Here is the problem and I'm a bit stuck in my proof.

We have an uncountable field $k$ and a transcendental element $t$ over $k$ so we consider the field $k(t)$ and want to prove that $\dim(k(t)/k)$ is uncountable.

The idea is to consider the vectors $\frac{1}{t-\lambda}$ for $\lambda \in k$ and to show that they are linearly independent over $k$.

Let us take $c_1,c_2,...,c_n$ elements of $k$ such that : $$\sum_{i=1}^{n}\frac{c_i}{t-\lambda_i}=0$$

So we get that: $$\sum_{i=1}^{n}c_i\prod_{j\neq i}(t-\lambda_j)=0$$

But $t$ being transcendental this polynomial in $t$ must be $0$ and all the coefficients must be $0$.

From here I don't know how to conclude so any help would be very welcomed.

  • 1
    $\begingroup$ You write : '... $t$ being transcendental ...must be 0". This is not clear since there could be cancellations. And that is why you need to evaluate it at different points $\lambda_i$'s. $\endgroup$ Commented May 26, 2017 at 0:42

1 Answer 1



$$\alpha = \sum_{i=1}^{n} c_{i} \prod_{j \neq i} (t- \lambda_{j}) \in k[t] \subset k(t)$$

As a $k$-algebra, $k[t]$ is isomorphic to the polynomial ring $k[X]$ in one variable, and $k[X]$ is universal in the following sense: for any $k$-algebra $A$ and $a \in A$, there is a morphism of $k$-algebras $k[X] \to A$ sending $X$ to $a$. Applying this to our situation with $A = k$, there is a morphism of $k$-algebras $k[t] \to k$ which sends $t$ to $a$ for any $a \in k$. In particular, for each $i$, there is a morphism $\varphi_{i} \colon k[t] \to k$ satisfying $\varphi_{i}(t) = \lambda_{i}$.

Since $\alpha$ is the zero element of $k[t]$, we must have $\varphi_{i}(\alpha) = 0$ for each $i = 1, \ldots, n$. But we see that for each $i$,

$$\varphi_{i}(\alpha) = c_{i} \prod_{j\neq i} (\lambda_{i} -\lambda_{j}) = 0$$ Since $\lambda_{i} \neq \lambda_{j}$ for each $i \neq j$, we find $c_{i} = 0$ for each $i$, as desired.

  • $\begingroup$ AH thanks, yes I just needed to evaluate in the $\lambda_i$'s to get the conclusion. $\endgroup$
    – WrabbitW
    Commented Apr 21, 2017 at 23:05
  • $\begingroup$ @WrabbitW: no problem, glad I could help! $\endgroup$ Commented Apr 21, 2017 at 23:07

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