Finding a basis for a field I have a polynomial f(x) = $x^3+x^2+1$ in $\mathbb{Z}_5[x]$ and it is given that F = $\mathbb{Z}_5[x]$/$<f(x)>$ = $\mathbb{Z}_5(\alpha)$ where $\alpha =x+<f(x)>$. 
I want to find a basis for F over  $\mathbb{Z}_5$. Since the polynomial has degree 3, i know that vector will consist of 3 elements and that it will be on the form {1,$\alpha$,$\alpha^2$}. 
When  $\alpha =x+<f(x)>$, does this mean that  $\alpha =x^3+x^2+x+1$ and that this is the second element of the vector? 
Similarly $\alpha^2$ will be $(x^3+x^2+x+1)^2 $ ? But this seems very weird. 
Obviously i'm very unsure and any answers will be appreciated. 
 A: No: $\alpha$  is the set of all polynomials of the form $x+f(x)(x^3+x^2+1), \quad f(x)\in \mathbf Z_5[x]$.
Similarly  $\alpha^2$  is the set  of all polynomials of the form $x^2+f(x)(x^3+x^2+1), \quad f(x)\in \mathbf Z_5[x]$.
A: The question is asking for a basis for a vector space over a field.  Here, the field is $\mathbb{Z}_5$ and the vector space is $F=\mathbb{Z}_5[x]/\langle f(x) \rangle$, where $f(x)=x^3+x^2+1$.  
First, observe that the polynomial $f(x)$ is irreducible (because it has degree 3, and so if it were reducible, it would have a linear factor, but substituting values from $\mathbb{Z}_5$ into $f(x)$ does not give a zero).  It follows that the quotient ring $F$ is a field.  The elements of this quotient ring $F$ are cosets of the form $p(x) + \langle f(x) \rangle$, where $\langle f(x) \rangle$ is the set of all multiples of the polynomial $f(x)$.  For example, the coset $0+\langle f(x) \rangle$ is the additive identity in the field $F$ and $1+\langle f(x) \rangle$ is the multiplicative identity in the field $F$. Addition and multiplication in $F$ are done modulo $f(x)$.  So, the product of two elements $a(x) + \langle f(x) \rangle$ and $b(x) + \langle f(x) \rangle$ is $c(x) + \langle f(x) \rangle$ where $c(x)=a(x)b(x) \mod f(x)$.  Thus, if $\alpha$ is the coset $x+\langle f(x) \rangle$, then $\alpha^2$ is the coset $x^2 + \langle f(x) \rangle$. The coset $\alpha^3$ is  $x^3 + \langle f(x) \rangle = -x^2-1 + \langle f(x) \rangle= 4x^2+4+\langle f(x) \rangle \in \mathbb{Z}_5[x]$, because the coset representatives may be reduced modulo $f(x)$.  Similarly, you can compute $\alpha^4$ by doing long division of $x^4$ by $f(x)$ and calculating the remainder $r(x)$.  Then $\alpha^4 = r(x) + \langle f(x) \rangle$, where $r(x)$ is some polynomial of degree at most 2 with coefficients from $\mathbb{Z}_5$. 
Observe that the map  $\phi: \mathbb{Z}_5 \rightarrow F$ that takes $c$ to $c+\langle f(x) \rangle$ is a ring homomorphism and injective.  Hence, $\mathbb{Z}_5$ is isomorphic to a subfield of $F$.  In this sense, $F$ contains the subfield $\mathbb{Z}_5$ and hence $F$ is an extension field of $\mathbb{Z}_5$.  It can shown that every extension field is a vector space over the original field. In our example, $F$ is a vector space over $\mathbb{Z}_5$. This means that $F$ satisfies the vector space axioms. For example, if $c \in \mathbb{Z}_5$ is a scalar and $u = a(x) + \langle f(x) \rangle$ and $v = b(x) + \langle f(x) \rangle$ are vectors, then the distributive law $c(u+v)=cu+cv$ of scalar multiplication over vector addition holds.  
The cosets in the quotient ring are of the form $a(x) + \langle f(x) \rangle$, where the coset representatives $a(x)$ can be chosen to be the unique polynomial which is the remainder upon dividing by $f(x)$. So the cosets are of the form $a(x) + \langle f(x) \rangle = a_0+a_1x+a_2x^2 + \langle f(x) \rangle$, where $a_i \in \mathbb{Z}_5$ and the coset representatives are the polynomials in $\mathbb{Z}_5[x]$ of degree at most 2. There are $5^3$ cosets in $F$. It can be seen that the dimension of the vector space $F$ over $\mathbb{Z}_5$ is 3, and a basis for this vector space is $1 + \langle f(x) \rangle$, $x + \langle f(x) \rangle$, $x^2 + \langle f(x) \rangle$.  Observe that any coset (ie vector) $a_0+1+a_1 x + a_2 x^2 + \langle f(x) \rangle$ in $F$ can be represented as a linear combination $a_0 (1+\langle f(x) \rangle) + a_1 (x+\langle f(x) \rangle) + a_2 (x^2 + \langle f(x) \rangle)$ of the three basis elements, where the coefficients $a_i$ of the linear combination are the scalars in $\mathbb{Z}_5$. 
A: I will provide a bit more general answer as it seems to me that it might shed some light in what is going on.
I'll assume that you know that if you have a field extension $F\supset K$, and $\alpha\in F$ algebraic over $K$, then $K(\alpha) = K[\alpha] \cong K[x]/\langle f\rangle$, where $f$ is minimal polynomial of $\alpha$ over $K$. Also, I'll assume that you know that $\{1,\alpha,\ldots,\alpha^{n-1}\}$ is $K$-base of $K(\alpha)$ in that case, where $n=\deg f$.

Claim.
  Let $K$ be a field, and $f$ a monic irreducible polynomial in $K[x]$
  of degree $n$. Let $\alpha = x + \langle f\rangle$. Then
  $\{1,\alpha,\ldots,\alpha^{n-1}\}$ is a $K$-base of $K[x]/\langle f\rangle$.

Set $F = K[x]/\langle f\rangle$. It is obviously a field of extension of $K$, since $K$ is included by $u\mapsto u +\langle f\rangle$. Also, note that $\alpha = x + \langle f\rangle$ is a root of $f$ in $F$ because $f(\alpha) = f(x) + \langle f\rangle = 0$. Thus $\alpha\in F$ is algebraic over $K$, it's minimal polynomial is $f$ (since it is monic and irreducible), and $K[\alpha] \cong K[x]/\langle f\rangle = F$. The claim follows.
