Break: Reducible Polynomial Attack

Module 02 | Breaking Weak Parameters

Using a reducible polynomial instead of an irreducible one creates zero divisors and destroys field structure.

Why This Matters

In Module 02 we learned the fundamental construction:

\text{GF}(p^n) = \mathbb{F}_p[x] / (f(x))

This yields a field (every nonzero element invertible) if and only if $f(x)$ is irreducible over $\mathbb{F}_p$ .

But what if someone accidentally (or maliciously) uses a reducible polynomial? The quotient ring is no longer a field. It has zero divisors, nonzero elements whose product is zero. Any cryptographic scheme that assumes field arithmetic (division, inverses) will break catastrophically.

The Scenario

Suppose we want to build a field with $5^2 = 25$ elements. The correct approach is:

\text{GF}(25) = \mathbb{F}_5[x] / (f(x))

where $f(x)$ is an irreducible degree-2 polynomial over $\mathbb{F}_5$ .

But someone picks $f(x) = x^2 + 3x + 2$ without checking irreducibility.

Spoiler: Over $\mathbb{F}_5$ , this polynomial factors as $(x + 1)(x + 2)$ . The quotient ring has zero divisors and is NOT a field.

In [ ]:

# === Step 1: Show that x^2 + 3x + 2 factors over GF(5) ===
R.<x> = PolynomialRing(GF(5))

f = x^2 + 3*x + 2
print(f'f(x) = {f}')
print(f'Is f(x) irreducible over GF(5)? {f.is_irreducible()}')
print(f'Factorization: {f.factor()}')
print()

# Check roots explicitly
for a in GF(5):
    val = f(a)
    root_marker = '  <-- ROOT!' if val == 0 else ''
    print(f'  f({a}) = {val}{root_marker}')

print()
print('f(x) has roots at x = 4 (= -1 mod 5) and x = 3 (= -2 mod 5).')
print('So f(x) = (x + 1)(x + 2) = (x - 4)(x - 3) over GF(5).')

Step 2: Find Zero Divisors

Build the quotient ring $\mathbb{F}_5[x]/(x^2 + 3x + 2)$ and find elements whose product is zero even though neither element is zero.

Since $f(x) = (x+1)(x+2)$ , the images of $(x+1)$ and $(x+2)$ in the quotient ring should multiply to zero (because their product equals $f(x) \equiv 0$ in the quotient).

In [ ]:

# === Step 2: Zero divisors in the quotient ring ===
R.<x> = PolynomialRing(GF(5))
f = x^2 + 3*x + 2

S.<a> = R.quotient(f)
print(f'Quotient ring: F_5[x] / ({f})')
print(f'Elements: {list(S)}')
print(f'Number of elements: {len(list(S))}')
print()

# The zero divisors: (a + 1) and (a + 2)
zd1 = a + 1
zd2 = a + 2
product = zd1 * zd2

print(f'(a + 1) = {zd1} , is this zero? {zd1 == S.zero()}')
print(f'(a + 2) = {zd2} , is this zero? {zd2 == S.zero()}')
print(f'(a + 1) * (a + 2) = {product} . ZERO!')
print()
print('Two nonzero elements multiply to zero. This ring has ZERO DIVISORS.')
print('It is NOT a field.')

In [ ]:

# === Find ALL zero divisors systematically ===
R.<x> = PolynomialRing(GF(5))
S.<a> = R.quotient(x^2 + 3*x + 2)
elems = list(S)

print('All zero-divisor pairs (nonzero elements whose product is 0):')
print()
zero_divisors = set()
for i, e1 in enumerate(elems):
    for e2 in elems[i:]:
        if e1 != S.zero() and e2 != S.zero() and e1 * e2 == S.zero():
            print(f'  {e1} * {e2} = 0')
            zero_divisors.add(e1)
            zero_divisors.add(e2)

print(f'\nZero divisors: {sorted(zero_divisors, key=str)}')
print(f'Count: {len(zero_divisors)} out of {len(elems) - 1} nonzero elements')

Step 3: How Zero Divisors Break a Toy Scheme

Consider a toy "encryption" scheme in the quotient ring:

Key: a nonzero element $k$ in the ring
Encrypt: $c = m \cdot k$ (multiply message by key)
Decrypt: $m = c \cdot k^{-1}$ (multiply ciphertext by inverse of key)

This works perfectly in a field (every nonzero $k$ has an inverse). But if $k$ is a zero divisor, $k^{-1}$ does not exist, and decryption is impossible. Worse: distinct messages can encrypt to the same ciphertext (information loss).

In [ ]:

# === Toy encryption scheme: works in a field, breaks with zero divisors ===
R.<x> = PolynomialRing(GF(5))
S.<a> = R.quotient(x^2 + 3*x + 2)

# Key is a zero divisor
key = a + 1
print(f'Key k = {key} (a zero divisor!)')
print()

# Try to encrypt several messages
print('Encrypting all nonzero messages with c = m * k:')
ciphertexts = {}
for m in S:
    if m == S.zero():
        continue
    c = m * key
    print(f'  Encrypt({m}) = {m} * {key} = {c}')
    if str(c) not in ciphertexts:
        ciphertexts[str(c)] = []
    ciphertexts[str(c)].append(str(m))

print()
print('=== Collisions ===')
for c_val, msgs in ciphertexts.items():
    if len(msgs) > 1:
        print(f'  Ciphertext {c_val} came from messages: {msgs}')
        print(f'  --> Decryption is AMBIGUOUS! Information is lost.')

print()
print('Now try to compute k^(-1)...')
try:
    key_inv = key^(-1)
    print(f'  k^(-1) = {key_inv}')
except Exception as e:
    print(f'  ERROR: {e}')
    print(f'  The inverse does not exist! Decryption is impossible.')

