## Platonic Solids via Euler Characteristic

A platonic solid is a regular, convex polyhedron. Informally, it is a two-dimensional, convex shape made by gluing together a number of copies of a regular polyhedron so that the same arrangements meet at every vertex. There are only five such solids, and the ancient world was fascinated with them. Plato himself associated them with earth, wind, air, fire, and the arrangement of the constellations.

The Platonic solids with their associated elements.

Euclid gave simple proof classifying the Platonic solids, but there is also a nice way to predict the Platonic solids via Euler Characteristic. That is, via Euler’s famous formula $V-E+F = 2$, where $V,E,F$ are the number of vertices, edges, and faces of the polyhedra.

Let our polygons have $p$-sides, with $q$ meeting at a vertex. We know $V-E+F = 2$. At the same time, $2E = pF$, because every edge belongs to two faces, so that if we go through each face, count its edges, and add them all up, we will have double-counted edges. Moreover, we also know that $2E = qV$, because to each vertex we can count the number of edges coming out of it, and adding this all together gives $qV$, but we will have double-counted edges because every edge belongs to two vertices. This gives

$V - E + F = 2 \Longrightarrow \frac{2E}{q} - E + \frac{2E}{p} = 2 \Longrightarrow\frac{2}{q} -1 + \frac{2}{P} = \frac{2}{E}$

We can divide by $2$ to obtain

$\frac{1}{q} + \frac{1}{p} = \frac{1}{E} + \frac{1}{2}$

Since $\frac{1}{E}$ is positive, we see that

$\frac{1}{q} + \frac{1}{p} > \frac{1}{2}$

Moreover, we know that $p,q \geq 3$, we only get $5$ possibilities, $(p,q) = (3,3),(4,3),(3,4),(5,3),(3,5)$. In fact, each of these gives a platonic solid, so we have found them all. The convenient short-hand for describing a regular, convex polyhedron we have given is known as a Schläfli symbol.

Can you try extending this technique to predict the regular, convex polytopes of higher dimensions?

## Convolutions, the Steinhaus Theorem, and Measurable Automorphisms (Part II)

In this post, we will use the Steinhaus Theorem to understand group homomorphisms from $\mathbb{R}^d$ into $\mathbb{C}$, which will hopefully shed let on the uncountably many elements of $\mbox{Aut}(\mathbb{C})$. Recall

Definition: A homomorphism $f: \mathbb{R}^d \to \mathbb{C}$ is a map with the property that $f(x+y) = f(x) + f(y)$ for $x,y \in \mathbb{R}^d$.

Theorem: (i) Measurable homomorphisms are continuous, and (ii) $f$ is a measurable homomorphism iff it takes the form $f(x_1, \cdots, x_d) = x_1z_1 + \cdots + x_dz_d$ for $x_1, \cdots, x_d \in \mathbb{R}$ and some complex coefficients $z_1, \cdots, z_d$.

Proof: (i) We shall use the $\epsilon-\delta$ definition of coninuity. Let $z$ be in the image of $f$, and let $D$ be some disk about $z$. The preimage $f^{-1}(D) = f^{-1}(D -z + z)$ can be thought of as $f^{-1}(D-z) + f^{-1}(z)$. That is, a copy of $f^{-1}(D-z)$ around each point in the preimage of $f^{-1}(z)$. If we can show that for each copy of $f^{-1}(D-z)$ around some point $f^{-1}(z)$ there exist an open neighborhood of its center that maps within $D$, we will have verified the $\epsilon-\delta$ definition of continuity.

