How to think about the object $A\otimes_kk'$. Let $k'/k$ be a finite extension of fields, and let $A$ be a finitely generated commutative $k'$-algebra.  Through $k\hookrightarrow k'$ we can consider $A$ to be a finitely generated commutative $k$-algebra, and then we can consider the $k'$-algebra $A\otimes_kk'$.
Are $A$ and $A\otimes_kk'$ isomorphic as $k'$-algebras?  If not, can we consider $A$ as a subalgebra of $A\otimes_kk'$ via $a\mapsto a\otimes 1$?  I believe this map has the right inverse $a\otimes c\mapsto ca$.  Are there conditions we can impose which make the two $k'$-algebras isomorphic?
In general, I'm having trouble thinking about the object $A\otimes_kk'$.  For $A$, I just think of some quotient of a polynomial ring over $k'$.  Perhaps my question reduces to asking about the object $k'\otimes_kk'$?
 A: The basic idea is that $k' \otimes_k -$ changes the coefficients from the ground field $k$ to a new ground field $k'$, but this will change how ideals used to define quotient $k$-algebras factorize. 
As an exercise, let's contemplate the structure of $\mathbb{C} 
\otimes_{\mathbb{R}} \mathbb{C}$, as a commutative $\mathbb{C}$-algebra. We have an exact sequence of modules (= vector spaces) over $\mathbb{R}$: 
$$0 \to (x^2 + 1)\mathbb{R}[x] \to \mathbb{R}[x] \to \mathbb{R}[x]/(x^2+1) \cong \mathbb{C} \to 0$$ 
and tensoring by $\mathbb{C}$ preserves this exact sequence. We get 
$$\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{C}[x]/(x^2 + 1).$$ 
(More precisely: the functor $\mathbb{C} \otimes_{\mathbb{R}} -: \text{CAlg}_{\mathbb{R}} \to \text{CAlg}_{\mathbb{C}}$ is left adjoint to the forgetful functor $\text{CAlg}_\mathbb{C} \to \text{CAlg}_\mathbb{R}$ from commutative $\mathbb{C}$-algebras to commutative $\mathbb{R}$-algebras, and being a left adjoint, it will preserve quotient algebra constructions.) 
But notice the principal ideal $(x^2 + 1)$ in $\mathbb{C}[x]$ factorizes as $(x+i)(x-i)$. From the Chinese remainder theorem, one has 
$$\mathbb{C}[x]/(x+i)(x-i) \cong \mathbb{C}[x]/(x+i) \times \mathbb{C}[x]/(x-i) \cong \mathbb{C} \times \mathbb{C}$$ 
so that is the structure of the tensor product. You should work this out as explicitly as you can, by exhibiting nontrivial idempotent elements in the tensor product which sum to $1$ but whose product is $0$; these represent the special elements $(1, 0)$ and $(0, 1)$ in $\mathbb{C} \times \mathbb{C}$. 
This is incredibly useful to know. For example, take a rational prime $p$ and consider the ideal it generates in the ring of Gaussian integers $\mathbb{Z}[i]$. Consider the structure of the quotient ring $\mathbb{Z}[i]/(p)$. We have 
$$\mathbb{Z}[i]/(p) \cong \mathbb{Z}[x]/(x^2 + 1, p) \cong \mathbb{F}_p[x]/(x^2 + 1)$$ 
where $\mathbb{F}_p = \mathbb{Z}/(p)$ is the field with $p$ elements. (In effect, we are computing $\mathbb{Z}(p) \otimes_\mathbb{Z} \mathbb{Z}[i]$.) Note that if $(x^2 + 1)$ splits over $\mathbb{F}_p$, i.e., if $x^2 = -1 \pmod p$ has two distinct solutions, then this tensor product has the structure $\mathbb{F}_p \times \mathbb{F}_p$. The kernels of the composite projection maps 
$$\mathbb{Z}[i] \to \mathbb{Z}[i]/(p) \cong \mathbb{F}_p \times \mathbb{F}_p \stackrel{\pi_i}{\to} \mathbb{F}_p,$$ 
where $\pi_i$, $i=1, 2$, are the two projection maps, are principal ideals $(a+b i)$, $(a - b i)$ which gives the factorization $p = (a + bi)(a - bi)$ in the Gaussian integers. In this way, the condition that $-1$ has a square root modulo $p$ (which is the case if $p \equiv 1 \pmod 4$ is equivalent to the expressibility of $p$ as a sum of two squares. 
Bruno has described another application, closely connected with the normal basis theorem in Galois theory. 
A: In general, $A$ and $A\otimes_k k'$ are not isomorphic as $k'$ algebras. Also, remark that $A \otimes_k k'$ has two a priori distinct $k'$ algebra structures, one coming from that of $A$ and the other coming from that of $k'$.
You are right to consider the case $A=k'$, because the trouble starts there. Remark that, unless $k=k'$,  $k' \otimes_k k'$ does not even have the same dimension over $k$ as $k'$...
Something interesting happens when $k'/k$ is finite Galois, with Galois group $G$. In that case, there is a canonical isomorphism of $k$-algebras $k'\otimes_k k' \cong \prod_{\sigma \in G} k'$ defined by taking $x\otimes y$ to the tuple $(x\sigma(y))_{\sigma \in G}$.
In étale cohomology, a finite Galois extension $k'/k$ is analogous to a Galois covering space (of the one-point scheme $\text{Spec }k$). If $E \to X$ is a finite Galois covering of topological spaces, with Galois group $G$, the object analogous to $k'\otimes_k k'$ is $E\times_X E$. This space breaks up into $|G|$ components, which are permuted transitively by the action of $G$ on $E\times_X E$ induced from the monodromy action of $G$ on $E$.
