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I am presented with $f(x) = 2x^3 + 3x^2 + 1$ $\in \mathbb{Z_5}[x]$ and need to explain why $F = \frac{\mathbb{Z_5}[x]}{f(x)}$ is a field and also find how many elements are in F.

So far I have shown that $f(x)$ is an irreducible polynomial and I also know that,

If $f(x)$ is an irreducible polynomial in $\mathbb{Z_5}[x]$, then the factor ring $\frac{\mathbb{Z_5}[x]}{f(x)}$ is also a field.

Basically I am not sure how to properly find the factor ring F and also determine how many elements are in it.

Thanks in advance

MathsRookie
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4 Answers4

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Well, in general, if a field $L$ is an extension field of some field $K$, then $L$ is also a $K$-vector space.

If $f$ is an irreducible polynomial of degree $n$ over, say, ${\Bbb Z}_p$, then the quotient field ${\Bbb Z}_p[x]/\langle f\rangle$ has $p^n$ elements and is a vector space over ${\Bbb Z}_p$ of dimension $n$.

Wuestenfux
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Hint: Use the division algorithm to find canonical representatives for each residue class $g(x) + (f(x))$.

lhf
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Outline:

You know $\Bbb Z_5[x]/I$ where $I=\langle f(x) \rangle $ is a field using the mentioned result. Now, the task is to find $|\Bbb Z_5[x]/I |$. Here $\Bbb Z_5[x]/I$ can be considered as an entension of $\Bbb Z_5$. Also $$\dim_\Bbb {Z_5} (\Bbb Z_5[x]/I)=3=\deg f$$

Call $\{v_1,v_2,v_3\}$ the basis of $\Bbb Z_5[x]/I$ over $\Bbb Z_5$

Now arbitrarily pick $f(x)+I \in \Bbb Z_5[x]/I$. Then $$f(x)+I=a_1v_1+a_2v_2+a_3v_3\;;\;a_i \in \Bbb Z_5$$

There are $5 \times 5 \times 5$ choices to choose the coefficients, so $$\text{number of such $f(x)+I$ }=5^3=|\Bbb Z_5[x]/I|$$

Chinnapparaj R
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Note that two polynomials $g(x)$ and $h(x)$, both of degree 2 or less, will represent different elements in $F$. This is due to the fact that their difference $g(x)-h(x)$ being of degree less than 3, cannot be a multiple of the cubic polynomials $f(x)=2x^3+3x^2+1$, and so belong to different cosets.

Now a polynomial $h(x)$ of degree 3 or more can be divided by $f(x)$ to get quotient $q(x)$ and remainder $r(x)$ such that $h(x)=q(x)f(x) + r(x)$

Clearly $r(x)$ and $h(x)$ will represent the same coset. This shows that the field $F$ in your question simply consists of polynomials of degree less than $3$. This is easy to count as each coefficient has only a finite number of choices.

P Vanchinathan
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