TITLE: Prime ideals in quotients of polynomial ring over finite field
QUESTION [5 upvotes]: Reading a book, i found this argument ($\mathbb{F}_2$ is the field with 2 elements):
consider the quotient ring:
$$\mathbb{F}_2[x]/(x+1)^2$$
Then it has only one prime ideal, namely the following:
$$(x+1)\mathbb{F}_2[x]/(x+1)^2$$
I've some questions:

what does the notation $(x+1)\mathbb{F}_2[x]/(x+1)^2$ means, i.e. what are the elements of this ideal?
how can i see that this ideal is prime?
how can i exclude that other primes are there, different from that?

Thanks for any help

REPLY [7 votes]: This is one of those cases where knowing less makes everything a lot simpler. Let $R$ be a commutative ring, and $I$ an ideal in $R$.
If $J$ is another ideal in $R$, the ideal $JR/I$ means the image of the ideal $J$ in $R/I$, given by the natural projection $p:R \to R/I$.
As rschwieb has pointed out below, $JR/I$ is actually just $(J+I)/I$.
In this case, $J=(x+1), R = \mathbb{F}_2[x], I = (x+1)^2$.
There is a one-to-one correspondence between ideals in $R$ containing $I$ and ideals of $R/I$, sending $A \to p(A)$ for $A$ an ideal in $R$, and $B \to p^{-1}(B)$, for $B$ an ideal in $R/I$. This is also a correspondence between prime ideals. Have you seen this result?
Thus what remains to be checked is that the ideal $(x+1)$ is the only prime ideal in $\mathbb{F}_2[x]$ containing $(x+1)^2$. Luckily, $\mathbb{F}_2[x]$ doesn't contain that many elements of small degree, so this can be checked manually.