# Need help understanding this moderately basic rings/ideals proof

I was doing some practice abstract algebra questions off the internet since I have a quiz coming up soon. However, I am not very skilled at abstract algebra. In fact, I did very average in my group theory class, so I am struggling in my ring theory one. Can someone please help explain what is happening in this proof? I'm very sorry if it's extremely straightforward, I just think I need some time to get used to the way of thinking that's required to solve these questions.

Let $$R_1$$ and $$R_2$$ be commutative rings with identities and let $$R = R_1 × R_2$$. The question asks to show that every ideal $$I$$ of $$R$$ is of the form $$I = I_1 × I_2$$ with $$I_1$$ an ideal of $$R_1$$ and $$I_2$$ an ideal of $$R_2$$.

The proof/solution given goes like this:

Let $$I$$ be an ideal of $$R$$, and let $$(a, b) ∈ I$$. Then $$(a, b)·(1, 0) = (a, 0)$$, and $$(a, b)·(0, 1) = (0, b)$$, so $$I = I_1 × I_2$$, where $$I_1$$ is the set of $$x ∈ R_1$$ such that $$(x, y) ∈ I$$ for some $$y$$. Similar for $$I_2$$.Then it is easy to show that these sets are ideals.

I don't completely understand this though. Not even just a specific part, but I guess how $$I = I_1 \times I_2$$ is implied from what comes before it. In my head, I keep (kind of) feeling like this means that $$a \in I_1$$ and $$b \in I_2$$, but I don't get how/why? Or if I'm wrong, could someone correct that too? :/

Again, really sorry if this is too basic. I genuinely feel dumb in this class sitting with people I can't compete with, so I can't even ask them for help, and I couldn't think of anywhere else I could find an explanation :/

• $a\in I_1$ because $I_1$ consists of all elements that are first coordinates of an element of $I$, and $(a,b)\in I$. $b\in I_2$ because $I_2$ consists of all elements that are the second coordinate of an element of $I$, and $(a,b)\in I$. Feb 8, 2022 at 19:38
• Here is the most well-trodden collection of solutions to the question you are working on, although I see here the nature of your question does not make it a duplicate. Feb 8, 2022 at 21:27

Because $$I$$ is an ideal, if you left- or right-multiply an element of $$I$$ by an element of $$R$$, then the result is in $$I$$.

Let \begin{align*} I_1 &= \{x\in R_1\mid \text{there exists }y\in R_2\text{ such that }(x,y)\in I\}\\ I_2 &= \{y\in R_2\mid \text{there exists }x\in R_1\text{ such that }(x,y)\in I\}. \end{align*} That is, $$I_1$$ is the image of $$I$$ under the projection $$\pi_1\colon R\to R_1$$ on the first coordinate, and $$I_2$$ is the image of $$I$$ under the projection $$\pi_2\colon R\to R_2$$. Because these are the images of an ideal under a surjective group homomorphism, we know that $$I_1$$ is an ideal of $$R_1$$ and $$I_2$$ is an ideal of $$R_2$$ (isomorphism theorems).

Now, by construction, $$I\subseteq I_1\times I_2$$ (verify!).

To show that $$I_1\times I_2\subseteq I$$, let $$a\in I_1$$. Then there exists $$b\in R_2$$ such that $$(a,b)\in I$$. Then $$(1_{R_1},0)(a,b) = (a,0)\in I$$, so $$I_1\times\{0\}\subseteq I$$.

Now prove that likewise $$\{0\}\times I_2\subseteq I$$.

Conclude that $$I_1\times I_2\subseteq I$$. This gives the equality.

Sometimes you only need to follow the definition and ignore others. In this case, it can be proved easily from the definition.

Suppose $$I$$ is an ideal of $$R_1\times R_2$$. Let $$(a,b),(c,d)\in I$$ and $$(r_1,r_2)\in R_1\times R_2$$. Then $$(a,b)-(c,d)=(a-c,b-d)\in I \quad\text{implies}\quad a-c\in \pi_1(I)\quad\text{and}\quad b-d\in \pi_2(I)$$ Where $$\pi_i(I)$$ is the $$i^{th}-$$projection of $$I$$. Also $$(r_1,r_2)(a,b)=(r_1a,r_2b)\in \pi_1(I)\times \pi_2(I)$$ Let $$I_1=\pi_1(I)$$ and $$I_2=\pi_2(I)$$. Then $$I_1$$ and $$I_2$$ are ideals of $$R_1$$ and $$R_2$$.