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Let $R$ a ring (with $1\in R$). I want to show that $R$ is a field $\iff$ it has exactly two ideals.

For $\implies $ it's obvious. Indeed, if $R$ is a field and $I\neq \{0\}$ an ideal, then, $I=A$ since $$x\in I\implies xx^{-1}\in I$$ and thus $1\in I$. In particular, if $x\in A$, then $x\cdot 1\in I$ and thus $x\in I$.

For the converse I don't really know how to do. My idea was the following one:

I show that if $R$ is not a field then it has more that two ideals. Indeed, let $R$ be not a field. Then, there is $x\in R$ that has no inverse. In particular, $(x)\notin \{R,\{0\}\}$. Indeed, if $(x)=(0)$ then $x=0$ which is a contradiction. If $(x)=R$, then $1\in (x)$ and thus there is $n$ s.t. $x^n=1$ what prove that $x$ is invertible and is also a contradiction.

Is it correct ? Is there a more direct proof ? i.e. like this:

Let $A$ and $\{0\}$ the two ideal of $R$. How can I show that all element of $A\backslash \{0\}$ are invertible ? By contradiction, if not, the $$xy\neq 1$$ for all $y\in R$. Therefore, $(x)$ doesn't contain $1$, and thus, $R$ has at least three ideal, which is a contradiction.

Is it correct ?

user26857
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user352653
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    You also need commutativity. A division ring has only one (left/right/bilateral) ideal, but it might not be commutative. –  Oct 21 '16 at 10:00
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    @G.Sassatelli And there are simple rings that aren't division rings. – egreg Oct 21 '16 at 10:12
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    $1\in (x)$ doesn't mean that there is an $n$ such that $x^n=1$. It does, however, mean that there is a $y\in R$ such that $xy=1$. Otherwise your ideas are good and your proof is correct, although the language could use some work. – Arthur Oct 21 '16 at 10:50
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    @AlexM. This is not a duplicate, since he's not asking for a proof with nothing to show for it, he's asking about verification of his own proof. – Arthur Oct 21 '16 at 10:52
  • @G.Sassatelli: $R$ is indeed commutative, but I don't see where I needed it in my proof... thank you. – user352653 Oct 21 '16 at 20:08

2 Answers2

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Your proof is essentially correct. I would have only stated it a bit differently (and more succintly), as follows.

Let $0 \ne x \in R$ and consider the ideal $(x)$. Clearly, $(x) \ne 0$, therefore necessarily $(x) = R$. This means that there exist $y \in R$ such that $yx = 1$. There also exist $z \in R$ such that $xz = 1$ (because $(x)$ is a bilateral ideal). Multiplying $yx = 1$ by $z$ on the right, we get $y(xz) = z$, i.e. $y = z$, so the left and right inverses coincide, therefore $x$ is invertible. We have therefore proven that every non-zero element is inversible, so $R$ is a field.

Alex M.
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    This assumes commutativity. For example in the matrix ring $M_2(K)$, $K$ a field, the ideal generated by an element $x=\pmatrix{0&1\cr0&0\cr}$ is the entire ring. Yet $x$ has neither left not right inverse. In fact, $M_2(K)$ has only the two trivial ideals, yet it is not a field. The claim is wrong without the commutativity assumption. – Jyrki Lahtonen Oct 21 '16 at 10:13
  • @JyrkiLahtonen: You are right, but this remark had already been made by G. Sasatelli before I wrote my answer (see the comments below the question), so I haven't included it in my post. – Alex M. Oct 21 '16 at 10:32
  • Oops. I did not check the time-stamps, sorry. – Jyrki Lahtonen Oct 21 '16 at 10:44
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Your proof is incorrect. By a small detail, but important:

If $(x)=R$, then $1\in(x)$ and so there exists $y\in R$ with $xy=1$.

Maybe your writing of $x^n=1$ is just a typo, but…

You are also assuming $R$ is commutative, otherwise the statement is false. There are (noncommutative) rings having exactly two (two-sided) ideals, but are not even division rings. For instance the matrix rings over a field (or a division ring, more generally).

egreg
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