Math 380, Spring 2018, Assignment 4

I was at the mathematical school, where the master taught his pupils after a method scarce imaginable to us in Europe. The proposition and demonstration were fairly written on a thin wafer, with ink composed of a cephalic tincture. This the student was to swallow upon a fasting stomach, and for three days following eat nothing but bread and water. As the wafer digested the tincture mounted to the brain, bearing the proposition along with it.

- Jonathan Swift, Gulliver's Travels

Read:

1. Section 1.5.
2. Section 2.1.

Carefully define the following terms, then give one example and one non-example of each:

1. Monic polynomial (in $\mathsf{k}[x]$).
2. $\mathrm{gcd}(f,g)$ (where $f,g\in\mathsf{k}[x]$).

Carefully state the following theorems (you do not need to prove them):

1. Theorem concerning the division algorithm.
2. Classification of ideals in $\mathsf{k}[x]$ ("Every ideal in $\mathsf{k}[x]$ is generated by...").
3. Theorem concerning monic generators of ideals in $\mathsf{k}[x]$.
4. Theorem relating $\mathrm{gcd}(f,g)$ to common divisors of $f$ and $g$.
5. Theorem relating $\left\langle f,g\right\rangle$ to $\left\langle g,r\right\rangle$ when $f=gq+r$.

Carefully describe the following algorithms:

1. Division algorithm (to compute quotient and remainder when $f$ is divided by $g$).
2. Euclid's algorithm (to compute $\mathrm{gcd}(f,g)$).
3. Algorithm to replace any system of finitely many univariate equations by a single univariate equation.
4. Algorithm to factor polynomials over $\mathbb{C}$. (Note: this algorithm is conceptually simple but should NOT be used in engineering applications; it is numerically unstable. See books on numerical analysis for more practical algorithms.)

Solve the following problems:

1. Section 1.5, problems 1, 3, 4, 11, and 12.
2. Prove that for any $f_1,\dots,f_s\in\mathsf{k}[x]$, one has $\left\langle f_1,f_2,f_3,\dots,f_s\right\rangle = \left\langle \mathrm{gcd}(f_1,f_2),f_3,\dots,f_s\right\rangle$. (Hint: use the ideal containment criterion from the previous assignment.)
3. Optional (but interesting): section 1.5, problem 17. (You may need to use the results of problems 13, 14, and 15.)

--------------------End of assignment--------------------

Solutions:

1.5.1: It is known that every nonconstant polynomial in $\mathbb{C}[x]$ has a root in $\mathbb{C}$. Let $a_1$ be such a root. By the factor theorem, $f$ is divisible by $x-a_1$ with remainder zero, that is $f=(x-a_1)q$ where $q$ is the quotient. If $q$ is a nonconstant polynomial, repeat this argument with $q$ instead of $f$, otherwise $q$ is the constant c in the desired form. Thus every univariate polynomial of degree $n>0$ over the complex numbers may be written $f=c(x-a_1)\cdots(x-a_n)$ where $a_1,\ldots,a_n,c\in\mathbb{C}$ and $c\neq0$.

1.5.3: We wish to prove that $I=\langle x,y\rangle\subseteq k[x,y]$ is not a principal ideal, that is, that it is not generated by one element. Suppose it were. Let $f$ be the single generator, i.e., $\langle x,y\rangle=\langle f\rangle$. Obviously $f$ cannot be a constant. Then every polynomial in $\langle x,y\rangle$ is equal to $fg$ for some $g\in k[x,y]$. Consider $x\in\langle x,y\rangle$. Then $x=fg$ for some $g$. Taking the degree of both sides we have $\text{deg}(x)=1=\text{deg}(fg)=\text{deg}(f)+\text{deg}(g)$. The only way this can be satisfied is if $f$ or $g$ is a constant and the other is degree 1. Since $f$ cannot be a constant, $g$ is. Say $g=c$, then $x=cf\implies f=\frac{x}{c}$. Since $f$ generates $\langle x,y\rangle$, it must also be that $y=fh$ for some $h\in k[x,y]$. Then $y=\frac{x}{c}h$, but this clearly cannot be satisfied, so we have a contradiction. Thus $\langle x,y\rangle$ is not a principal ideal.

1.5.4: $\text{gcd}(f,g)$ is the monic generator of the ideal $\langle f,g\rangle$, so, if $h=\text{gcd}(f,g)$, then $\langle h\rangle=\langle f,g\rangle$. By definition, this means that $h\in\langle f,g\rangle$, so there exists $A,B\in k[x]$ such that $Af+Bg=h$.

