What is a polynomial over a commutative ring?
Note 1: ETH::DiskMat
Deck: ETH::DiskMat
Note Type: Horvath Classic
GUID:
modified
Note Type: Horvath Classic
GUID:
AR?8CyMux0
Before
Front
Back
What is a polynomial over a commutative ring?
A polynomial \(a(x)\) over a commutative ring \(R\) in the indeterminate \(x\) is a formal expression of the form: \[ a(x) = {{c2::a_d x^d + a_{d-1}x^{d-1} + \cdots + a_1 x + a_0}} = \sum_{i=0}^d a_i x^i \] for some non-negative integer \(d\), with \(a_i \in R\).
The set of polynomials in \(x\) over \(R\) is denoted \(R[x]\).
After
Front
What is a polynomial over a commutative ring?
Back
What is a polynomial over a commutative ring?
A polynomial \(a(x)\) over a commutative ring \(R\) in the indeterminate \(x\) is a formal expression of the form: \[ a(x) = a_d x^d + a_{d-1}x^{d-1} + \cdots + a_1 x + a_0 = \sum_{i=0}^d a_i x^i \] for some non-negative integer \(d\), with \(a_i \in R\).
The set of polynomials in \(x\) over \(R\) is denoted \(R[x]\).
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Back | <p>A polynomial \(a(x)\) over a commutative ring \(R\) in the indeterminate \(x\) is a formal expression of the form: \[ a(x) = |
<p>A polynomial \(a(x)\) over a commutative ring \(R\) in the indeterminate \(x\) is a formal expression of the form: \[ a(x) = a_d x^d + a_{d-1}x^{d-1} + \cdots + a_1 x + a_0 = \sum_{i=0}^d a_i x^i \] for some non-negative integer \(d\), with \(a_i \in R\).</p> <p>The set of polynomials in \(x\) over \(R\) is denoted \(R[x]\).</p><p><br></p> |
Note 2: ETH::DiskMat
Deck: ETH::DiskMat
Note Type: Horvath Classic
GUID:
modified
Note Type: Horvath Classic
GUID:
r3)589~fN6
Before
Front
What is the cardinality of \(F[x]_{m(x)}\)?
Back
What is the cardinality of \(F[x]_{m(x)}\)?
Lemma 5.34: Let \(F\) be a finite field with \(q\) elements and let \(m(x)\) be a polynomial of degree \(d\) over \(F\). Then: \[|F[x]_{m(x)}| = q^d\]
Explanation: Each polynomial has \(d\) coefficients (from \(0, \dots, d - 1\)), and each coefficient can be any of \(q\) elements from \(F\).
After
Front
What is the cardinality of \(F[x]_{m(x)}\)?
Back
What is the cardinality of \(F[x]_{m(x)}\)?
Lemma 5.34: Let \(F\) be a finite field with \(q\) elements and let \(m(x)\) be a polynomial of degree \(d\) over \(F\). Then: \[|F[x]_{m(x)}| = q^d\]
Explanation: Each polynomial of \(\deg d - 1\) has \(d\) coefficients (from \(0, \dots, d - 1\)), and each coefficient can be any of \(q\) elements from \(F\).
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Back | <p><strong>Lemma 5.34</strong>: Let \(F\) be a finite field with \(q\) elements and let \(m(x)\) be a polynomial of degree \(d\) over \(F\). Then: \[|F[x]_{m(x)}| = q^d\]</p> <p><strong>Explanation</strong>: Each polynomial has \(d\) coefficients (from \(0, \dots, d - 1\)), and each coefficient can be any of \(q\) elements from \(F\).</p> | <p><strong>Lemma 5.34</strong>: Let \(F\) be a finite field with \(q\) elements and let \(m(x)\) be a polynomial of degree \(d\) over \(F\). Then: \[|F[x]_{m(x)}| = q^d\]</p> <p><strong>Explanation</strong>: Each polynomial of \(\deg d - 1\) has \(d\) coefficients (from \(0, \dots, d - 1\)), and each coefficient can be any of \(q\) elements from \(F\).</p> |