Anki Deck Changes

Note 1: ETH::DiskMat

Deck: ETH::DiskMat
Note Type: Horvath Cloze
GUID: PI2pek)9t%

modified

Before

Front

In a group, \(a^0\) is defined as the identity element \(e\).

Back

In a group, \(a^0\) is defined as the identity element \(e\).

After

Front

In a group, \(a^0\) is defined as the identity element \(e\).

Back

In a group, \(a^0\) is defined as the identity element \(e\).

Field-by-field Comparison

Field	Before	After
Text	<p>In a group, \({{c1::a^0}}\) is defined as the {{c2::identity element~~}} \({{c3::e}}\)~~.</p>	<p>In a group, \({{c1::a^0}}\) is defined as the {{c2::identity element \(e\)}}.</p>

Tags: ETH::1._Semester::DiskMat::5._Algebra::3._The_Structure_of_Groups::4._The_Order_of_Group_Elements_and_of_a_Group

Note 2: ETH::DiskMat

Deck: ETH::DiskMat
Note Type: Horvath Classic
GUID: q[i&,qVWc9

modified

Before

Front

What is modular congruence in a field?

Back

What is modular congruence in a field?

\[a(x) \equiv_{m(x)} b(x) \quad \overset{\text{def}}{\Leftrightarrow} \quad m(x) \ | \ (a(x) - b(x))\]

After

Front

What is modular congruence in a polynomial field?

Back

What is modular congruence in a polynomial field?

\[a(x) \equiv_{m(x)} b(x) \quad \overset{\text{def}}{\Leftrightarrow} \quad m(x) \ | \ (a(x) - b(x))\]

Field-by-field Comparison

Field	Before	After
Front	<p>What is modular congruence in a field?</p>	<p>What is modular congruence in a polynomial field?</p>

Tags: ETH::1._Semester::DiskMat::5._Algebra::8._Finite_Fields::1._The_Ring_F[x]ₘ₍ₓ₎

Note 3: ETH::DiskMat

Deck: ETH::DiskMat
Note Type: Horvath Classic
GUID: xkr{;Hl0wh

modified

Before

Front

How do you find the GCD of two polynomials?

Back

How do you find the GCD of two polynomials?

To find \(\gcd(a(x), b(x))\):

Find a common factor \((x - \alpha)\) using the roots method:
Try all possible elements of the field to find roots
If \(\alpha\) is a root of both, then \((x - \alpha)\) is a common factor
Use division with remainder to reduce to smaller polynomials
Repeat the process on the smaller polynomials
Important: Don't just find all roots and multiply! A root might be repeated (e.g., \((x+1)^2\)), and you'd miss the higher multiplicity

Example: For \(a(x) = (x+1)(x+1)(x+2)\), the GCD with another polynomial might be \((x+1)^2\), not just \((x+1)\).

After

Front

How do you find the GCD of two polynomials?

Back

How do you find the GCD of two polynomials?

To find \(\gcd(a(x), b(x))\):

Find a common factor \((x - \alpha)\) using the roots method:
Try all possible elements of the field to find roots
If \(\alpha\) is a root of both, then \((x - \alpha)\) is a common factor
Use division with remainder to reduce to smaller polynomials
Repeat the process on the smaller polynomials
After they have no roots anymore, try all monic polynomials up to degree d/2 to find irreducible factors.
Important: Don't just find all roots and multiply! A root might be repeated (e.g., \((x+1)^2\)), and you'd miss the higher multiplicity

Example: For \(a(x) = (x+1)(x+1)(x+2)\), the GCD with another polynomial might be \((x+1)^2\), not just \((x+1)\).

Field-by-field Comparison

Field	Before	After
Back	<p>To find \(\gcd(a(x), b(x))\):</p> <ol> <li>Find a common factor \((x - \alpha)\) using the <strong>roots method</strong>:</li> <li>Try all possible elements of the field to find roots</li> <li>If \(\alpha\) is a root of both, then \((x - \alpha)\) is a common factor</li> <li>Use <strong>division with remainder</strong> to reduce to smaller polynomials</li> <li>Repeat the process on the smaller polynomials</li> <li><strong>Important</strong>: Don't just find all roots and multiply! A root might be repeated (e.g., \((x+1)^2\)), and you'd miss the higher multiplicity</li> </ol> <p><strong>Example</strong>: For \(a(x) = (x+1)(x+1)(x+2)\), the GCD with another polynomial might be \((x+1)^2\), not just \((x+1)\).</p>	<p>To find \(\gcd(a(x), b(x))\):</p> <ol> <li>Find a common factor \((x - \alpha)\) using the <strong>roots method</strong>:</li> <li>Try all possible elements of the field to find roots</li> <li>If \(\alpha\) is a root of both, then \((x - \alpha)\) is a common factor</li> <li>Use <strong>division with remainder</strong> to reduce to smaller polynomials</li> <li>Repeat the process on the smaller polynomials</li><li>After they have no roots anymore, try all monic polynomials up to degree d/2 to find irreducible factors.</li> <li><strong>Important</strong>: Don't just find all roots and multiply! A root might be repeated (e.g., \((x+1)^2\)), and you'd miss the higher multiplicity</li> </ol> <p><strong>Example</strong>: For \(a(x) = (x+1)(x+1)(x+2)\), the GCD with another polynomial might be \((x+1)^2\), not just \((x+1)\).</p>

Tags: ETH::1._Semester::DiskMat::5._Algebra::6._Polynomials_over_a_Field::3._Analogies_Between_Z_and_F[x],_Euclidean_Domains_*

Note 4: ETH::DiskMat

Deck: ETH::DiskMat
Note Type: Horvath Cloze
GUID: zDSrp9w@De

modified

Before

Front

In a group, the equation \(a * x = b\) has a unique solution \(x\) for any \(a\) and \(b\) (So does the equation \(x * a = b\)).

Back

In a group, the equation \(a * x = b\) has a unique solution \(x\) for any \(a\) and \(b\) (So does the equation \(x * a = b\)).

After

Front

In a group, the equation \(a * x = b\) (and \(x*a = b\)) has a unique solution \(x\) for any \(a\) and \(b\).

Back

In a group, the equation \(a * x = b\) (and \(x*a = b\)) has a unique solution \(x\) for any \(a\) and \(b\).

Field-by-field Comparison

Field	Before	After
Text	In a group, the equation \(a * x = b\) has {{c1:: a unique solution \(x\)}} for any \(a\) and \(b\)~~ {{c1:: (So does the equation \(x * a = b\))}}~~.	In a group, the equation \(a * x = b\) (and \(x*a = b\)) has {{c1:: a unique solution \(x\)}} for any \(a\) and \(b\).

Tags: ETH::1._Semester::DiskMat::5._Algebra::2._Monoids_and_Groups::3._Inverses_and_Groups