only share a factor (itself) with either 0 or polynomials of greater or equal degree. Since the derivative $f'$ always has lesser degree, $f'$ has to be zero in order for $f$ and $f'$ to share a factor.