Note that $D-z$ is a disk around the origin. Let us rename it $2K$ for the sake of simplicity in notation, where $K$ is a disc about the origin of half the radius. Consider the countable collection $\{K+p+iq : p,q \in \mathbb{Q} \}$. This collection covers $\mathbb{C}$, so that

$\bigcup_{p,q} f^{-1}(K_{p,q}) = \mathbb{R}^d$

If for all $p,q$ we had $m(f^{-1}(K_{p,q})) = 0$, then we would have

$m(\mathbb{R}^d) = m(\bigcup_{p,q} f^{-1}(K_{p,q})) \leq \sum_{p,q} f^{-1}(K_{p,q}) = 0$

which, I assure you, is profoundly absurd. Thus we can find $K_{p,q} = K+{p,q}$ whose inverse has positive measure. Note, as before, that the inverse of $K+{p,q}$ is $f^{-1}(K) + f^{-1}(p+iq)$. Consider

$f^{-1}(K) + f^{-1}(p+iq) - f^{-1}(K) - f^{-1}(p+iq)$

By the Steinhaus theorem this must contain an open neighborhood of the origin. But this is just the set of points that map into $K-K$. Since $K$ is a disc about the origin, we have $K-K = 2K = D-z$. Thus, an open neighborhood about the origin maps to $D-z$. Call this open neighborhood $O$. Consider the collection $O+f^{-1}(z)$ that contains $f^{-1}(D)$. Then this is the union of open sets, and hence open, and if the preimage is restricted to within this open set, then the image of $f$ is within $D$, confirming continuity.

(ii) If $f$ is of the form described, then it is clearly continuous, and hence must be measurable. Conversely, suppose that $f$ is a measurable homomorphism, so continuous by the prior part of the exercise. Then let $z_i = f(e_i)$, where $e_i$ is the $i$th standard basis vector for $\mathbb{R}^d$. To show linearity in all coordinates, it suffices to show linearity in each coordinate at a time, so without loss of generality it suffices to consider $g = f_1$, the first coordinate of $f$. We have

$g(1) = z_1$

Let $n \in \mathbb{N}$, we have

$g(n) = g(1) + \cdots g(1) = z_1 + \cdots + z_1 = nz_1$Let $p/q$ be a rational. We have

$pz_1 = g(p) = g(q(p/q)) = g(p/q) + \cdots + g(p/q) = qg(p/q) \Longrightarrow g(p/q) = \frac{p}{q} z_1$

Lastly, if $x$ is an arbitrary real, let $r_n \to x$ be a sequence of rationals converging to it. By continuity of $g$,

$g(x) = g(\lim_{n} r_n) = \lim_{n} g(r_n) = \lim_{n} r_{n}z_1 = xz_1$

With this theorem in hand, we can now understand the automorphisms of $\mathbb{C}$ (and indeed much more). Any such automorphism is also a homomorphism from $\mathbb{C}$ to $\mathbb{C}$ as vector spaces, and hence can be viewed as a map from $\mathbb{R}^2$, which is also a $2$-dimensional real vector space. In general, an automorphism $\psi$ of $\mathbb{C}$ as a field must send $0$ to $0$ and $1$ to $1$. By the above theorem, if the automorphism is continuous it is linearly determined by where it sends $i$. But $\psi(i)^2 = \psi(i^2) = \psi(-1) = -1$, so that $\psi(i)$ must be a square root of $-1$. If the image of $i$ is $i$, we get the identity automorphism, and if the image of $i$ is $-i$, then we get the conjugation automorphism.

Thus, any other automorphism of $\mathbb{C}$ cannot be continuous, and hence cannot be measurable. So, if you were looking for a concrete way to think of these automorphisms, you are out of luck! These maps are ridiculous and you might do well by stopping to think about them.

## Convolutions, the Steinhaus Theorem, and Measurable Automorphisms (Part I)

This post, and the one following following it, are based on a series of exercises in Tao’s “An Introduction to Measure Theory” that I found particularly interesting.

Our initial question is concerned with field automorphisms of $\mathbb{C}$. Certainly there are two obvious ones: the trivial automorphism, and the conjugation automorphism. Any more? Well, assuming choice, we have the following proposition.

Proposition (AC) : There exist uncountably many field automorphisms of $\mathbb{C}$.

