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For any ring $R$, let $R[X]$ denote the ring of polynomials with coefficients from $R$ and indeterminate $X$.

Which of the following pair of rings are isomorphic:

(a) $ \mathbb{R}[X] $ and $\mathbb{C}[X]$

(b) $\mathbb{Q}[X]/(X^2-X)$ and $\mathbb{Q} \times \mathbb{Q}$.

Here are my approaches for both problem:

(a) In $ \mathbb{C}[X] $ every polynomial has a root which is not the case for $ \mathbb{R}[X] $. So they are not isomorphic.

I'm not sure whether this property will make the rings no-isomorphic.

(b) Observation: $\mathbb{Q}[X]/(X)$ and $\mathbb{Q}[X]/(X-1)$ are isomorphic to $\mathbb{Q}$ because both of the quotient rings represent the constant polynomials of $\mathbb{Q}[X] $ which are $\mathbb{Q}$.

So by Chinese Remainder Theorem, $$\mathbb{Q}[X]/(X^2-X)=\mathbb{Q}[X]/(X(X-1)) \simeq \mathbb{Q}[X]/(X) \times \mathbb{Q}[X]/(X-1) \simeq \mathbb{Q} \times \mathbb{Q}$$ as the ideals $(X)$ and $(X-1)$ are co-maximal.

Just want to verify if the proofs are correct. If I'm missing out something please point out it. If the proofs are entirely wrong then I would greatly appreciate if you can give a hint .

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    Your arguments look fine to me. – 2017-02-24
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    "Every polynomial **in** $\mathbb C[X]$ has a root" is not something that is about the _ring structure_ of $\mathbb C[X]$ alone, so you cannot be sure it is preserved by isomorphisms. – 2017-02-24
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    Your argument for (a) doesn't make any sense. What do you mean? How does that prove that they are not isomorphic? Every polynomial in $ \mathbb R[X] $ certainly has a root in $ \mathbb C $... Your argument for (b) is fine. – 2017-02-24
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    @Starfall ,Not every polynomial in $\mathbb{R}[X] $ doesn't have a root in $\mathbb{R} $ . For example take $p(x)=x^2+1$ then the equation $p(x)=0$ does not have any root in $\mathbb{x}$ . – 2017-02-24
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    For (a), perhaps it will be useful to think about how to show that $\mathbb{C}$ and $\mathbb{R}$ are not isomorphic as rings. As others have mentioned, your argument for (b) is correct. – 2017-02-24
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    @user26857 You should prove that an isomorphism preserves degrees. – 2017-02-24
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    @user26857 Why should the degree of the image of $X^2+1$ be at least $2$, under the assumed isomorphism? – 2017-02-24

1 Answers 1

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Your argument for $\mathbb{R}[X]$ and $\mathbb{C}[X]$ is incorrect. You can however say that the latter ring has an element whose square is $-1$, which the former hasn't.

The second part is good. Another approach is by considering the ring homomorphism $\varphi\colon\mathbb{Q}[X]\to\mathbb{Q}\times\mathbb{Q}$ defined, for $f\in\mathbb{Q}[X]$, by $$ \varphi(f)=(f(0),f(1)) $$ It's easy to see that the kernel is exactly $(X^2-X)$:

  1. $X(X-1)\in\ker\varphi$
  2. If $f\in\ker\varphi$ then $f(X)=(X^2-X)q(X)+aX+b$; since $f(0)=0$, we have $b=0$; since $f(1)=0$, we have $a=0$.

Finish by proving that $\varphi$ is surjective.