1.5.11: (a) As noted in class, every nonconstant polynomial has a root so $\mathbb{V}(f)\neq \emptyset$ iff $f$ is not a constant.

(b) If $\text{gcd}(f_1,\ldots,f_s)=1$, we have $\langle f_1,\ldots,f_s\rangle=\langle 1 \rangle$ and so $\mathbb{V}(f_1,\ldots,f_s)=\mathbb{V}(1)=\emptyset$. If $\mathbb{V}(f_1,\ldots f_s)=\emptyset$, then $\mathbb{V}(\text{gcd}(f_1,\ldots,f_s))=\emptyset$. This means that $\text{gcd}(f_1,\ldots,f_s)$ is a polynomial in $\mathbb{C}[x]$ which doesn't have a root in $\mathbb{C}$. Therefore it is a constant polynomial. By the definition of the gcd, it is monic, so $\text{gcd}(f_1,\ldots,f_s)=1$.

(c) Determine $\text{gcd}(f_1,\ldots,f_s)$. If this is not equal to $1$ then the affine variety $\mathbb{V}(f_1,\ldots,f_s)$ is not empty.

1.5.12: (a) By inspection of the factored form of $f$, the polynomial vanishes at the points $\{a_1,\ldots a_l\}$ and doesn't vanish at any others. Thus $\mathbb{V}(f)=\{a_1,\ldots, a_l\}$.

(b) We wish to show $\mathbb{I}(\mathbb{V}(f))=\langle f_\text{red}\rangle$. We proceed by mutual inclusion. $\mathbb{I}(\mathbb{V}(f))$ is the set of all polynomials which vanish at $\{a_1,\ldots,a_l\}$. We can clearly see that $f_\text{red}$, as defined, vanishes at all of these points, so $f_\text{red}\in\mathbb{I}(\mathbb{V}(f))\implies \langle f_\text{red}\rangle\subseteq\mathbb{I}(\mathbb{V}(f))$. Now let $g\in\mathbb{I}(\mathbb{V}(f))$. We want to show that $g\in\langle f_\text{red}\rangle$. Since $g\in\mathbb{I}(\mathbb{V}(f))$, $g$ has, among its roots, $\{a_1\ldots,a_l\}$. As it is a univariate polynomial over $\mathbb{C}$, we may write $g$ in the fully factored form \[ g=c(x-b_1)^{n_1}\cdots(x-b_s)^{n_s}\;. \] But it must be that $l$ of the roots are the $a_i$, so we can arrange to write $g$ as \[ g=c(x-a_1)^{n_1}\cdots(x-a_l)^{n_l}(x-b_{s-l})^{n_{s-l}}\cdots(x-b_s)^{n_s}\;. \] By inspection, then, we can see that $g=h f_\text{red}$ for an appropriately choosen $h\in\mathbb{C}[x]$, and so $g\in\langle f_\text{red}\rangle\implies \mathbb{I}(\mathbb{V}(f))\subseteq\langle f_\text{red}\rangle$.

2: $\text{gcd}(f_1,f_2)$ is the unique monic generator of $\langle f_1,f_2\rangle$. Consequently, $\langle \text{gcd}(f_1,f_2)\rangle=\langle f_1,f_2\rangle\implies\text{gcd}(f_1,f_2)\in\langle f_1,f_2\rangle\implies\text{gcd}(f_1,f_2)\in\langle f_1,\ldots,f_s\rangle\implies\langle \text{gcd}(f_1,f_2),f_3,\ldots,f_s\rangle\subseteq\langle f_1,\ldots,f_s\rangle$. Since $\text{gcd}(f_1,f_2)$ is a divisor of $f_1$, $f_1=h_1 \text{gcd}(f_1,f_2)$. Likewise $f_2=h_2 \text{gcd}(f_1,f_2)$. This shows that $f_1,f_2\in\langle \text{gcd}(f_1,f_2)\rangle\implies\langle f_1,f_2\rangle\subseteq\langle\text{gcd}(f_1,f_2)\rangle$. We can add the generators $f_3,\ldots,f_s$ to both ideals without changing the inclusion, so we have $\langle f_1,\ldots,f_s\rangle\subseteq\langle\text{gcd}(f_1,f_2),f_3,\ldots,f_s\rangle$.