How do I find the ideal $I+J$ and quotient $R/(I+J)$?

Question

This is a homework problem:

Consider the polynomial ring $R=\mathbb Z_2[x_0,x_1,\dots,x_n]$. Let $I=\langle x_0x_1\cdots x_n\rangle$ and $J=\langle x_0+x_1,x_0+x_2,\dots,x_0+x_n\rangle$. Find $I+J$. Hence find $R/(I+J)$.

I first tried thinking about a more general statement. If $R$ is a commutative ring with unity and $I=\langle f_1,\dots,f_i\rangle$ , $J=\langle g_1,\dots, g_j\rangle$ are ideals of $R$ then what can we say about the generators of $I+J$ in terms of the generators of $I$ and $J$? But even in the case if both are principal - $I=\langle f\rangle$ and $J=\langle g\rangle$ I am not sure how to proceed. I am convinced it is not $\langle a+b\rangle$. How would I proceed to tackle such a problem?

Thank you.

Answer 1

In your general case, we have $I+J=\langle f_1,\ldots,f_i,g_1,\ldots,g_j\rangle$.

In this case, that means we have the ideal $$I+J=\langle x_0+x_1,\ldots,x_0+x_n,x_0x_1\cdots x_n\rangle$$Now note that $$x_0x_1\cdots x_n+(x_0+x_1)x_0x_2x_3\cdots x_n= x_0^2x_2x_3\cdots x_n\in I+J$$(remember that our coefficients are in $\Bbb Z_2$, so $+$ and $-$ are the same). Keeping this up, we get $x_0^{n+1}\in I+J$. Also note that this process is reversible, so $$ x_0x_1\cdots x_n\in \langle x_0+x_1,\ldots,x_0+x_n,x_0^{n+1}\rangle\implies I+J= \langle x_0+x_1,\ldots,x_0+x_n,x_0^{n+1}\rangle$$ We are now ready to tackle $R/(I+J)$. Note that when calculating a quotient by an ideal with more than one generator, you may divide out by one generator at the time. Each of the generators of $I$ has the effect of saying "In the quotient, $x_0$ and $x_i$ are the same" (again, coefficients in $\Bbb Z_2$; the usual generator for that statement is $x_0-x_i$). That means that you will eventually reach the point $R/(I+J)\approx \Bbb Z_2[x_0]/\langle x_0^{n+1}\rangle$, which is as far as we can go.

To be fair, we didn't really need to prove that $x_0^{n+1}$ substitutes for $x_0x_1\cdots x_n$ as a generator of $I+J$. We could have just started the division, and after the first one say "We are now in a quotient where $x_1$ and $x_0$ are the same, so $x_0x_1\cdots$ is the same as $x_0^2x_2\cdots x_n$", and begin replacing it at that level instead. You would still get the same result in the end. However, you were told to find $I+J$, and I think that the generator substitution makes for a nicer expression for that ideal.

Answer 2

In my opinion, the exercise's approach is going about the problem backwards. (or, the exercise just wanted you to do something trivial to obtain $I+J$ rather than do any calculation)

The ring $R/J$ is so simple that it is much easier to compute

$$ R / (I+J) \cong (R / J) / I $$

and then if you want to, find $I+J$ from that.

On the right hand side, by $(R/J)/I$, I really mean to take the quotient by the ideal $\pi(I)$, where $\pi$ is the projection $R \to R/J$. It's simple to describe: $\pi(I)$ is the ideal generated by $\pi(i)$ where $i$ is the generator of $I$.

This is, in my opinion, the simpler way to express the relevant isomorphism theorem; the more usual way is by

$$ R / (I+J) \cong (R/J) / ((I+J) / J) $$

While this still involves $I+J$ on the right hand side, we only need the trivial thing: if $I$ is generated by a set $S$ and $J$ is generated by a set $T$, then $I+J$ is generated by the set $S \cup T$, so we have

$$I + J = \langle x_0x_1\cdots x_n, x_0+x_1,x_0+x_2,\cdots,x_0+x_n\rangle $$

and note that all of the generators of $J$ vanish after modding out by $J$. That is, $(I+J)/J = \pi(I)$.