Extend a rational number field $\mathbb{Q}$ by using a transcendental number? Here denoting a set of real transcendental numbers $\mathbb{T}$, what can we then say about the structure
$$
\mathbb{Q}(t) = \left\{\, \sum_{k=0}^{+ \infty} a_k t^k\mathrel{}\middle|\mathrel{} a_k \in \mathbb{Q} \,\right\},
\quad \text{for some } t \in \mathbb{T}.
$$
Now it is not a ring, is it? What do we have to add to make it a ring (or a field)?
Is $\mathbb{Q} \cup \mathbb{T}$ a ring (or a field)?
 A: Sure, what you write as $\mathbb{Q}(t)$ is a ring - it's the ring of formal power series with rational coefficients. Usually it's denoted $\mathbb{Q}[[t]]$. 
Any element of the form $$\sum_{k=0}^\infty a_k t^k$$ is invertible if and only if $a_k \neq 0$; if $a_0=0, a_1 \neq 0$, the inverse will take the form $$\sum_{k=-1}^\infty b_k t^k$$ This justifies the fact that the field of fractions of $\mathbb{Q}[[t]]$ (denoted $\mathbb{Q}((t))$) is the ring of formal Laurent series:
$$\sum_{k=-n}^\infty a_k t^k$$ for some integer $n$.

EDIT: I'm no longer satisfied with this answer. What I've written above is just fine if you consider $\mathbb{Q}[[t]]$ just as a ring of formal power series, not trying to evaluate any element of it (by replacing $t$ by the relevant transcendental number). However, many of these series will not converge - pick, say, $a_k = \pm 1$ for something that goes to $\pm \infty$, or $a_k = (-1)^k$ for something that oscillates. The elements of this field don't live in the real numbers - they live on their own, as formal power series. Now, if you restricted this field to those series that do converge, I find it likely that you'll get all of $\mathbb{R}$; just pick a sequence $a_k$ as follows: Have $a_0$ agree with your target real number for the first digit, have $a_0+ta_1$ agree for the first two digits, and so on... and then, we'll have many different power series that go to the same real number! But then you can, for instance, set an equivalence relation, so that if they go to the same real number they're considered the same power series, and then we're just back to $\mathbb{R}$ again.
The question you really seem to want answered, though, is "Can we have a proper subfield of $\mathbb{R}$ that contains both $\mathbb{Q}$ and a specific transcendental $t$?" To which the answer is yes, and this field is denoted $\mathbb{Q}(t)$ (this is not your notation - it refers to the field of fractions of polynomials in $t$). We get this by forming the polynomials in $t$ (unlike the above power series, these can always be evaluated), and then making fractions with them; I described the explicit process for forming this field in an earlier revision, but it's really not that important here. Now, we do have $\mathbb{Q}(t) \subsetneq \mathbb{R}$ - and similarly for any finite set of transcendentals, we have $\mathbb{Q}(t_1,t_2,...,t_n)\subsetneq \mathbb{R}$.
A: The union $\Bbb{Q\cup T}$ is not a ring, nor a field. The reason is that $\pi$ and $\frac{\sqrt2}{\pi}$ are in $\Bbb T$, but their product is not in $\Bbb{Q\cup T}$.
