Math4302 Modern Algebra (Lecture 35)

If $f(x)$ is irreducible, then $I=(f(x))$ is maximal in $F[x]$, and $E=F[x]/(f(x))$, then $E$ is a field.

$f(x)$ has a zero in $E$: $x+I$.

Let $f(x)=x^2+1\in \mathbb R[x]$, $I=(x^2+1)\in \mathbb R[x]$.

$E\coloneqq \mathbb R[x]/I=(x^2+1)\simeq \mathbb C$ is a field.

Let $\phi$ be the function that take $a+bi\mapsto (a+bx)+I$.

• Addition is trivial
• For multiplication:
  • $(aa'-bb'+(a'b+a'b)x)+I=aa'+bb'x^2+(aa'+bb')x+I$, note that $bb'x^2-(-bb')=bb(x^2+1)\in I$, so the two cosets are equal.

Therefore, $\phi$ is injective.

Then we need to show that $\phi$ is bijection:

• $\phi$ is injective: if $\phi(a+bi)=0$, then $(a+bx)+I=I$, so $a+bx\in I=(x^1+1)=\{p(x)(x^2+1)|p(x)\in \mathbb R[x]\}$
• $\phi$ is surjective: if $g(x)+I$ is in $\mathbb R[x]/I$, then divide $g(x)$ by $x^2+1$ to get $g(x)=q(x)(x^2+1)+r(x)$, then $\deg r(x)\leq 1$. So $g(x)-r(x)\in I$, therefore $g(x)+I=r(x)+I$.

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$f(x)=x^2+xZ+1\in \mathbb Z_2[x]$ ahs no zero in $\mathbb Z_2[x]$, so it is irreducible.

Let $I=(x^2+x+1)\subset\ \mathbb Z_2[x]$ being the maximal ideal.

$E=\mathbb Z_2[x]/I$ is a field. By the theorem belows, we have $4$ cosets

• $0+I$
• $1+I$
• $x+I$
• $(x+1)+I$

For higher degree polynomials, $(x^3)+I=1+I$, since $x^3=(x+1)(x^2+x+1)+1=x^3+1$.

The additive group is Klein 4, the multiplicative (removing 0) is $\mathbb Z_3$.

Existnce: By division algorithm, if $p(x)+I\in F[x]/I$, then divide $p$ by $f$: $p(x)=f(x)q(x)+r(x)$, $r=0$ or $\deg r<\deg f$

Uniqueness: If $g_1(x)+I=g_2(x)+I$, with $\deg g_1,g_2<\deg f$, or ($g_1(x)=0$ or $g_2(x)=0$), then $g_1(x)-g_2(x)\in I$, so $f(x)|g_1(x)-g_2(x)$. $g_1(x)-g_2(x)$ can only be zero.

So $g_1(x)=g_2(x)$.