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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 :/

Deirdre
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2 Answers2

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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.

Arturo Magidin
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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 $R_2$.

Eugene Zhang
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