This is part of Exercise 2.1.5(2) of Springer's, "Linear Algebraic Groups (Second Edition)". It is likely to be a simple question.
Fix an algebraically closed field $k$. Let $n\in \Bbb N$.
The Question:
Show that the subgroup $\Bbb U_n$ of all unipotent upper triangular matrices of the linear algebraic group $\Bbb{GL}_n(k)$ is closed (w.r.t. the Zariski topology).
The Details:
A definition of a linear algebraic group is given in this previous question of mine. I use some of the notation established there, here:
In Springer's book:
$\require{AMScd}$
2.1.1. An algebraic group is an algebraic variety $G$ which is also a group such that the maps defining the group structure $\mu: G\times G\to G$ with $\mu(x,y)=xy$ and $i:x\mapsto x^{-1}$ are morphisms of varieties. [. . .] If the underlying variety is affine, $G$ is a linear algebraic group.
and
2.1.2. Let $G$ be a linear algebraic group and put $A=k[G]$. [. . .] [T]he morphisms $\mu$ and $i$ are defined by an algebra homomorpism $\Delta: A\to A\otimes_k A$ (called comultiplication) and an algebra isomorphism $\iota: A\to A$ (called antipode). Moreover, the identity element is a homomorphism $e:A\to k$. Denote by $m:A\otimes A\to A$ the multiplication map (so $m(f\otimes g)=fg$) and let $\epsilon$ be the composite of $e$ and the inclusion map $k\to A$.
The group axioms are expressed by the following properties:
(associativity) the homomorphisms $\Delta \otimes {\rm id}$ and ${\rm id}\otimes \Delta$ of $A$ to $A\otimes A\otimes A$ coincide;${}^{\dagger}$
(existence of inverse) $m\circ(\iota\otimes{\rm id})\circ \Delta =m\circ({\rm id}\otimes \iota)\circ \Delta =\epsilon;$
(existence of identity element) $(e\otimes {\rm id})\circ \Delta =({\rm id}\otimes e)\circ \Delta ={\rm id}$ (we identify $k\otimes A$ and $A\otimes k$ with $A$).
The properties can also be expressed by commutative properties of the following diagrams:
$$\begin{CD}A @>\Delta>> A\otimes A\\
@V \Delta V V @VV {\rm id}\otimes\Delta V\\
A\otimes A @>>\Delta \otimes {\rm id} > A\otimes A\otimes A,
\end{CD}$$$$\begin{CD}A\otimes A @>\iota\otimes{\rm id}>> A\otimes A\\
@A \Delta AA @VV m V\\
A @>>\epsilon > A\\
@V \Delta VV @AA m A\\
A\otimes A @>> {\rm id}\otimes \iota > A\otimes A,
\end{CD}$$$$\begin{CD} A @<e\otimes{\rm id}<< A\otimes A\\ @A {\rm id}\otimes e AA {_{\rlap{\ {\rm id}}}\style{display: inline-block; transform: rotate(30deg)}{{\xleftarrow[\rule{2em}{0em}]{}}}} @AA \Delta A\\ A\otimes A @<<\Delta < A. \end{CD}$$
For a definition of the Zariski topology and relevant concepts, see this question of mine.
Quoting page 23 of Springer's book:
View the set $\Bbb M_n$ of all $n\times n$-matrices as $k^{n^2}$ in the obvious manner. For $X\in\Bbb M_n$ let $D(X)$ be its determinant. Then $D$ is a regular function on $\Bbb M_n$. The general linear group $\Bbb{GL}_n$ is the principal open set $\{ X\in\Bbb M_n\mid D(X)\neq 0\}$, with matrix multiplication as group operation.
We have $A=k[T_{ij}, D^{-1}]_{1\le i,j\le n}$ with $D=\det(T_{ij})$. Now $\Delta$ is given by
$$\Delta T_{ij}=\sum_{h=1}^n T_{ih}\otimes T_{hj}$$
and $\iota T_{ij}$ is the $(i, j)$-entry of the matrix $(T_{ab})^{-1}$. The identity $e$ sends $T_{ij}$ to $\delta_{ij}$ (Kronecker symbol). [. . .] Since $\Bbb M_n$ is an irreducible variety so is $\Bbb{GL}_n$.
Define $\Bbb U_n$ to be the linear algebraic group of unipotent upper triangular matrices over $k$, understood as a subgroup of $\Bbb{GL}_n$; its elements are of the form
$$\begin{pmatrix}
1 & * & * & \dots & *\\
0 & 1 & * & \dots & * \\
0 & 0 & \ddots & \ddots & \vdots \\
\vdots & & & & * \\
0 & 0 & \dots & 0 & 1
\end{pmatrix};$$
that is, an upper triangular matrix whose diagonal entries are all $1$.
A subgroup of a linear algebraic group is closed if it is closed with respect to the Zariski topology.
Thoughts:
We have $\Bbb U_1=\{(1)\}$, the trivial group, which is closed as it is finite.
By inspection,
$$\begin{align}\Bbb U_n&=
\left\{
\begin{pmatrix}
1 & t\\
0 & 1
\end{pmatrix}
\,\middle |\, t\in k\right\}\\
&\cong k[X_1]
\end{align}$$
is closed.
My guess/intuition is that
$$\Bbb U_n\cong k[\{ X_{ij}\mid 1\le i<j\le n\}].$$
I think this because the given $T_{ij}$-entries for $1\le i<j\le n$ do not influence the determinant at all and so they are free to vary, and the rest are constants.
However, I don't think this is enough to show that $\Bbb U_n$ is closed. I don't know how to prove my guess, even if it is on the right track.
$\dagger$: I think this should be "$({\rm id}\otimes \Delta)\circ\Delta$ and $(\Delta\otimes{\rm id})\circ\Delta$ [. . .] coincide."
Best Answer
Your guess is correct and it gives the "absolute" description of $\mathbb{U}_n$. However, to see it is a closed subgroup, we need the "relative" description of $\mathbb{U}_n\subset \mathbb{G}\mathbb{L}_n(k)$. In fact this is simple, too: $\mathbb{U}_n$ is defined by the equations $$X_{ii}=1\ (1\leq i\leq n),\ X_{ij}=0\ (i> j)$$ inside $\mathbb{G}\mathbb{L}_n(k)$. It would be clear that $\mathbb{U}_n$ is Zariski-closed.