Proof:  Any automorphism of $\mathbb{C}$ fixes $\mathbb{Q}$. Let $T$ be a transcendence basis for $\mathbb{C}/\mathbb{Q}$ (this exists by Zorn’s Lemma, which is equivalent to choice).

Claim: $T$ is uncountable.

Proof of claim: We know that

$|\mathbb{C}| = 2^{\aleph_{0}} >$  $\aleph_0 = |\mathbb{Q}|$

Suppose that $T$ was countable. Then every element in $\mathbb{C}$ would be the root of a polynomial over $\mathbb{Q} (T)$. However, since $\mathbb{Q} (T)$ is countable, this gives countably many polynomials, each with finitely many roots, so that $|\mathbb{C}| = \aleph_{0}$, which is impossible, proving the claim.

Now, we claim that there are uncountably many automorphisms of $\mathbb{Q}(T)$. This is because $\mathbb{Q}(T)$ is isomorphic to the function field $\mathbb{Q}(X_{i})_{i \in |T|}$, and any permutation of the indices of the variables induces an automorphism of this function field.

Lastly, we need to show that we can extend these uncountably many elements of $\mbox{Aut}(\mathbb{Q}(T))$ to elements of $\mbox{Aut}(\mathbb{C})$. However,$\mathbb{C}/\mathbb{Q}(T)$ is algebraic, so this follows from standard results in Galois theory. $\square$.

This result is quite interesting. Not only are there more automorphisms than the trivial automorphism and conjugation, there are uncountably many automorphisms! If so, what exactly do these field automorphisms look like? How should we best go about picturing them? To answer that question, we take a detour into the theory of Lebesgue measure and integration.

Recall the construction of the convolution: Let $f,g: \mathbb{R}^d \to \mathbb{C}$ be Lebesgue measurable functions such that $f$ is absolutely integrable and $g$ is essentially bounded. Then the convolution $f \ast g: \mathbb{R}^d \to \mathbb{C}$ is defined by

$f \ast g(x) = \int_{\mathbb{R}^d} f(y)g(x-y) dy$

Proposition: The convolution is well-defined, bounded and continuous.

Proof: Let $g$ be essentially bounded by $M$. Note that

$|f \ast g(x)| = \left| \int_{\mathbb{R}^d} f(y)g(x-y) dy \right|$

$\leq \int_{\mathbb{R}^d} |f(y)g(x-y)| \leq M \int_{\mathbb{R}^d} |f(y)| = M ||f||_{L^1} < \infty$

so that the convolution is both well defined and bounded. To see that it is continuous, note that

$|f \ast g(x+h) - f \ast g(x)| = \left| \int_{\mathbb{R}^d} f(y)g(x+h-y) dy - \int_{\mathbb{R}^d} f(y)g(x-y) dy \right|$

Since the Lebesgue integrable is translation invariant, we can translate the first integral by $y \to y + h$. Then we have

$\left| \int_{\mathbb{R}^d} f(y)g(x+h-y) dy - \int_{\mathbb{R}^d} f(y)g(x-y) dy \right|$

$= \left| \int_{\mathbb{R}^d} f(y+h)g(x-y) dy - \int_{\mathbb{R}^d} f(y)g(x-y) dy \right|$

$=\left| \int_{\mathbb{R}^d} g(x-y)(f(y+h) - f(y)) dy \right| \leq M \int_{\mathbb{R}^d} |f(y+h) - f(y)|$

But if we let $h \to 0$, then since translation is continuous in $L^1$ (a proof of which can be found in Tao’s text on Measure Theory, Prop. 1.6.13)

$M \int_{\mathbb{R}^d} |f(y+h) - f(y)| \to 0$

With the machinery of convolutions, we now prove the Steinhaus theorem.

Theorem (Steinhaus): Let $E \subset \mathbb{R}^d$ be a Lebesgue measurable set of positive measure. Then the set $E - E := \{x-y: x,y \in E\}$ contains an open neighborhood of the origin.

