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Sorry to ask this, I know it's not really a maths question but a definition question, but Googling didn't help. When asked to show that elements in each are irreducible, is it the same?

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There is no difference, because $\mathbb{Q}[\sqrt{-d}]$ is already a field. – M Turgeon May 17 '12 at 18:59
up vote 10 down vote accepted

Here is the general definition of the two notations: given a field $K$, a field $L$ that contains $K$, and an element $a\in L$, then $$K[a]=\{f(a)\mid f\in K[x]\}=\{c_0+c_1a+\cdots+c_na^n\mid c_i\in K, n\geq0\}$$ and $$K(a)=\left\{\tfrac{f(a)}{g(a)}\;\middle\vert\; f,g\in K[x]\text{ where }g(a)\neq 0\right\}=\text{the field of fractions of }K[a].$$ An alternative characterization is that $K[a]$ is the smallest sub$\!$*ring* of $L$ that contains the element $a$ as well as all of $K$, and that $K(a)$ is the smallest such sub$\!$*field*.

It is a straightforward theorem that $K[a]=K(a)$ if and only if $a$ is algebraic over $K$, which means that there is some non-zero polynomial $f\in K[x]$ such that $f(a)=0$. Because $\sqrt{-d}$ is algebraic over $\mathbb{Q}$ - it is a root of the polynomial $x^2+d=0$, which is in $\mathbb{Q}[x]$ - we therefore have that $\mathbb{Q}(\sqrt{-d})=\mathbb{Q}[\sqrt{-d}]$.

In contrast, $\pi$ is transcendental over $\mathbb{Q}$, and $\mathbb{Q}(\pi)$ contains elements like $\frac{1}{\pi}$ and $\frac{3}{\pi^2 + 1}$, while $\mathbb{Q}[\pi]$ does not.

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The notation $\rm\:R[\alpha]\:$ denotes a ring-adjunction, and, analogously, $\rm\:F(\alpha)\:$ denotes a field adjunction. Generally if $\alpha$ is a root of a monic $\rm\:f(x)\:$ over a domain $\rm\:D\:$ then $\rm\:D[\alpha]\:$ is a field iff $\rm\:D\:$ is a field. The same is true for arbitrary integral extensions of domains. See this post for a detailed treament of the quadratic case.

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In general, if you have a field extension $L/K$, and $\alpha\in L$, we define $K[\alpha]$ to be the subring of $L$ generated by $K$ and $\alpha$, whereas we define $K(\alpha)$ to be the subfield of $L$ generated by $K$ and $\alpha$. Now, it sometimes happens that $K[\alpha]$ is already a field, in which case we have $K[\alpha]=K(\alpha)$. This is what happens with $\mathbb{Q}[\sqrt{-d}]$ and $\mathbb{Q}(\sqrt{-d})$.

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