How come algebraic closure is not unique? I was getting confused trying to understand why an algebraic closure of a field is not unique. If I consider any rational polynomial in one variable, then the roots may not be in $\mathbb{Q}$ but aren't they always in $\mathbb{C}$? And I there in only on $\mathbb{C}$..? I think I'm missing something very simple here, but I am getting confused with this so I would appreciate any explanation. Thank you. 
 A: All algebraic closures of a given field are isomorphic, but that doesn't mean there is only one.
The rings $\Bbb R[x]$ and $\Bbb R[y]$ aren't the same ring. They are very similar, but not identical. Even more different are the rings $\Bbb R[x]/(x^2 + 1)$ and $\Bbb R[y]/(y^2 + 2)$. They are both algebraic closures of $\Bbb R$, they both represent the complex numbers, and they are isomorphic. But they are very much two distinct rings.
A: While I like Arthur's answer (+1), I'd like to add a perspective on what it actually means for a construction or object to be unique, and why it fails for algebraic closures.
I tend to find categories helpful for formalizing intuitive ideas, so I'll give a categorical formulation of uniqueness.
What is uniqueness?
Suppose we have a category $C$, and a property $P$ that may or may not apply to the objects of $C$, but is a "sane" property in the sense that if $X$ and $Y$ are isomorphic in $C$, then $P(X)=P(Y)$ (i.e. the property is either true for both $X$ and $Y$ or false for both of them).
Then we say that an object $X$ of $C$ is characterized by property $P$ up to unique isomorphism if $P(X)$ and for any other object $Y$ satisfying $P$, there is a unique isomorphism $\alpha : X\to Y$. In particular, this means that for $X$ to be characterized by $P$ up to unique isomorphism, there can only be one isomorphism $X\to X$, which must be the identity map. That is, such objects can have no nontrivial automorphisms.
How does this apply to algebraic closures?
We have to be careful with what exactly our category $C$ is. An algebraic closure $K$ of a field $k$ is an algebraic field extension of $k$ such that every polynomial in $k[x]$ has a root in $K$. Thus $C$ here should be the category of field extensions of $k$. I.e., the objects of $C$ are fields $L$ equipped with ring homomorphisms $\iota_L: k\to L$, and the morphisms of $C$ between field extensions $L/k$ and $M/k$ are ring homomorphisms $\phi : L\to M$ such that $\phi\iota_L = \iota_M$.
To show that the algebraic closure of $\Bbb{R}$ is not unique (up to unique isomorphism), it suffices to notice that as a field extension of $\Bbb{R}$, $\Bbb{C}$ has a nontrivial automorphism over $\Bbb{R}$, namely complex conjugation, $z\mapsto \bar{z}$.
An example where we do have uniqueness
Let $G$ be a group, let $C$ be the category of abelian groups $A$ equipped with morphisms $\phi_A : G\to A$ (I will call these as abelian groups under $G$), and $P(A)$ be the property that for any other abelian group under $G$, $B$, there is a unique morphism in $C$ $\psi :A\to B$. (This property is often simply called being initial in $C$). Note that for $\psi$ to be a morphism in the category of abelian groups under $G$, we must have that $\psi\phi_A = \phi_B$.
If $A$ satisfies $P$, we say that it is the abelianization of $G$, and it is unique up to unique isomorphism (since part of the definition is that it has a unique morphism to any other object in the category, and if the other object also has this property, then these unique morphisms must be inverses). 
It's well known that for any group $G$ we can find such an abelian group, namely it is the quotient of $G$ by its commutator subgroup.
Note on the example
The method in the example is a very common method of defining objects unique up to unique isomorphism. We define some auxiliary category and then define our desired object to be an initial (or terminal) object in the category, and then go about constructing such an object (to prove one actually exists).
Side note
You may also find the answers to this similar question helpful.
