The problem is this: 1) Prove that $f = x^3+x+1$ is irreducible over $\mathbb Z_5$. 2) In field K = $\mathbb Z_5/(f)$ $c = x+(f)$. Find in $K$ element $(c^2 + c + 1)^{-1}$ and write it in the form $kc^2+lc+m, k,l,m \in \mathbb Z_5$.
$$$$I know how to prove that $f$ is irreducible. But I don't know how to proceed. $\mathbb Z_5/(f)$ is a ring, so I can work with $c$. How do I find this element?
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Hint: By definition of $K,$ we have $c^3+c+1=0,$ and so any polynomial in $c$ can be expressed in terms of powers of $c$ less than or equal to $2.$ Moreover, the elements of $K$ are uniquely expressible as $\Bbb Z_5$-linear combinations of $1,c,c^2.$
Since $f$ is irreducible over $\Bbb Z_5,$ then $K$ is a field, and $c^2+c+1$ is non-zero, so has a multiplicative inverse of the form $kc^2+lc+m.$ That is, there exist unique $k,l,m\in\Bbb Z_5$ such that $$1=(c^2+c+1)(kc^2+lc+m)=kc^4+(k+l)c^3+(k+l+m)c^2+(l+m)c+m.$$ Regrouping gives us $$1=(kc+k+l)c^3+(k+l+m)c^2+(l+m)c+m,$$ whence, recalling that $c^3=-c-1,$ we have $$1=-(kc+k+l)(c+1)+(k+l+m)c^2+(l+m)c+m.$$ Can you rewrite the right-hand side as a standard quadratic in $c$ and take it from there?
I assume you mean $K = \mathbb{Z}_{5}[x]/(f)$. Since $\mathbb{Z}_{5}$ is a field, $\mathbb{Z}_{5}[x]$ is a Euclidean domain. Note that $f$ is irreducible of degree $3$, so it is relatively prime to any polynomial of lower degree. Hence, there exist polynomials $\alpha(x), \beta(x) \in \mathbb{Z}_{5}[x]$ such that $\alpha(x)(x^{2}+x+1) + \beta(x)f(x)=1$. In the quotient field $K$, reducing both sides modulo $f$ shows that $(\alpha(x) + f(x))(c^{2}+c+1)=1$, i.e. the coset of $\alpha(x)$ in $K$ is the inverse of $c^{2}+c+1$. Use the Euclidean algorithm to compute $\alpha(x)$.