TUT ALG3 7

Problem 1

Let $R\subseteq S$ be integral domains and let $f(x)={\sum }_{i=0}^n{a}_i{x}^i\in R[x]$. Define $f^{\prime }(x)={\sum }_{i=1}^n{a}_{i}i{x}^{i-1}$. Let $a\in S$.

Lemma 333.1(Chain rule).

If $f(x)=g(x)h(x)$, then $f^{\prime }(x)=g(x)h^{\prime }(x)+g^{\prime }(x)h(x)$.

Proof.

$$\begin{array}{rl}f(x)& =\sum _{i=0}^{m+n}\left(\sum _{j=0}^i{g}_j{h}_{i-j}\right)x^i\\ f^{\prime }(x)& =\sum _{i=0}^{m+n-1}\left(\sum _{j=0}^{i+1}{g}_j{h}_{i+1-j}\right)(i+1)x^i.\end{array}$$ $$\begin{array}{rl}g(x)h^{\prime }(x)+g^{\prime }(x)h(x)& =\left(\sum _{i=0}^n{g}_i{x}^i\right)\left(\sum _{i=0}^{m-1}{h}_{i+1}(i+1)x^i\right)+\left(\sum _{i=0}^{n-1}{g}_{i+1}(i+1)x^i\right)\left(\sum _{i=0}^m{h}_i{x}^i\right)\\ & =\sum _{i=0}^{m+n-1}\left(\sum _{j=0}^i{g}_j{h}_{i+1-j}(i+1-j)\right)x^i+\sum _{i=0}^{m+n-1}\left(\sum _{j=0}^i{g}_{j+1}{h}_{i-j}(j+1)\right)x^i\\ & =\sum _{i=0}^{m+n-1}\left(\sum _{j=0}^i{g}_j{h}_{i+1-j}(i+1-j)\right)x^i+\sum _{i=0}^{m+n-1}\left(\sum _{j=1}^{i+1}{g}_j{h}_{i+1-j}(j)\right)x^i\\ & =\sum _{i=0}^{m+n-1}\left(\sum _{j=0}^{i+1}{g}_j{h}_{i+1-j}\right)(i+1)x^i.\end{array}$$ □

Exercise 333.2(Exercise).

Show that $a$ is a multiple root of $f$ iff $f(a)=0$ and $f^{\prime }(a)=0$.

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Suppose $f(a)=0$. Then, $f(x)=(x-a)g(x)=xg(x)-ag(x)$.

$$\begin{array}{r}{f}^{\prime }(x)=g(x)+x{g}^{\prime }(x)-a{g}^{\prime }(x)=g(x)+(x-a)g^{\prime }(x).\end{array}$$

Thus, if $(x-a)|g(x)$ then $(x-a)|f^{\prime }(x)$, and vice versa.

Exercise 333.3(Exercise).

Let $R$ be a field and suppose $\gcd (f,f^{\prime })=1$ in $S$. Show that $f$ has no multiple roots in $S$.

Suppose $f(a)=0$ for some $a\in S$. Let $u(x),v(x)\in S[x]$ be such that $u(x)f(x)+v(x)f^{\prime }(x)=1$. If $f^{\prime }(a)=0$, substituting $a$ into the previous equation yields $0=1$, a contradiction. Thus, by Exr 2, $a$ is not a multiple root of $f$.

Exercise 333.4(Exercise).

Let $R$ be a field and suppose $f(x)$ is irreducible in $R[x]$ and suppose $f$ has a root in $S$. Show that $f$ has no multiple roots in $S$ iff $f^{\prime }\ne 0$.

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Problem 5

Exercise 333.5(Exercise).

Let $R=\mathbb{Z}[x_1,x_2,x_3,x_4]$. Determine which of the following polynomials are irreducible:

1. $f=x_1^{p+1}-x_2^p$
2. $g=x_1{x}_4-x_2{x}_3$
3. $h=x_1^p{x}_2-x_3^p$
4. $fg+g^p$

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Problem 6

Exercise 333.6(Exercise).

