Irreducible over the field Prove that $x^2 +x+4$ is irreducible over $F,$ the field of integers $\mod 11$ and
prove directly that $F[x]=(x^2 + x + 4)$ is a field having $121$ elements.
What is this question exactly asking? I don't really need a detail proof but I do need an idea of what the question is asking for. 
I know how to do the first part of the question but the second part I don't understand. How can I have an intuitive idea of what is going on in terms of ideals and integral domain?
 A: An element of a domain is irreducible if it doesn't factor into a product of more than one nonunit, and in particular a polynomial in $F[x]$, where $F$ is a field, is irreducible if it does not factor into a product of more than one nonconstant polynomial. The "field of integers mod $11$" refers to the ring of integers mod $p$ where $p=11$. For primes $p$, it is simple to prove $\Bbb Z/p\Bbb Z$ is a domain, and it is also easy to prove that finite domains are fields, so these rings are sometimes denoted $\Bbb F_p$.
If a quadratic polynomial factors into a product of nonconstant polynomials, it must be a product of linear factors, i.e. $x^2+ax+b=(x+c)(c+d)$. Thus, a quadratic polynomial is reducible if and only if it has a root in the base field. It is straightforward to check if $x^2+x+4$ has a root over the field $F=\Bbb Z/11\Bbb Z$; either check all cases or see if the discriminant is a square (note, the quadratic formula works in any ring in which $2$ is invertible).
An ideal $I\triangleleft R$ of a ring $R$ is an additive subgroup which is closed under ambient multiplication, i.e. we have $I+I=\{i+j:i,j\in I\}=I$ and $RI=\{ri:r\in R,i\in I\}\subseteq I$. This is the ring analogue of the group-theoretic concept of normal subgroups. If $f:G\to H$ is a homomorphism then the kernel $\ker f$ is a normal subgroup of $G$, and conversely every normal subgroup $N\triangleleft G$ is the kernel of the projection homomorphism $G\to G/N$. Similarly, the kernel (i.e. preimage of $0$) of any homomorphism out of a ring $R$ will be an ideal of $R$, and every ideal $I\triangleleft R$ is the kernel of the homomorphism $R\to R/I$. The notation $R/I$ stands for the quotient ring.
Note: if you were not already familiar with the concepts in the above paragraph, then you are not yet equipped to tackle this problem. In order to study field theory you need to start out knowing the basics about rings. Therefore, if necessary, please take the time to become acquainted. Just like with groups, $R/I$ is the set of cosets of $I$, subsets of $R$ of the form $r+I=\{r+i:i\in I\}$ for various $r\in R$. This set of cosets is equipped with its own addition and multiplication operations given easily enough by $(a+I)+(b+I)=(a+b)+I$ and $(a+I)(b+I):=ab+I$. One can check that, just like with groups, we can also define $R/I$ via equivalence relations. This makes $R/I$ a ring.
If $R$ is a ring (suppose it's commutative for the sake of skipping details about left / right / two- sidedness), the notation $(a)$ means $aR:=\{ar:r\in R\}$, the set of all multiples of $a\in R$. It is easy to check such a set is an ideal of $R$. Such ideals are called principal ideals; they are the ones "generated by one element." Thus, if $n\Bbb Z$ is the set of multiples of $n\in\Bbb Z$, then the "integers mod $n$" is precisely the ring $\Bbb Z/n\Bbb Z$ (for example).
If $R$ is a ring, the notation $R[x]$ means the set of polynomials in a formal variable "$x$" whose coefficients are from $R$. This set has the obvious addition and multiplication rules, and so $R[x]$ is itself a ring, called the polynomial ring. Thus, $\Bbb F_{11}[x]/(x^2+x+4)$ is the quotient ring you are being asked about. You want to prove this ring is a domain, and finite, hence a field, hence has size $11^2$ (the claims progress in that order, logically speaking). A standard fact covered in texts is that $F[x]$ is always a PID (principal ideal domain) when $F$ is a field, which is to say a domain in which every ideal is principal, and another standard exercise given to students is to prove that if $R$ is a PID and $a\in R$ irreducible, then $R/(a)$ is a domain.
Thus, showing $x^2+x+4$ is irreducible over $\Bbb F_{11}$ (by showing it has no roots, which suffices because the polynomial is quadratic) implies $\Bbb F_{11}[x]/(x^2+x+4)$ is a domain. It is easy to show that it is a vector space over $\Bbb F_{11}$ with basis $\{1,x\}$, therefore has $|{\Bbb F}_{11}\times\Bbb F_{11}|=11^2$ elements and hence is finite; since it is a finite domain it must be a field itself. 
As a general fact, if $p(x)\in F[x]$ is a polynomial and $F$ a field, then $F[x]/(p(x))$ is a vector space over $F$ with basis $\{1,x,x^2,\cdots,x^{n-1}\}$, where $n=\deg p(x)$. The fact that the quotient ring is a vector space is clear. Showing the given basis is in fact such requires a student to show that any coset $g(x)+(p(x))$ with $\deg g(x)\ge n$ can be rewritten as $h(x)+(p(x))$ with $\deg h(x)<\deg g(x)$; one can use this to reduce any coset of $(p(x))$ to the form $ax^{n-1}+\cdots+bx+c+(p(x))$ hence the stated basis is in fact such. As a hint, given $g(x)$, one may write $g(x)=a(x)p(x)+b(x)$ via the division algorithm.
