MHB How do you construct a field with 27 elements?

[Chris L T521]

Thanks to those who participated in last week's POTW! Here's this week's problem!

-----

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.

-----

 

[Chris L T521]

This week's question was correctly answered by Sudharaka. You can find his solution below.

Spoiler

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