TUT ALG3 5

Problem 1

Exercise 303.1(Exercise).

Let $R=\mathbb{C}[x,y]$. Show that $H:=(xy-3y^2,x^2+4y^2-1,13y^3-y)=(y,x-1)\cap (13y^2-1,x-3y)\cap (y,x+1)$.

To show $H\subseteq I\cap J\cap K$, it suffices to show each generator of $H$ is in $I$, $J$, and $K$. $xy-3y^2$ and $13y^2-y$ are trivially in $I\cap J\cap K$. $x^2+4y^2-1=(x-1)(x+1)+y(4y)=(x-3y)(x+3y)+(13y^2-1)$, so $x^2+4y^2-1\in I\cap J\cap K$.

Observe that the ideals $I:=(y,x-1)$, $J:=(13y^2-1,x-3y)$, and $K:=(y,x+1)$ are pairwise comaximal:

$$\begin{array}{lllccc}I+J=R& :& 3y-(x-1)& +& x-3y& =1\\ J+K=R& :& x-3y& +& 3y-(x+1)& =-1\\ K+I=R& :& x+1& +& 1-x& =2.\end{array}$$

By Theorem 111.2, $I\cap J\cap K=IJK$.

To show $H\supseteq IJK$, we now just have to crunch the numbers:

$$\begin{array}{lll}y(13y^2-1)y& =& (13y^3-y)y\\ y(13y^2-1)(x-1)& =& (13y^3-y)(x-1)\\ y(x-3y)y& =& (xy-3y^2)y\\ y(x-3y)(x+1)& =& (xy-3y^2)(x+1)\\ (x-1)(13y^2-1)y& =& (13y^3-y)(x-1)\\ (x-1)(13y^2-1)(x+1)& =& (x^2+4y^2-1)(13y^2-1)-(13y^3-y)(4y)\\ (x-1)(x-3y)y& =& (xy-3y^2)(x-1)\\ (x-1)(x-3y)(x+1)& =& (x^2+4y^2-1)(x-3y)-(xy-3y^2)(4y).\end{array}$$

Exercise 303.2(Exercise).

Which of the ideals $(y,x-1)$, $(13y^2-1,x-3y)$, and $(y,x+1)$ are maximal in $R$?

$$\begin{array}{r}\frac{\mathbb{C}[x,y]}{(y,x-1)}\cong \frac{\mathbb{C}[x]}{(x-1)}\cong \mathbb{C}\cong \frac{\mathbb{C}[x]}{(x+1)}\cong \frac{\mathbb{C}[x,y]}{(y,x+1)},\end{array}$$

so $(y,x-1)$ and $(y,x+1)$ are maximal in $R$. $(13y^2-1,x-3y)$ is maximal iff¹

$$\begin{array}{r}\frac{\mathbb{C}[x,y]}{(13y^2-1,x-3y)}\cong \frac{\mathbb{C}[y]}{(13y^2-1)}\end{array}$$E1

is a field, which happens iff $(13y^2-1)$ is maximal in $\mathbb{C}[y]$, which happens iff $(13y^2-1)$ is prime in $\mathbb{C}[y]$, which happens iff $13y^2-1$ is irreducible in $\mathbb{C}[y]$, which is false. Thus, $(13y^2-1,x-3y)$ is not maximal in $R$.

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Problem 2

Exercise 303.3(Exercise).

Let $F$ be a field. Prove that the ring $F[x,x^{-1}]$ of Laurent polynomials is a PID.

Let $R=F[x,x^{-1}]$. Note that $F[x]\subseteq R$ is a subring. Let $I$ be an ideal in $R$. $I\cap F[x]$ cannot be $\varnothing$: for every $h(x)\in I$, there exists $k\ge 0$ such that $x^{k}h(x)\in F[x]$. Let $f(x)$ be the minimum degree polynomial in $I\cap F[x]$.

Let $g(x)\in I$ be arbitrary. Let $k\ge 0$ be the smallest integer such that $x^{k}g(x)\in F[x]$. Since $F[x]$ is a Euclidean domain, we can write:

$$x^{k}g(x)=f(x)q(x)+r(x),$$

where $q(x),r(x),\in F[x]$ and $\text{deg}(r(x))<\text{deg}(f(x))$. $r(x)=x^{k}g(x)-f(x)q(x)\in I\cap F[x]$, and since $f(x)$ is the minimum degree polynomial in $I\cap F[x]$, we must have $r(x)=0$. Thus,

$$g(x)=x^{-k}q(x)f(x),$$

and $I=(f(x))$.

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Problem 3

Exercise 303.4(Exercise).

Show that $x^2+x+1$ has a root in ${\mathbb{F}}_p$ $⟺$ $p\equiv 1\mathrm{mod}3$, where $p>3$.