Proof: Without loss of generality, $E$ is bounded, else we can replace it with some $E' = E \cap [-N,N]$ that has positive measure (such an $N$ must exist, by the upward monotone convergence theorem for measurable sets). Then $E' \subset E' \subset E - E$, so if the former contains an open set about the origin, so must the latter.

Consider the convolution

$1_{E} \ast 1_{-E} (x) = \int_{R^d} 1_{E}(y)1_{-E}(x-y) dy$

This is well defined, since the functions in question are both $L^1$ and essentially bounded (by the boundedness of $E$).Note that

$1_{E} \ast 1_{-E}(0) = m(E) > 0$

For an arbitrary $x$, there exists $y$ for which $1_{E}(y)1_{-E}(x-y) \neq 0$ iff $y \in E$ and $y - x \in E$, so that $y - (y-x) = x \in E$. So if $x$ is not in $E-E$, then $1_{E} \ast 1_{-E}(x) = 0$. Consider the set

$S = (1_{E}\ast 1_{-E})^{-1}((0,\infty))$

that is, the set of $x$ for which the integral defining the convolution is positive. We have shown that $S \subset E-E$ and that $0 \in S$. Since the convolution is continuous, and $(0,\infty)$ is an open set, $S$ is an open set containing the origin inside of $E-E$.

It may seem that our initial question in field theory, and our latter discussion of real analysis, are completely unrelated. This will be resolved in the following post, where we will use the Steinhaus theorem to better understand the nature of these bizarre field automorphisms of $\mathbb{C}$ produced by the axiom of choice.

## Dehn Surgery and Decayed Knots

I gave a talk this Wednesday on Dehn Surgery and Decayed Knots, summarizing some of the reading I’ve been doing as part of an REU with Prof. Liam Watson. It was also my 22nd birthday.

## Distinguishing Numbers of Graphs & Groups

For my the final report of Math 191, I wrote the following report on  Distinguishing Numbers. Check it out!

## An Interesting Problem from Enderton

The following is a problem from Enderton’s “A Mathematical Introduction to Logic”:

10. Say that a set $\Sigma_{1}$ of well-formed formulas (wffs) is equivalent to a set $\Sigma_{2}$ of wffs if for any wff $\sigma$, we have $\Sigma_{1} \models \sigma$ iff $\Sigma_{2} \models \sigma$. A set $\Sigma$ is independent iff no member of $\Sigma$ is tautologically implied by the remaining members of $\Sigma$.

Parts (a) and (b) of the problem ask us to show that any finite set of wffs has an independent equivalent subset, but that this might fail for infinite sets of wffs. Both of these are very easy, but the interesting part is (c).

(c)  Let $\Sigma$ be any infinite sequence on wffs. Show that some independent equivalent set exists.

Solution: WLOG, $\Sigma$ is consistent, otherwise we can replace it with the independent, equivalent set consisting of a single false statement. We are going to construct a set $\Sigma'$ as follows. Firstly, enumerate all the well-formed formulas, and go through the list one element at a time. At each element $\sigma$, make the following considerations:

1. Does $\Sigma \models \sigma$? If not, skip $\sigma$.

2. Does adding $\sigma$ to $\Sigma'$ affect the independence of $\Sigma'$? If not, add $\sigma$ to $\Sigma'$. Otherwise, proceed to step 3.

3. Why does adding $\sigma$ to $\Sigma'$ interfere with the independence of $\Sigma'$? If it is because we can already prove $\sigma$ from $\Sigma'$, then $\sigma$ is of no use to us, so do not add it to $\Sigma'$. Otherwise, it is because we can prove some $\alpha \in \Sigma'$ from the other elements of $\Sigma' \cup \sigma$. In a sense, we might say that $\sigma$ is strictly stronger than some of the statements in $\Sigma'$. In this case, proceed to the last step.

