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Choose the real embeddings ${\sigma }_i$ and one from each complex pair ${\tau }_j$. The logarithmic embedding of number field units isIt is a group homomorphism from multiplication to addition. A unit has absolute norm one, so the coordinate sum is zero and its image lies in a Euclidean hyperplane of dimension $r+s-1$.

Here is the required discreteness argument. There are only finitely many algebraic integers of $K$ whose conjugate absolute values are bounded by a fixed constant: their monic minimal polynomials have degree at most $n$, and each integer coefficient is an elementary symmetric function of bounded roots, hence belongs to a fixed finite range. There are finitely many such polynomials and finitely many roots of each. A bounded set of logarithms bounds all conjugate absolute values, so every compact region in the logarithmic hyperplane meets the unit image in finitely many points. Thus the image is a discrete additive subgroup.

Use the allowed Euclidean-group theorem: a discrete subgroup of a finite-dimensional real vector space is free abelian on finitely many vectors, with rank at most the dimension. Therefore $\lambda ({\mathcal{O}}_K^{\times })$ has rank at most $r+s-1$. Its kernel is finite by the same bounded-conjugate argument, now with every absolute value one. Each element of a finite multiplicative group is a root of unity, and conversely every root of unity has all logarithms zero. Thus the kernel is exactly the finite group of roots of unity, and exactly the torsion subgroup of units.

Lift generators of the free logarithmic image to units; together with the finite kernel they generate every unit. HenceThis proves the requested upper bound without invoking the existence half of the Dirichlet unit theorem. The field $\mathbb{Q}(\sqrt{11})$ is a subfield of $\mathbb{R}$, and its only real roots of unity are $\boxed{1,-1}$.

For a number field with $r$ real embeddings ${\sigma }_i$ and $s$ conjugate pairs of complex embeddings ${\tau }_j$, use the logarithmic embedding of number field unitsIt is a group homomorphism from multiplication to addition. A unit has field norm $\pm 1$, so the coordinates sum to zero. Deleting one coordinate identifies this hyperplane with ${\mathbb{R}}^{r+s-1}$ and gives the desired map.

If its image coordinates remain in a bounded set, the omitted coordinate is bounded too, and every conjugate of $u$ has bounded absolute value. The coefficients of its monic minimal polynomial are integers and elementary symmetric expressions in at most $[K:\mathbb{Q}]$ bounded conjugates. They are therefore bounded integers. Only finitely many such polynomials, and finitely many roots of them, exist. Thus the inverse image of every bounded set is finite. In particular the kernel is finite and the image is discrete. This argument includes the zero-dimensional target case; it does not yet assert the full-rank part of Dirichlet unit theorem.

First suppose $|{\sigma }_i(\alpha )|=1$ for every complex embedding. The coefficients of the monic polynomialare rational algebraic integers and hence integers. They are uniformly bounded in $m$, because each is an elementary symmetric sum of numbers of modulus one. Only finitely many such integer polynomials can occur, so only finitely many algebraic integers ${\alpha }^m$ occur. Two powers coincide; since $\alpha \ne 0$, this gives ${\alpha }^{r-s}=1$. Thus $\alpha$ is a root of unity. This is Kronecker theorem on algebraic integers in the unit disk.

The second printed assertion is literally false for $\alpha =0$. For the intended statement with $\alpha \ne 0$, use the Minkowski embedding: algebraic integers of $K$ form a lattice, so only finitely many have all conjugates of modulus at most $2$. A nonzero algebraic integer with every conjugate of modulus at most one hasso every conjugate has modulus one and the first part applies. Consequently every non-root in that finite box has ${\max }_i|{\sigma }_i(\alpha )|>1$. Choose $c>1$ no larger than the least of these finitely many maxima, taking $c=2$ if there are no non-roots. Then every nonzero $\alpha \in {\mathcal{O}}_K$ satisfying $|{\sigma }_i(\alpha )|<c$ for all $i$ is a root of unity.

Dirichlet unit theorem states that for signature $(r_1,r_2)$,and the logarithmic embedding of number field units maps the free part to a full lattice in the hyperplane where the weighted coordinate sum is zero.

For a real quadratic field, the roots of unity are $\{\pm 1\}$ and the logarithms of positive units form a nonzero discrete subgroup $a\mathbb{Z}\subset \mathbb{R}$. Its smallest positive element is $\log ϵ$ for a unit $ϵ>1$. Given any positive unit $u>1$, choose $n$ so that $1\le u{ϵ}^{-n}<ϵ$; minimality forces the quotient to be $1$. Applying signs and inverses shows

Since $11\equiv 3(\mathrm{mod}4)$, the ring of integers of a quadratic field is $\mathbb{Z}[\sqrt{11}]$. A unit $a+b\sqrt{11}$ satisfies the Pell equation $a^2-11b^2=\pm 1$. The elementhas norm $1$. To prove it is the smallest unit above one without invoking the continued-fraction algorithm, suppose $1<a+b\sqrt{11}<10+3\sqrt{11}$. Replacing by its inverse or negative if needed makes $b>0$, and the conjugate relation implies $b\sqrt{11}<(u+u^{-1})/2<10$, so $b\le 3$. For $b=1,2,3$, the numbers $11b^2\pm 1$ are respectivelyOnly the last list contains a square, namely $100=10^2$, which yields the endpoint itself. Hence no smaller unit exists and

Choose the real embeddings ${\sigma }_1,\dots ,{\sigma }_r$ and one embedding ${\tau }_j$ from each pair of complex-conjugate embeddings. The logarithmic embedding of number field units isIt is a group homomorphism from multiplication to addition.

If $u$ lies in its kernel, every conjugate of $u$ has modulus one. Under the Minkowski embedding, every power $u^n$ therefore lies in one fixed bounded subset. Each $u^n$ belongs to ${\mathcal{O}}_K$, whose Minkowski image is discrete, so that bounded subset contains only finitely many such lattice points. Hence $u^a=u^b$ for some $0\le a<b$, andThus $u$ is a root of unity. Conversely, every root of unity clearly has all conjugates of modulus one and lies in the kernel. This proves the kernel of the logarithmic unit embedding is precisely $\mu (K)$.

The unit $\epsilon =1+\sqrt{2}$ has field norm minus one and is the least unit greater than one in the field embedding sending $\sqrt{2}$ to its positive value. It is a fundamental unit. The Dirichlet unit theorem therefore gives

Writing ${\epsilon }^n=a_n+b_n\sqrt{2}$, direct multiplication givesFor $n=0,1,2,3,4$, the pairs $(a_n,b_n)$ are $(1,0),(1,1),(3,2),(7,5),(17,12)$, andIn particular, $\eta ={\epsilon }^2=3+2\sqrt{2}$ has field norm one. Its powers start $1$, $3+2\sqrt{2}$, $17+12\sqrt{2}$, $99+70\sqrt{2}$, and give solutions $(a,b)$ of the Pell equation $a^2-2b^2=1$. This turns finding one unit into generating infinitely many exact integer solutions.

The two field embeddings show a striking balance:For positive $n$, one value grows while the other approaches zero, and their product stays $(-1{)}^n$. Taking logarithms of their absolute values converts this multiplication into equally spaced points, the simplest example of the logarithmic embedding of number field units.^[ref]