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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.

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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.

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  • $\begingroup$ 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 $\mathbb{F}_{11}$? $\endgroup$ – Tri Nguyen Feb 18 at 23:52
  • $\begingroup$ 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 $A$. $\endgroup$ – Servaes Feb 18 at 23:55
  • $\begingroup$ This was very helpful. Thank you! The book skipped a couple of steps and that threw me off. $\endgroup$ – Tri Nguyen Feb 19 at 0:03
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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 $\mathbb{F}_{11}$.

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

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