4. Suppose that at this point in the construction

$\Sigma' = \{ \alpha_{1}, \cdots, \alpha_{n}\}$

Instead of adding $\sigma$ to $\Sigma'$,  add

$\sigma' := \alpha_{1} \wedge \cdots \wedge \alpha_{n} \to \sigma$

We claim that adding $\sigma'$ to $\Sigma'$ does not affect its independence. This is easily seen because:

(a) The other elements of $\Sigma'$ cannot prove $\sigma'$, as that would imply that they could prove $\sigma$, contrary to hypothesis.

(b) We cannot use $\sigma'$ to help prove some other $\alpha_{i} \in \Sigma'$, as if we do not assume every $\alpha_{i}$ to begin with, $\sigma'$ is meaningless. To put it precisely, if we can use $\alpha_{1}, \cdots, \alpha_{n-1}, \sigma'$ to prove $\alpha_n$, then we can get $\sigma$ from only $\alpha_{1}, \cdots, \alpha_{n-1}$ and $\sigma'$. Since

$\sigma' := \alpha_{1} \wedge \cdots \wedge \alpha_{n} \to \sigma$

and since $\alpha_{n}$ is presumed independent from the other elements of $\Sigma'$, we can only get $\sigma$ from $\alpha_{1}, \cdots, \alpha_{n-1}$ and $\sigma'$ if we already knew $\sigma$, contrary to hypothesis.

Now, we complete this construction to obtain $\Sigma'$. By construction, and by Compactness, $\Sigma'$ is independent, as it is independent at every finite step. Moreover, it is equivalent to $\Sigma$, because if $\Sigma \models \sigma$, then either:

1. $\sigma$ is already in $\Sigma'$.

2. $\sigma$ can be proven from elements that were already in $\Sigma'$.

3. $\sigma$ isn’t in $\Sigma'$, but a statement of the form

$\sigma' := \alpha_{1} \wedge \cdots \wedge \alpha_{n} \to \sigma$

is, with the $\alpha_{i} \in \Sigma'$, so that $\Sigma' \models \sigma$.

## Probability That Two Randomly Chosen Integers Are Coprime

Theorem 1: Euler Product Formula

$\displaystyle\sum_{n} \frac{1}{n^{s}} = \displaystyle\prod_{p \mbox{ prime}} \frac{1}{1 - p^{-s}}$

Proof:  We know that $(1-x)^{-1} = 1 + x + x^{2} + x^{3} + \cdots$. Let $x = p^{-s}$. Then

$(1 - p^{-s})^{-1} = 1 + p^{-s} + p^{-2s} + \cdots$.  Thus

$\prod_{p} (1 - \frac{1}{p^{s}})^{-1} = \prod_{p} (1 + p^{-s} + p^{-2s} + \cdots)$

We now write

$\prod_{p} [1 + p^{-s} + p^{-2s} + \cdots] = (1 + 2^{-s} + 2^{-s} + \cdots)(1 + 3^{-s} + 3^{-2s} + \cdots) \cdots (1 + p^{-s} + p^{-2s} + \cdots) \cdots$

Multiplying out these parenthesis, we see by the Fundamental Theorem of Arithmetic that we get

$= 1 + 2^{-s} + 3^{-s} + 4^{-s} + 5^{-s} = \sum_{n} \frac{1}{n^{s}}$

So what is the probability that any two integers are coprime? Let $n$ and $m$ be integers. For a fixed prime $p$ the probability that any number is divisible by $p$ is $\frac{1}{p}$, as every $p$th number is a multiple of $p$. The probability that two numbers are both divisible by $p$ is $\frac{1}{p^2}$. The probability that at least one is not divisible by $p$ is $1 - \frac{1}{p^{2}}$. For any two primes, the events that $n,m$ are both divisible by either prime are mutually independent. Thus, to find out the probability that any prime divides both $n$ and $m$, we need take the product

$\prod_{p}(1 - \frac{1}{p^{2}}) = \big( \prod_{p} \frac{1}{1-p^{2}}\big)^{-1}$

By Theorem 1, this is just $\frac{1}{\zeta(2)}$. Pick your favorite proof that

$\zeta(2) = \frac{\pi^{2}}{6}$.

