Tensor product of a number field $K$ and the $p$-adic integers

Imagine that $K=\mathbb{Q}[\alpha]$ for some algebraic number $\alpha$. Let $m(x)$ be the minimal polynomial of $\alpha$. Then we know that $K\cong \mathbb{Q}[x]/\langle m(x)\rangle$. As $\mathbb{Q}_p$ is an extension field of the rationals, the polynomial $m(x)$ (irreducible in $\mathbb{Q}[x]$) may factor into a product of irreducible factors $$ m(x)=\prod_i m_i(x), $$ where $m_i(x)$ are irreducible polynomials in $\mathbb{Q}_p[x]$. Because $m(x)$ is separable, the polynomials $m_i(x)$ are distinct. Therefore the Chinese Remainder Theorem says that $$ \mathbb{Q}_p[x]/\langle m(x)\rangle\cong\bigoplus_i \mathbb{Q}_p[x]/\langle m_i(x)\rangle. $$ Here on the right hand side the summands $\mathbb{Q}_p[x]/\langle m_i(x)\rangle$ are all extension fields of $\mathbb{Q}_p$, because the polynomials $m_i(x)$ are irreducible. OTOH $$ \mathbb{Q}_p[x]/\langle m(x)\rangle\cong\mathbb{Q}[x]/\langle m(x)\rangle\otimes\mathbb{Q}_p\cong K\otimes\mathbb{Q}_p. $$ Undoubtedly you can guess the rest - the fields $\mathbb{Q}_p[x]/\langle m_i(x)\rangle$ are the fields $K_{p,i}$.

Others will have fuller, more accurate, and more efficient explanations than mine, but let me give it a try.

Let $\mathscr O$ be the ring of algebraic integers of $K$. Then $p\mathscr O=\prod_i\mathfrak p_i^{e_i}$, where $\mathfrak p_1,\ldots,\mathfrak p_g$ are the prime ideals of $\mathscr O$ dividing $p$. The $\mathfrak p_i$-adic completion of $K$ is a field $K_{p,i}$ that’s of degree $n_i$ over $\mathbb Q_p$, and we have $n_i=e_if_i$ where $e_i$ is as above and $f_i$ is the residue-class field degree at $p_i$, and furthermore $\sum_in_i=[K\colon\mathbb Q]$.

Then the map $K\to\bigoplus_iK_{p,i}$ by sending an element to the $g$-tuple of all its images in the various completions extends to $K\otimes\mathbb Q_p$. All the remaining details and verifications are left to the reader.

$K \otimes \mathbb{Q}_p$ is a $\mathbb{Q}_p$ algebra, where $\mathbb{Q}_p$ is embedded in $K \otimes \mathbb{Q}_p$ via $x \mapsto 1 \otimes x$. The fact that $\mathbb{Q}_p$ isn't contained in $K$ doesn't matter. The tensor product is over $\mathbb{Q}$, their common base field.

In general, for any 2 fields $K$ and $F$ which are extensions of a common field $E$, we may regard $K \otimes_E F$ as an $F$-algebra in the same way.