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Let $R = K[x], K$ a field. Define for $a \space \epsilon \space K$ the ideal $ I_a := (x-a)$ in $K[x]$ and see $I_a$ as an $R$ module. Using that for $ a \space \epsilon \space K$ and $ b \space \epsilon \space K$, $I_a$ and $I_b$ are isomorphic, prove that the quotient modules $R/I_a$ and $R/I_b$ are not isomorphic as $R$ -modules.

How should I prove this? I have tried doing it with the first isomorphism theorem for modules, constructing a function $ \theta $ with $ ker(\theta) = I_a$ but I don't seem to get anywhere. Does anyone know a proper solution to this question? Thanks.

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    Have you tried the morphism $K[x]\to K$ with $f\mapsto f(a)$? – 2017-01-13
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    Yes I've tried it, the $ker$ is indeed $I_a$ but does this have $R/I_b$ as image? I mean, what seems logic in my eyes is doing a map to $R/I_b$ to show that it is isomorphic to $R/ker(\theta)$ with $ker(\theta)$ being $I_a$ but I don't seem to find a map with $R/I_a$ as image. – 2017-01-13
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    I don't think that this map helps you since it makes $R/I_b$ and $R/I_a$ isomorphic to $K$ via the isomorphism theorems. But this is only as $K$ modules. – 2017-01-13
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    Okay, then how should I do it? I'm really confused now, sorry. – 2017-01-13
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    Sorry, I didn't want to confuse you. I was just trying to elaborate on the point you made in the second paragraph about $ker(\theta)=I_a$. I'm not sure about it either. – 2017-01-13
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    What if I suppose they are isomorphic and construct an isomorphism, would I come to a contradiction? Maybe I could try to to this. – 2017-01-13

2 Answers 2

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Suppose you have a isomorphism $f:R/I_b \to R/I_a$ as $R$ modules. Observe that $x=b$ in $R/I_b$ and $x=a$ in $R/I_a$. Then you have

$$ bf(1)=f(b)=f(x)=xf(1)=af(1) $$

Now recall that the module structure in $R/I_a$ comes from the projection map $\pi$. Now by assumption $a\neq b$. And this means $(a-b)f(1)=0$ implies that $f(1)\in I_a$ since $a-b \notin (x-a)=ker(\pi)$ but this means $f(1)=0 $ in $R/I_a$ i.e. $f(1)\in ker(f)$ contradicting $f$ being an isomorphism.

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    Ok , I understand most of this. Only, why is $x = b$ in $R/I_b$ and the same for $a$. – 2017-01-13
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    This is true since $R/I_a=K[x]/(x-a)$ and therefore the equivalence class of $x$ is just $a$ since $x-a=0$ in $R/I_a$. Same goes for $R/I_b$. – 2017-01-13
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    YES! I get it now, thank you sir. – 2017-01-13
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If they were isomorphic, they would have the same annihilator. Unfortunately $$\operatorname{Ann}_R R/I_a=I_a, \enspace\text{whereas}\quad\operatorname{Ann}_R R/I_b=I_b.$$ Note, however, they're isomorphic as $k$-vector spaces.

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    Wow! This is a pretty simple way to look at it, thanks! – 2017-01-13
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    Why are the annihilators _equal_ and not only isomorphic? – 2017-01-13
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    @Alphonse You basically do the same as in my answer to see this. When there is a isomorphism you can use it to see that everything that kills one module also kills the other one. – 2017-01-13
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    @Alphonse: it's more general than this precise situation: if $f\colon M\simeq N$, and if $aM=0$, then for any $n\in N$, we have $n=f(f^{-1}(n))$, so $an=f(af^{-1}(n))=f(0)=0$. – 2017-01-13