Then the probability that any two numbers are coprime is $\frac{6}{\pi^{2}} \approx 61 \%$.

From this, it is easy to see that the probability that any $k$ numbers are coprime is $\frac{1}{\zeta(k)}$.

## A Proof that 2 is Prime

The following excerpt is taken directly from Carl E. Linderholm’s, “Mathematics Made Difficult”.

Mathematically, it is indeed inconclusive that nobody has ever succeeded in finding any numbers that divide $2$ exactly, other than the numbers $1$ and $2$ which obviously do this; this does not at all show that no such numbers exist. On the other hand, it may be quite easy to show that there are no such numbers. In mathematics, if one approach does not work, one must always try another; according to the principle catus multifariam deglubitur.

I answer that the number $2$ is prime. It is easily seen that the only numbers between $0$ and $2$, including $0$ but excluding $2$, are $0$ or $1$. Thus the remainder left by any number on division by $2$ is either $0$ or $1$. Hence the quotient ring

$\mathbb{Z}/2\mathbb{Z}$

where $2\mathbb{Z}$ is the ideal in $\mathbb{Z}$ generated by $2$, has only the elements $[0]$ and $[1]$, where these are the images of $0$ and $1$ under the canonical quotient map. Since $[1]$ must be a unit of this ring, every element of this ring except $[0]$ is a unit, and the ring is a field. As such, it has no zero-divisors other than $[0]$. But looking back now at $\mathbb{Z}$, this shows that if $ab =2$, then one of $a,b$ is an element $2\mathbb{Z}$, i.e., is an even number. In other words, we have either $a = 2m$ or $b = 2m$; say the former. By cancellation, $mb = 1$, so that both $m$ and $b$ are units. (Or one may argue that since $\mathbb{Z}/2\mathbb{Z}$ is a field, the ideal $2\mathbb{Z}$ is maximal in $\mathbb{Z}$, and hence prime, which implies that the generator is prime.)

The fact that the number $2$ is prime is useful in many ways. It prevents us, for instance, from indulging in the time-wasting habit, to which the ignorant are so prone, of attempting to discover a proper factor for the number $2$. It shows that it would be very unwise to program a computer to keep on trying numbers one after the other until the computer should have found a divisor of $2$ distinct from $1$ and $2$. It saves the trouble of looking for a rectangle with integral sides and area $2$

## Primary Ideals in a PID

This post is about classifying primary ideals in a PID. Nothing too exciting, to be honest. You can take this problem in a few different directions, such as noticing that a PID is a UFD, in which case the primary ideals are generated by powers of primes. In that light, I’m going to offer a different solution, which I think is rather pleasant to read.

Claim: Let $R$ be a PID. Then $\frak{A} < R$ is primary if and only if $\sqrt{\frak{A}}$ is maximal.

First Direction: Assume $\frak{A}$ is primary. Then $\sqrt{\frak{A}}$ is prime.

Proof: Let $xy \in \sqrt{\frak{A}}$. Then $x^n y^n \in \frak{A}$. Thus either $x^n$ or $y^{nk}$ is in $\frak{A}$, so that either $x$ or $y$ is in $\sqrt{\frak{A}}$.

Moreover, prime ideals in a PID are maximal.

Proof: Let $\frak{B} = (b)$ be a prime ideal in $R$, a PID. Suppose that $\frak{B} \subset \frak{C} = (c)$. Then we can write $b = ac$. This implies that $ac \in \frak{B}$. Since $\frak{B}$ is prime, either $a$ or $c$ is in $\frak{B}$. If is the latter, then $\frak{B} = \frak{C}$. If it is the former, then we can write $a = db$. This gives us

$b = bdc ---> b(1-dc) = 0$

Since $R$ is a domain, $1= dc$. Then $c$ is a unit. This implies that $\frak{C} = R$.

