# Product ring isomorphism from example 11.6.3 in Artin's Algebra

I am currently reading chapter 11.6 in Artin's Algebra on Product Rings. There's a proposition that says if $$e$$ is an idempotent element of a ring $$S$$ and $$e' = 1 -e$$ then $$S \cong eS \times e'S$$. I am struggling with an example I'm hoping to get help:

Example 11.6.3: Let $$R'$$ be obtained by adjoining an element $$\delta$$ to $$\mathbb{F}_{11}$$ with the relation $$\delta^2-3=0$$. Its elements are the $$11^2$$ linear combinations $$a+b\delta$$, with $$a,b \in \mathbb{F}_{11}$$ and $$\delta^2=3$$. This is not a field since $$\mathbb{F}_{11}$$ contains two square roots $$\pm5$$ of $$3$$. The elements $$e = \delta -5$$ and $$e'=-\delta-5$$ are idempotents in $$R'$$. Therefore $$R'$$ is isomorphic to the product $$eR'\times e'R'$$.

(So far I am able to understand until here. But I don't understand the rest). Parentheses mine below.

Since the order of $$R'$$ is $$11^2$$, $$|eR|=|e'R'|=11$$ (I don't see why). The rings $$eR'$$ and $$e'R'$$ are both isomorphic to $$\mathbb{F}_{11}$$ (bigger why), and $$R'$$ is isomorphic to the product ring $$\mathbb{F}_{11} \times \mathbb{F}_{11}$$. (This is given from the last two sentences).

What should I do to see the above? I know that $$eR'$$ and $$e'R'$$ are ideals of $$R'$$. If their orders are $$11$$ then I could try to construct an isomorphism for the second sentence. But how do I see that they have $$11$$ element each in $$eR'$$ and $$e'R'$$?

Thank you so much.

As noted, the ring $$R'$$ is isomorphic to $$eR'\times e'R'$$, and $$|R'|=11\times11$$. It follows that $$|eR'|\times|e'R'|=|R'|=11\times11,$$ So $$|eR'|=11^a$$ and $$|e'R'|=11^b$$ for some nonnegative integers $$a$$ and $$b$$ with $$a+b=2$$.

Of course $$0,e\in eR'$$ and $$0,e'\in e'R'$$, where $$e\neq0$$ and $$e'\neq0$$, so $$|eR'|>1$$ and $$|e'R'|>1$$. Hence $$|eR'|=|e'R'|=11.$$

It follows that both $$eR'$$ and $$e'R'$$ are isomorphic to $$\Bbb{F}_{11}$$, because that is the unique ring (up to isomorphism) of $$11$$ elements. More generally, the unique ring (up to isomorphism) of $$p$$ elements is $$\Bbb{F}_p$$ whenever $$p$$ is prime.

• Thank you so much! Just to confirm that I get it correct, since $eR'$, $e'R'$ are ideals of $R'$, their order must divide the ring $R'$ and hence they can only be $1, 11,$ or $11^2$. Then from the fact that $R' \cong eR' \times e'R'$, we have $|R'| = |eR'|\cdot |e'R'|$. And then we can deduce the order of each is $11$ since neither is the zero ring. And since they are rings, there's only one of order 11 and so they are isomorphic to $\mathbb{F}_{11}$? – Tri Nguyen Feb 18 '19 at 23:52
• Yes, this is exactly right. Though in general, if you know that $A\cong B\times C$ then also $|A|=|B|\times|C|$, so this in fact proves that the orders of $B$ and $C$ divide the order of $A$. – Servaes Feb 18 '19 at 23:55
• This was very helpful. Thank you! The book skipped a couple of steps and that threw me off. – Tri Nguyen Feb 19 '19 at 0:03

For the cardinality you note that $$11^2=\vert R'\vert = \vert eR'\vert \cdot \vert r R'\vert$$. There are not many factorisations of $$11^2$$ as $$11$$ is prime. Both $$eR'$$ and $$rR'$$ are finite integral domains and hence fields. All fields of order $$11$$ are isomorphic to $$\mathbb{F}_{11}$$.

• Thank you! I am able to understand this now. – Tri Nguyen Feb 19 '19 at 0:03