# Problem of the Week #21 - August 20th, 2012

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#### Chris L T521

##### Well-known member
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Thanks to those who participated in last week's POTW!! Here's this week's problem!

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Problem: Let $D$ be an integral domain. A polynomial $f(x)$ from $D[x]$ that is neither the zero polynomial nor a unit in $D[x]$ is said to be irreducible over $D$ if, whenever $f(x)$ is expressed as a product $f(x)=g(x)h(x)$, with $g(x),h(x)\in D[x]$, then $g(x)$ or $h(x)$ is a unit in $D[x]$.

(a) Suppose that $f(x)\in\mathbb{Z}_p[x]$ and is irreducible over $\mathbb{Z}_p$, where $p$ is prime. If $\deg f(x)=n$, prove that $\mathbb{Z}_p[X]/\langle f(x)\rangle$ is a field with $p^n$ elements.

(b) Construct a field with 27 elements.

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#### Chris L T521

##### Well-known member
Staff member
This week's question was correctly answered by Sudharaka. You can find his solution below.

Theorem 1: Let $$F[x]$$ be a polynomial field. An ideal $$\langle p(x)\rangle\neq\{0\}$$ of $$F[x]$$ is maximal if and only if $$p(x)$$ is irreducible over $$F$$.

(Reference: Theorem 31.6, A First Course in Abstract Algebra by J.B.Fraleigh, 3rd Edition, Page 282)

Theorem 2: Let $$R$$ be a commutative ring with unity. Then $$M$$ is a maximal ideal of R if and only if $$R/M$$ is a field.

(Reference: Theorem 29.4, A First Course in Abstract Algebra by J.B.Fraleigh, 3rd Edition, Page 257)

It is given that $$f(x)$$ is irreducible over the field $$\mathbb{Z}_p$$. Therefore by Theorem 1 it follows that, $$\langle f(x)\rangle$$ is a maximal ideal of $$\mathbb{Z}_p[x]$$. Then by Theorem 2 it follows that $$\displaystyle\frac{\mathbb{Z}_p[X]}{\langle f(x)\rangle}$$ is a field.

Take any element $$\displaystyle g(x)+\langle f(x)\rangle\in\frac{\mathbb{Z}_p[X]}{\langle f(x)\rangle}$$ where $$\deg g(x)=m$$. By the division algorithm for polynomial fields there exist polynomials $$h(x)$$ and $$r(x)$$ in $$\mathbb{Z}_p[X]$$ such that,

$g(x)=f(x)h(x)+r(x)\mbox{ where }\deg r(x)<\deg f(x)=n$

$\therefore g(x)+\langle f(x)\rangle=f(x)h(x)+r(x)+\langle f(x)\rangle$

But, $$f(x)h(x)\in\langle f(x)\rangle$$. Therefore,

$g(x)+\langle f(x)\rangle=r(x)+\langle f(x)\rangle\mbox{ where }\deg r(x)<n$

$\therefore g(x)+\langle f(x)\rangle=a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_{1}x+a_{0}+\langle f(x)\rangle$

For each $$a_{i}\mbox{ where }i=\{0,1,\cdots,n-1\}$$ there are $$\left|\mathbb{Z}_{p}\right|=p$$ choices to choose from. Therefore,

$\left|\frac{\mathbb{Z}_p[X]}{\langle f(x)\rangle}\right|=p^{n}$

Q.E.D

Let $$f(x)=x^3$$ and $$p=3$$. Then the field $$\displaystyle\frac{\mathbb{Z}_{3}[X]}{\langle x^3\rangle}$$ has $$3^3=27$$ elements.

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