Math4302 Modern Algebra (Lecture 34)

$\impliedby$: Proved last time.

$\implies$: If $I\subsetneq J\subseteq R$, where $J$ is an ideal.

Then we pick $a\in J$, $a\notin I$. We have $a+I\neq I$.

So $a+I$ is non-zero element in $R/I$, so it has a multiplicative inverse $b+I$: $(a+I)(b+I)=1+I$.

So $ab+I=1+I$, so $ab-1\in I\subset J$, so $1=ab-i\in J$, since $a\in J$, and $i\in J$.

So $J=R$.

$n\mathbb{Z}$ is a prime ideal if and only if $n$ is prime.

$n\mathbb{Z}$ prime when $n|ab$ implies $n|a$ or $n|b$.

$(a+I)(b+I)=0+I$ if and only if $a+I=I$ or $b+I=I$.

$\implies$: If $ab\in I$, then $ab+I=0+I$, so $(a+I)(b+I)=I$, which is the zero element in $R/I$.

Since $R/I$ is an integral domain, $(a+I)(b+I)=0+I$ implies $a+I=I$ or $b+I=I$.

$\impliedby$: If $(a+I)(b+I)=0+I$, then $ab\in I$, so $a\i nI$ or $b\in I$ (by definition of ideal), so $a+I=I$ or $b+I=I$.

Field is an integral domain, so every maximal ideal is a prime ideal.

In $\mathbb{Z}[x]$, we saw $M=\{f(x)\mid \text{constant term is even}\}$ is an ideal.

Let $I=\{f(x)\mid \text{constant term is 0}\}$.

Note that $I\subsetneq M$ (Therefore $I$ is not maximal), and $I$ is an ideal.

And note that $I$ is also a prime ideal (if any of the coefficients is 0, then the constant term is 0).

We can assume that $f(x)$ is irreducible. (we can write all the polynomial in $F[x]$ as product of irreducible polynomials.)

Consider the ideal generated by $f(x)$.

Since $f(x)$ is irreducible, $I$ is a maximal ideal in $F[x]$.

We have $F[x]/I$ is a field.

This is a field contain $F$. We have an inclusion map $F\subset F[x]/I$. $a\mapsto a+I$.

Think of $f(x)$ as a polynomial with coefficients in $E(x)$ and evaluate it at $x+I$.

If $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, then

\begin{aligned} f(x+I)&=(a_n+I)(x+I)^n+(a_{n-1}+I)(x+I)^{n-1}+\cdots+(a_1+I)(x+I)+(a_0+I)\\ &=(a_nx^n+I)+(a_{n-1}x^{n-1}+I)+\cdots+(a_1x+I)+(a_0+I)\\ &=(a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0)+I\\ &=f(x)+I\\ &=I \end{aligned} $$. Where $I$ is the zero element in $F[x]/I$. </details> <details> <summary>Example</summary> $f(x)=x^2+1\in \mathbb{R}[x]$ has no zero in $\mathbb{R}$. Let $I=(x^2+1)\subset \mathbb{R}[x]$. $I$ is maximal ideal in $\mathbb{R}[x]$, and consider $\mathbb{R}[x]/I$. $R\to \mathbb{R}[x]/I$. $r\mapsto r+I$. So that $x+I$ is a zero of $x^2+1\in E[x]$. --- We claim that $E\simeq \mathbb{C}$. The isomorphism is given by $\mathbb C\overset{\phi}{\to}\mathbb R/(x^2+1)$, $a+bi\mapsto (a+bx)+I$ $\phi((a+bi)+(a'+bi'))=(a+bx)+(a'+b'x)+I==\phi(a+bi)+\phi(a'+bi')$ $\phi((a+bi)\cdot (a'+bi'))=(a+bx)(a'+bx)+I=(a^2+b^2)+I=\phi(a+bi)+\phi(a'+bi')$ </details>