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I would like to get an example of a finite commutative ring $R$ and a subring $S$ of $R$ that is not an ideal.

I have tried working with $\mathbb Z_n$ and most of the examples I have tried end up being either both or none. Can we say that every subring of $\mathbb Z_n$ is also an ideal? (I think the fact that all subrings of $\mathbb Z$ look like $k\mathbb Z$ for some $k$ which are also the only ideals of $\mathbb Z$ probably factors in here)

Is a result like true in general? That is, is any ideal of a finite commutative ring also a subring?

Thank you.

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2 Answers 2

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Hint: Consider the situation when $R$ is a finite field.

Subhint:

What are the ideals in any field?

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  • $\begingroup$ If $R$ is a finite field then its only ideal are $\{0\}$ and $R$. So I need to see what the non-trivial subrings of $R$ are. But I'm still not able to see examples. For example, for all $\mathbb Z_p$ we still get all the ideals and subrings are just $\{0\}$ and $\mathbb Z_p$ right? $\endgroup$
    – R_D
    Commented Aug 30, 2021 at 6:22
  • $\begingroup$ You're correct on both claims. Finite fields of prime order will not be counterexamples! Do you know any other examples of finite fields? $\endgroup$
    – Stahl
    Commented Aug 30, 2021 at 6:28
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    $\begingroup$ @R_D As a very concrete example, start by considering $\Bbb{F}_3,$ the finite field with $3$ elements. Like in $\Bbb{R},$ there is no element $i\in\Bbb{F}_3$ such that $i^2 = -1,$ and we may formally add such an $i$ to $\Bbb{F}_3$: consider the set $\Bbb{F}_3[i] = \{a + bi\mid a,b\in\Bbb{F}_3\},$ with addition given by $(a + bi) + (a' + b'i) = (a + a') + (b + b')i$ and multiplication given by $(a + bi)(a' + b'i) = (aa' -bb') + (ab' + a'b)i.$ I leave it to you to verify that this is indeed a field, and in particular, a field with $9$ elements. ... $\endgroup$
    – Stahl
    Commented Aug 30, 2021 at 7:22
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    $\begingroup$ In fact, $\Bbb{F}_3[i]$ is an example of a splitting field -- it is the splitting field of $x^2 + 1\in\Bbb{F}_3[x].$ Can you find any nontrivial subrings of this field? More generally, for any prime number $p$ and integer $n > 0,$ there exists a unique (up to isomorphism) field with $p^n$ elements, which may be formed as the splitting field of some polynomial $f_{p,n}\in\Bbb{F}_p[x].$ $\endgroup$
    – Stahl
    Commented Aug 30, 2021 at 7:23
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    $\begingroup$ Yes! The subset $\{a+0i:a\in\mathbb F_3\}$ is a subring but not an ideal. Thank you! $\endgroup$
    – R_D
    Commented Aug 31, 2021 at 5:30
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Let $F_2$ be the field of two elements and $R=F_2\times F_2$ and $S=\{(0,0), (1,1)\}$.

$R$ has exactly four ideals, but none of them are $S$, the prime subring.


Is a result like true in general? That is, is any ideal of a finite commutative ring also a subring?

Depending on your definition of "subring" yes. If a subring is just an additive subgroup that is closed under multiplication (no identity necessary) then yes. If you require some identity, then not always but sometimes. If you require a subring to share the same identity as the containing ring, then "only in the trivial case where the ideal is the whole ring, and no otherwise."

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    $\begingroup$ I have this game I play by myself to see how many questions I can use this particular ring to answer. 🙄 $\endgroup$
    – rschwieb
    Commented Aug 31, 2021 at 13:42

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