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Note that $(x^2+x+1)$ is not a square in ${\mathbb{F}}_p$: if

$$(ax+b{)}^2=a^2{x}^2+2abx+b^2=x^2+x+1,$$

we must have $a^2=1$, $b^2=1$, and $2ab=1$. $\pm 1$ are solutions for the first two equations, and these are the only ones (see Proposition 114.14; a quadratic polynomial may have at most 2 roots). It is now easily seen that these three equations have no solutions in $a,b$ for $p>3$.

Since $x^3-1=(x-1)(x^2+x+1)$, $x^2+x+1$ has a root in iff $x^3-1$ has non trivial roots in ${\mathbb{F}}_p$ iff $x^3-1$ has three distinct roots. The roots of $x^3-1$ are solutions to $x^3\equiv 1\mathrm{mod}p$, which form a subgroup of order $3$ of ${\mathbb{F}}_p^{\times }$, which exists iff ${\mathbb{F}}_p^{\times }$ has an element of order $3$, which happens iff $3|p-1$ by Cauchy’s theorem, which is equivalent to $p\equiv 1\mathrm{mod}3$.

Exercise 303.5(Exercise).

$(p)$ is maximal in $\mathbb{Z}[\omega ]$ iff $p\equiv 2\mathrm{mod}3$.

$$\begin{array}{rl}& (p)\text{ is maximal in }\mathbb{Z}[\omega ]\\ ⟺& \frac{\mathbb{Z}[\omega ]}{(p)}\text{ is a field}\\ ⟺& \frac{{\mathbb{F}}_p[x]}{x^2+x+1}\text{ is a field}\\ ⟺& (x^2+x+1)\text{ is maximal in }{\mathbb{F}}_p[x]\\ ⟺& x^2+x+1\text{ is irreducible in }{\mathbb{F}}_p[x]\\ ⟺& x^2+x+1=0\text{ has no roots in }{\mathbb{F}}_p[x]\\ ⟺& p\equiv 2\mathrm{mod}3,\end{array}$$

where the last equivalence is obtained from Exercise 4.

Exercise 303.6(Exercise).

$p$ factors in $\mathbb{Z}[\omega ]$ iff $p=a^2+ab+b^2$ for some integers $a$ and $b$.

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Problem 4

Exercise 303.7(Exercise).

Let $f,g$ be polynomials in $\mathbb{C}[x,y]$ with no common factor. Show that the ring $R=\mathbb{C}[x,y]/(f,g)$ is a finite dimensional vector space over $\mathbb{C}$.

Let $\mathbb{C}(x)$ be the field of fractions of $\mathbb{C}[x]$. Use results from here to see that if $f$ and $g$ have no common factor in $\mathbb{C}[x,y]=\mathbb{C}[x][y]$, they have no common factor in $\mathbb{C}(x)[y]$. Thus, the gcd of $f$ and $g$ in $\mathbb{C}(x)[y]$ is $1$, and since $\mathbb{C}(x)[y]$ is a PID, there exist $p(x),q(x)\in \mathbb{C}(x)$ such that

$$\begin{array}{r}f(x,y)p(x)+g(x,y)q(x)=1.\end{array}$$

Multiply out by the denominators of $p$ and $q$ to obtain

$$f(x,y)p^{\prime }(x)+g(x,y)q^{\prime }(x)=h_1(x),$$

where $p^{\prime },q^{\prime },h_1\in \mathbb{C}[x]$. Thus, $h_1(x)$ is contained in the ideal $(f,g)$ generated by $f$ and $g$ in $\mathbb{C}[x,y]$. Similar reasoning furnishes a polynomial $h_2[y]\in \mathbb{C}[y]$ such that $h_2[y]\in (f,g)$.

Consider the ideal $(h_1(x),h_2(y))$ in $\mathbb{C}[x,y]$. Let the degree of $h_1(x)$ and $h_2(y)$ be $n$ and $m$ respectively. Note that the set $\{x^i{y}^j|1\le i\le n,1\le m\le j\}$ is a finite basis for $\mathbb{C}[x,y]/(h_1,h_2)$. Note that $(f,g)$ contains $(h_1,h_2)$. By the correspondence theorem, $(f,g)/(h_1,h_2)$ is an ideal of $\mathbb{C}[x,y]/(h_1,h_2)$. It follows that

Footnotes

1. To establish the isomorphism in (E1), consider the map $\phi :\mathbb{C}[x,y]\to \mathbb{C}[y]$ defined by $c\mapsto 3y$. Show that the kernel of this map is $(x-3y)$. Use first isomorphism theorem to conclude $\mathbb{C}[x,y]/(x-3y)\cong \mathbb{C}[y]$. Note that $(13y^2-1,x-3y)$ contains $\ker \phi$, and $\phi ((13y^2-1,x-3y))=(13y^2-1)$. Use (4) from the the correspondence theorem to get (E1). ↩