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Let $R$ be a commutative ring with $1$, if $R$ contains only finitely many ideals, then every nonzero element of $R$ is either a unit or a zero divisor. I know it's true. How about the converse? i.e. If every nonzero element of $R$ is either a unit or a zero divisor, then $R$ contains only finitely many ideals. Is it true? Can you give me an example about a commutative ring with $1$ that has infinitely many zero divisors? Thanks.

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    $\begingroup$ Example: $\mathbb{F}_2^\mathbb{N}$. Everything is $0$, the unit, or a zero divisor. $\endgroup$ – Daniel Fischer Jul 17 '13 at 19:11
  • $\begingroup$ Another example: math.stackexchange.com/questions/2101395: "Let $R$ be the ring of functions from $[0,1]$ to $\mathbb{R}$. Addition and multiplication are defined as usual: $(f+g)(x)=f(x)+g(x)$ and $(fg)(x)=f(x)g(x)$." $\endgroup$ – Watson Jan 17 '17 at 12:24
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The converse is not true. Consider the ring $R=\mathbb{C}[x,y]/(x^2,y^2,xy)$. Notice that every element has a representative of the form $a+bx+cy$. If $a=0$, this is a zero-divisor. If $a\ne 0$, then $a^{-1}-a^{-2}bx-a^{-2}cy$ is an inverse. So this ring satisfies the hypotheses of the question. Can you exhibit infinitely many ideals of this ring?

$R$ is also an example of an infinite commutative ring with infinitely many zero divisors.

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