In [ ]:

# === Contrast: the same scheme with an IRREDUCIBLE polynomial works perfectly ===
R.<x> = PolynomialRing(GF(5))

# x^2 + 2 is irreducible over GF(5) (check: no roots)
g = x^2 + 2
print(f'g(x) = {g}')
print(f'Irreducible? {g.is_irreducible()}')

T.<b> = R.quotient(g)
print(f'This quotient ring IS a field: {T.is_field()}')
print()

# Same key value, but now in a field
key_good = b + 1
print(f'Key k = {key_good}')
key_good_inv = key_good^(-1)
print(f'k^(-1) = {key_good_inv}')
print(f'k * k^(-1) = {key_good * key_good_inv}  (= 1, as expected)')
print()

# Encrypt and decrypt a message
msg = 3*b + 4
ct = msg * key_good
recovered = ct * key_good_inv
print(f'Message:     m = {msg}')
print(f'Ciphertext:  c = m * k = {ct}')
print(f'Decrypted:   c * k^(-1) = {recovered}')
print(f'Correct?     {recovered == msg}')

The Fix: Always Verify Irreducibility

Before using a polynomial $f(x)$ to build a quotient ring for cryptographic purposes:

Check irreducibility: Use f.is_irreducible() or test for roots (sufficient for degree 2 or 3).
Use known irreducible polynomials: Standards like AES specify the exact irreducible polynomial.
Never trust user-supplied polynomials without verification.

Property	Irreducible $f(x)$	Reducible $f(x)$
Quotient ring $\mathbb{F}_p[x]/(f(x))$	Field	Not a field
Zero divisors	None	Present
Every nonzero element invertible	Yes	No
Safe for crypto (inversion, division)	Yes	No, breaks schemes

Exercises

Another reducible polynomial: Try $f(x) = x^2 + 4x + 3$ over $\mathbb{F}_5$ . Factor it, build the quotient ring, and find all zero divisors.
Degree 3: Over $\mathbb{F}_2$ , the polynomial $x^3 + x^2 + x + 1 = (x+1)(x^2+1) = (x+1)^3$ is reducible. Build $\mathbb{F}_2[x]/(x^3 + x^2 + x + 1)$ and find zero divisors.
Counting zero divisors: In the ring $\mathbb{F}_5[x]/(x^2 + 3x + 2)$ , how many of the 25 elements are zero divisors? How many are units (invertible)? Hint: By the Chinese Remainder Theorem for rings, $\mathbb{F}_5[x]/((x+1)(x+2)) \cong \mathbb{F}_5 \times \mathbb{F}_5$ .
The analogy with integers: We know $\mathbb{Z}/15\mathbb{Z}$ has zero divisors because $15 = 3 \times 5$ . The analogue here is that $f(x)$ factors. Write down the complete parallel between $\mathbb{Z}/15\mathbb{Z}$ and $\mathbb{F}_5[x]/(x^2+3x+2)$ .

In [ ]:

# === Exercise workspace ===
R.<x> = PolynomialRing(GF(5))

# Exercise 1: Try f(x) = x^2 + 4x + 3 over GF(5)
f_ex1 = x^2 + 4*x + 3
print(f'Exercise 1: f(x) = {f_ex1}')
print(f'Irreducible? {f_ex1.is_irreducible()}')
print(f'Factorization: {f_ex1.factor()}')
print()

# TODO: Build the quotient ring and find zero divisors
# S.<a> = R.quotient(f_ex1)
# ...

Summary

	Irreducible polynomial	Reducible polynomial
Example	$x^2 + 2$ over $\mathbb{F}_5$	$x^2 + 3x + 2 = (x+1)(x+2)$ over $\mathbb{F}_5$
Quotient ring	$\text{GF}(25)$ , a field	$\mathbb{F}_5 \times \mathbb{F}_5$ . NOT a field
Zero divisors	None	$(x+1)$ and $(x+2)$ multiply to $0$
Inverses	Every nonzero element invertible	Some elements have no inverse
Crypto safety	Safe	Broken

Key takeaways:

A quotient ring $\mathbb{F}_p[x]/(f(x))$ is a field only when $f(x)$ is irreducible.
If $f(x) = g(x) \cdot h(x)$ , then $g(x)$ and $h(x)$ become zero divisors in the quotient.
Zero divisors break any scheme that relies on multiplicative inverses (encryption, signatures, key exchange).
Always verify irreducibility before using a polynomial to construct a field.
This is the polynomial analogue of the integer fact: $\mathbb{Z}/n\mathbb{Z}$ is a field only when $n$ is prime.

Back to Module 02: Rings, Fields, and Polynomials