Second Direction: Conversely, suppose that $\frak{A} = (a)$ is not primary. Thus there exists $x,y \in R$ such that $xy \in \frak{A}$, but $x \notin \frak{A}$$y^{n} \notin \frak{A}$. Then $y \notin \sqrt{\frak{A}}$.

Observe that $(a)$ is a proper subset of the ideal $(a) + (y)$. Moreover, $(a) + (y) \neq R$, for suppose that $1 \in (a) + (y)$. Then there exists $u,v \in R$ such that

$1 = ua + vy$

Multiplying by $x$,

$x = uxa + xyv$

However, the RHS is the sum of two elements in $(a)$, so is in $(a)$. Yet $x \notin (a)$ by hypothesis.

Furthermore, $\sqrt{(a)}$ is a proper subset of the ideal $\sqrt{(a) + (y)}$, as only the latter contains $y$. However, $\sqrt{(a) + (y)} \neq R$, for suppose that $1 \in \sqrt{(a) + (y)}$. Then $1^{n} = 1 \in (a) + (y)$, which is impossible. We conclude that $\sqrt{(a)} = \sqrt{\frak{A}}$ is not maximal. $\square$

## Towards Completeness and Decibility for ACF of Characteristic p

Last time, we showed that $ACF_{p}$ was $\kappa$-categorial for uncountable cardinalities $\kappa$. In this post, we make steps towards proving the completeness and decidability of $ACF_{p}$.

Recall Tarski’s Test : Let $M \subseteq N$ be $L$-structures. Then $M \preceq N$ if and only if, for every $L$-formula $\phi(x,y)$ for $x = (x_{1}, \cdots , x_{n})$ and $y$ a single variable, and every $a = (a_{1}, \cdots, a_{n}) \in M^{n}$ we have the following: if $N \models \exists y \phi(a,y)$ then there is $b \in M$ such that $N \models \phi(a,b)$.

Lemma: Suppose the theory $T$ has infinite models. Then for any $\kappa \geq |L| + \aleph_{0}$, $T$ has a model of size $\kappa$.

Proof: Let $N \models T$ be infinite. By inflating the model with extra elements if necessary, we can assume $|N| \geq \kappa$. Now, take $A \subseteq N$ with $|A| = \kappa$. Set $M_{0} = $, the model generated by $\kappa$. Note that $|M_{0}| = \kappa$, as extending from the set $A$ to a structure requires completion under functions and interpretation symbols in $L$, which cannot increase the cardinality, for $\kappa \geq |L| + \aleph_{0}$.

Now, add a witness to $M_{0}$ for every $L$-formula $\phi(x,y)$ and $a \in (M_{0})^{n}$, such that $N \models \exists y \phi (a,y)$. What we are trying to do is to fulfill the requirements of Tarski’s test, to get a model of size $\kappa$ elementarily equivalent with $N$. Call the resulting set (after adding witnesses), $A_{1}$, which may not the be universe of a substructure. However, we can generate a substructure from it, i.e. $M_{1} = $. Again, $|M_{1}| = \kappa$. However, we are not done here, as we only took witnesses for parameters in $M_{0}$, and now we need to consider paramaters in $M_{1}$. To avoid this problem, we interate this process countably many times.

$M_{0} \subseteq M_{1} \subseteq M_{2} \subseteq \cdots \subseteq N$

Note that $M_{i} = \kappa$. Now, put $M = \cup_{i} M_{i} , |M| = \kappa$. Then $M$ is the universe of a substructure of $N$, and by construction, $M$ satisfies the hypothesis of Tarski’s Test, so that $M \preceq N$. As a result, $M \models T$, and we have obtained the model we were looking for.