ideal of $\Bbb F2[x]/(x^2)$

Question

Let $R = \Bbb F2[x]/(x^2)$. Determine all the ideals in $R$.

I have problem about understanding the part that the Ideal of $R/1$ and have no ideal how to deal with this problem.

Answer 1

You said you know that as a set, $\mathbb{F}_2[x]/(x^2)=\{[0],[1],[x],[x+1]\}$. You should know that an ideal $I=R$ if and only if $1 \in R$. You should know also that every ideal $I$ contains $0$. Since the ring is 'small', let's use this to find all the ideals. [Before beginning, look at your ring carefully and note that $[x][x]=[x^2]=[0]$, $[x]+[x]=[2x]=[0]$, and $[1]+[1]=[2]=[0]$.]

You always have two trivial ideals: What are they$?$

Now suppose $I$ is an ideal which is not one of the trivial ideals, then one of $[x]\in I$ or $[x+1] \in I$. We first show that both cannot be in $I$:

Suppose $[x],[x+1] \in I$. Then $[x][x+1]=? \in I$, which by the above comments means $I=?$ [You fill in the $?$'s.]

Suppose $[x] \in I$. Then the ideal generated by $[x]$ is contained in $I$. What is this ideal$?$ [Hint: Just compute this. Then since $[x+1] \notin I$ by our preceding work, you have computed $I$.]

Now suppose $[x+1] \in I$. Then the ideal generated by $[x+1]$ is contained in $I$. But then similar to the work above, what element is in $I?$ What must $I$ be then$?$

Now celebrate since you're done! You've gone through every possible element that any ideal can contain; hence, you've completed all the possibilities!

Answer 2

The ideal $(1)$ is actually $R$. Indeed, $x\in(1)$ iff $x=1\cdot y$, namely $x=y$. The ideal $(0)$ is the only zero itself. Notice that this holds for any ring $R$, namely this is not a specific case.

Finally: $(x)$ contains only $0,x$, indeed $x\cdot x=x^2=0$ and $x\cdot (x+1)=x$; and the ideal $(x+1)$ contains $1$, indeed $(x+1)(x+1)=x^2+2x+1=1$, hence it coincides with $R$.