Defining isomorphism of rings

Question

I'm having some difficulty with the following problem:

Define an isomorphism of rings

$$f: \mathbb{Q} [T]/(T^2-d) \rightarrow \mathbb{Q}(\sqrt{d})$$ and an isomorphism

$$g: \mathbb{R} [T]/(T^2+1) \rightarrow \mathbb{C},$$ where $d$ is an element of $\mathbb{Z}\setminus\{0\}$.

I'm not quite sure how I'm supposed to find a fitting isomorphism. The hint says that I should define a ring homomorphism $f: \mathbb{Q} [T] \rightarrow \mathbb{C}$ with ${Q}(\sqrt{d})=\mathrm{Im}(f)$.

Thanks in advance for any help.

Answer 1

Consider the unique ring homomorphism $$ \varphi\colon \mathbb{Q}[T]\to\mathbb{Q}(\sqrt{d}) $$ such that $\varphi(a)=a$ for $a\in\mathbb{Q}$ and $\varphi(T)=\sqrt{d}$. Is this homomorphism surjective? What's its kernel?

For the other question, consider the similar $\gamma\colon\mathbb{R}[T]\to\mathbb{C}$ with $\gamma(T)=i$.

Remember also the homomorphism theorems.

Answer 2

Hint:

Look at the elements in each ring, and then you'll probably see a natural isomorphism.

• $R_1=\Bbb Q[T]/(T^2-d)$. In $R_1$ every polynomial that is contained inside the ideal $(T^2-d)$ serves as $0$. Another way to think about it is that you can replace any $T^2$ with $d$ (since $T^2-d\iff T^2=d$). Effectively this means that you can reduce any polynomial to a linear one, f. ex., $$T^4+T+3\\=(T^2)^2+T+3\\=d^2+T+3$$ This means that $$R_1=\{a_1T+a_0\mid a_1,a_0\in\Bbb Q\}$$ Compare this to the ring (field) $$\Bbb Q(\sqrt d)=\{a_1\sqrt d+a_0\mid a_1,a_0\in\Bbb Q\}$$ Notice that in order to represent each element of $\Bbb Q(\sqrt d)$ as $a_1\sqrt d+a_0$, we need $\sqrt d\not\in\Bbb Q$. If $\sqrt d\in\Bbb Q$ then $R_1$ is not an integral domain. Given $\sqrt d\not\in\Bbb Q$, can you see a natural way to construct the isomorphism $f:R_1\rightarrow \Bbb Q(\sqrt d)$?

• $R_2=\Bbb R[T]/(T^2+1)$, the idea is exactly the same for $R_2$ as with $R_1$

To show that $f$ (and $g$) is an isomorphism, you can either find an inverse homomorphism, or show that both:

• $\ker(f)=0$
• $f$ is surjective, that is, if I give you an element $y\in \Bbb Q(\sqrt d)$, then you can find an $x\in R_1$ that maps to it ($f(x)=y$).