Determine which of the following polynomials in $\mathbb{C}[x,y,z]$ are irreducible:

1. $p_1=y^5-z^4$
2. $p_2=z^8+2x{y}^3-x^{2}yz-z^3$
3. $p_3=2y^2{z}^6+x{z}^7-x^{3}y-2y^{2}z+3x{z}^2$
4. $p_4=y^4{z}^5-y^4+2x{y}^{2}z-x^2{z}^2$
5. $p_5=2x{y}^4{z}^4+y{z}^7+2x^2{y}^2-2x^{3}z-y{z}^2$
6. $p_6=x^2{y}^4{z}^3+3y^3{z}^5+2xy{z}^6-x^4-3y^3+4xyz$

Polynomial 1

Consider $p_1$ as a polynomial in $y$ with coefficients in $\mathbb{C}[z]$; Let $k$ denote the algebraic closure of $\mathbb{C}[z]$. Denote by $z^{4/5}$ a root of $p_1$ in $k$. It is easy to see that $e^{2\pi ij/5}{z}^{4/5}$ is a root of $p_1$ for $i=0,1,\dots ,4$. By Prp 114.14, these are all the roots of $p_1$ in $k$. Since no proper subset of the factors $\{(y-e^{2\pi ij/5}{z}^{4/5}):j=0,1,\dots ,4\}$ yields an element of $\mathbb{C}[z,y]$, it follows that $p_1$ is irreducible.

Polynomial 4

Consider $p_4$ as a polynomial in $x$ with coefficients in $\mathbb{C}[y,z]$.

$$p_4=(-z^2)x^2+(2y^{2}z)x+(y^4{z}^5-y^4).$$

Suppose $p_4=fg$. If ${\text{deg}}_x(f)=2$, $g$ must be a unit, since the gcd of the coefficients $-z^2,2y^{2}z,y^4{z}^5-y^4$ is $1$. The case ${\text{deg}}_x(f)={\text{deg}}_x(g)=1$ is possible iff the discriminant $4y^4{z}^2+4z^2{y}^4(z^5-1)=4z^2{y}^4{z}^5$ is a prefect square; it is clearly not.

Polynomial 2

Consider $p_2$ as a polynomial in $x$ with coefficients in $\mathbb{C}[y,z]$.

$$p_2=(-yz)x^2+(2y^3)x+(z^8-z^3).$$

The discriminant is $4y^6+4y{z}^4(z^5-1)$ is not a prefect square.

Polynomial 3

Consider $p_3$ as a polynomial in $x$ with coefficients in $\mathbb{C}[y,z]$.

$$p_3=(-y)x^3+(z^7+3z^2)x+(2y^2{z}^6-2y^{2}z).$$

Use Thm 118.2 with $p=z$.

Polynomial 5

Consider $p_5$ as a polynomial in $x$ with coefficients in $\mathbb{C}[y,z]$.

$$p_5=(-2z)x^3+(2y^2)x^2+(2y^4{z}^4)x+(y{z}^7-y{z}^2).$$

Use Thm 118.2 with $p=y$.

Polynomial 6

Let $\lambda$ denote an arbitrary unit.

Consider the map $\phi :\mathbb{C}[x,y,z]\to \mathbb{C}[t]$ defined by $\phi (x)=t^3$, $\phi (y)=t^4$, and $\phi (z)=t^5$ (this is the same map we encountered in Exr 301.1!). Observe that $\phi (p_6)=6t^{37}$; neat. Now, suppose $f,g\in \mathbb{C}[x,y,z]$ such that $p_6=fg$. Then, $6t^{37}=\phi (f)\phi (g)$. This forces $\phi (f)=\lambda t^a$, $\phi (g)=\lambda t^b$ for some $a,b⩾0$. It follows that $f$ must be a sum of monomials $m_1+\cdots +m_n$ such that $\phi (m_i)=\lambda t^a$ for each $i$; ditto for $g$.

Denote the set of monomials of $f$ by $\text{Mon}(f)$. Denote ”$\alpha \in \text{Mon}(f)$ and $\beta \in \text{Mon}(g)$” by $(\alpha ,\beta )$.

Since $y^3\in \text{Mon}(fg)$, WLOG we must have either $(y,y^2)$ or $(1,y^3)$.
Since $x^4\in \text{Mon}(fg)$, one of $(1,x^4)$, $(x,x^3)$, $(x^2,x^2)$, $(x^3,x)$, $(x^4,1)$ must be true.

If $(y,y^2)$, none of $1,x,x^2,x^3,x^4$ can be in $\text{Mon}(f)$, since none of them have image $\lambda t^4$ under $\phi$.
If $(1,y^3)$, then $(1,x^4)$ is forced, and $f$ must be a unit.

Thus, $p_6$ is irreducible.

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Problem 7

Exercise 333.7(Exercise).

Show that $f:=x^p-x-c$ is irreducible in $\mathbb{Z}/p\mathbb{Z}[x]$ iff it has no root in $\mathbb{Z}/p\mathbb{Z}$.

It is clear that $f$ is reducible if it has a root. Suppose