Math4302 Modern Algebra (Lecture 33)

The maximal ideal of $\mathbb{Z}$. $p\mathbb{Z}$ is maximal ideal if and only if $p$ is prime.

---

In $\mathbb Z[x]$

$$M=\{a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\mid a_i\in \mathbb Z,a_0\text{ is even}\}$$

$M$ is a maximal ideal.

• $M$ is clearly an ideal
• If $M\subsetneq I$ with $I$ an ideal, then there is some $g(x)=b^nx^n+b_{n-1}x^{n-1}+\cdots+b_1x+b_0\in I$, where $b_0$ is odd $f(x)=b^nx^n+b_{n-1}x^{n-1}+\cdots+b_1x+(b_0+1)\in M\subset I$, so $f(x)-g(x)\in I$ and $1\in I$, so $I=\mathbb Z[x]$.

All ideal in $\mathbb Z$ are principal.

$n\mathbb Z=(n)$

---

$M=\{a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\mid a_i\in \mathbb Z,a_0\text{ is even}\}$ is not principal.

We proceed by contradiction, suppose $M=(h(x))$, and $2\in M$ therefore $2=h(x)g(x)$ for $g(x)\in R$, therefore $h(x)$ must be constant event number.

But that would mean every coefficient of polynomial in $M$ is even. But consider $x+2$ should in $M$ but it is impossible to generate it with coefficient that are even.

---

Consider $\mathbb R[x,y]$ (The polynomials generated by 2 variables), $I=\{f(x,y):f(0,0)=0\}$

We need at least two polynomial containing $x$ and $y$ to generate $I$.

If $I=\{0\}$, then we are done.

Suppose $I\subset F[x]$ is an ideal and principal $=(0)$

Suppose $I\neq \{0\}$. Let $0\neq f(x)\in I$ be the polynomial with smallest degree.

We claim $I=(f(x))$.

If $g(x)\in I$ then we can divide $g$ by $f$

$$g(x)=f(x)q(x)+r(x)$$

Where $r(x)$ is the remainder $r(x)=0$ or $\deg r(x)<\deg f(x)$, which is impossible since $r=g-gf\in I$. This contradicts that $f$ is the smallest degree.

Therefore $g(x)=f(x)q(x)\in (f(x))$.

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

$(x^2+1)$ is maximal ideal in $\mathbb R[x]$. By the theorem $\mathbb R[x]/(x^2+1)$ is a field $\mathbb C$.

$(f(x))$ in $\mathbb C[x]$ is maximal ideal if and only if $\deg f(x)=1$

$\impliedby$ We know that $R/I$ is a commutative ring with unity $1+I$.

So it is enough to show every non-zero element $r+I$ (i.e. $r\notin I$) has a multiplicative inverse $s+I$.

So $(s+I)(r+I)=1+I$. So $sr+I=1+I$. equivalently $sr-1\in I$.

$I\subseteq J\coloneqq\{i+ar|i\in I,a\in R\}\subset R$

We claim that $J$ is an ideal

• $(i+ar)+(i'a'r)=(i+i')+(a+a')r\in J$,
• and $b(i+ar)=bi+(ba)r$, $bi\in I$ for $b\in R$.

Therefore since $I$ is maximal, $J=R$ so $I\in J$.

so $1=i+ar$ for some $i\in I$ and $a\in R$.

$ar-1=-i\in I$, let $s=a$, $s+I$ is the multiplicative inverse of $r+I$.